IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: Saladin on May 04, 2010, 12:44:39 pm
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Hey, I do Edexcel S1, please post all your doubts here!!!
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Here's this question! I was solving through the book exercises and then I found this question a bit confusing.
A farmer keeps a record of the weekly milk yield of his herd of seven cows over a period of 6 months. He finds that the mean yield is 24 liters. He buys another cow that he is told will produce 28 liters of milk in a week. Work out the effect this will have on the mean milk yield of his cows.
Thank you in advance.
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i just face difficulties in answering those questions requiring comments :(
dunno what to do about it!! and those type of questions are more common nowdays then in the old papers >:(
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Here's this question! I was solving through the book exercises and then I found this question a bit confusing.
A farmer keeps a record of the weekly milk yield of his herd of seven cows over a period of 6 months. He finds that the mean yield is 24 liters. He buys another cow that he is told will produce 28 liters of milk in a week. Work out the effect this will have on the mean milk yield of his cows.
Thank you in advance.
It will increase his yield to 24.5
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i just face difficulties in answering those questions requiring comments :(
dunno what to do about it!! and those type of questions are more common nowdays then in the old papers >:(
It is best just to understand what the question is essentially asking. What is the question trying to investigate...
Sometimes, it just comes with practice.
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It is best just to understand what the question is essentially asking. What is the question trying to investigate...
Sometimes, it just comes with practice.
yeah true that..i prefer the C2 its more challenging :)
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help plz
Given that and
find .
my answer was 0.251
but in the MS the answer is 0.0796
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Here is how you get it:
Therefore,
The Probability is
Therefore,
If I was using actual values, instead of statistical tables, I would have gotten your answer.
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help plz
Given that and
find .
my answer was 0.251
but in the MS the answer is 0.0796
Which paper did you get this from?
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help plz
Given that and
find .
my answer was 0.251
but in the MS the answer is 0.0796
As per da MS dey hve taken S.D frm as 10
but 10 is da variance, so da MS is wrong.While ur ans is correct.
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Which paper did you get this from?
jan 2004 q2
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As per da MS dey hve taken S.D frm as 10
but 10 is da variance, so da MS is wrong.While ur ans is correct.
ya , i also think the MS is wrong
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jan 2004 q2
Will check again, I dont think I am wrong though.
How did you get ur answer?
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Will check again, I dont think I am wrong though.
How did you get ur answer?
same as ur method
but at the end u have to subtract
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Will check again, I dont think I am wrong though.
How did you get ur answer?
paper link S1 jan 04 (http://www.srepapmaxeeeerf.org/A%20Level/Maths/Edexcel/S1/S1%202004-01.pdf)
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Da MS smetimes tends 2 b wrong.
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same as ur method
but at the end u have to subtract
oops, forgot that part. heheheh.
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okay guys i have a doubt....
of the customers visiting the stereo section of a large electronic store 75% on average make a purchase....find the least possible number of customers, given that the probability of all customers making a purchase is less than 5%
thanks in advance ;D
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ehm ehm one more doubt ;D ;D
A license plate has 3 letters and 3 digits in that order.A witness to a hit and run accident saw the first 2 letters and the last digit.If the letters and digits can be repeated,how many license plate should be checked before finding the culprit?
ANS: 2600
Thanks again ;D
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ehm ehm one more doubt ;D ;D
A license plate has 3 letters and 3 digits in that order.A witness to a hit and run accident saw the first 2 letters and the last digit.If the letters and digits can be repeated,how many license plate should be checked before finding the culprit?
ANS: 2600
Thanks again ;D
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26C1=26(choosing da last remaining letter)
10C1*10C1=100(choosing both digits,dey cn b repeated)
26*100=2600
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###***
26C1=26(choosing da last remaining letter)
10C1*10C1=100(choosing both digits,dey cn b repeated)
26*100=2600
aye aye i see...thanks immortal!
+1 rep! ;D
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U're welcome :)
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okay guys i have a doubt....
of the customers visiting the stereo section of a large electronic store 75% on average make a purchase....find the least possible number of customers, given that the probability of all customers making a purchase is less than 5%
thanks in advance ;D
P(x=n)<0.05
= (0.75)^n<0.05
take log on both sides
log(0.75)^n<log(0.05)
nlog(0.75)<log(0.05)
n(-0.28768)<-2.99573
when we convert negative to positive the inequality reverses
n>2.99573/0.28768
n>10.4134
n>11
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I have doubt ladies and gentlemen
November 2003 paper 6
Question 4 part 3
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19/20*19/20*1/20
Da first & second card cn b any expect da 1 he wants
da third one he gets da one he wants...afterward it does not matter..
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19/20*19/20*1/20
Da first & second card cn b any expect da 1 he wants
da third one he gets da one he wants...afterward it does not matter..
I thought of the same after i got my first answer wrong
I was confused because he bought a total of 5 cards right
so couldnt it be 3C1(19/20)^2(1/20) as he stopped looking at cards after the third one?
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I thought of the same after i got my first answer wrong
I was confused because he bought a total of 5 cards right
so couldnt it be 3C1(19/20)^2(1/20) as he stopped looking at cards after the third one?
I advice u 2 think simple :)
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I advice u 2 think simple :)
lol
right :)
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What's the formula booklet that the edexcel board will give us during the examination?? or did they stop doing that??
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What's the formula booklet that the edexcel board will give us during the examination?? or did they stop doing that??
i will post it soon
and yes you will be given it
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i will post it soon
and yes you will be given it
Okay then :) thank you :D
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need help in this question
jan 08
Q3
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need help in this question
jan 08
Q3
I'm on it now.
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need help in this question
jan 08
Q3
First you have to calculate the area under the bars.
(1*6)+(1*7)+(4*2)+(2*6)+(3*5.5)+(5*2)+(3*1.5)+(12*0.5)
=70
Therefore, now calculating the number of people per unit area.
Work out the area in the interval given.
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First you have to calculate the area under the bars.
(1*6)+(1*7)+(4*2)+(2*6)+(3*5.5)+(5*2)+(3*1.5)+(12*0.5)
=70
Therefore, now calculating the number of people per unit area.
Work out the area in the interval given.
thanx:)
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P(x=n)<0.05
= (0.75)^n<0.05
take log on both sides
log(0.75)^n<log(0.05)
nlog(0.75)<log(0.05)
n(-0.28768)<-2.99573
when we convert negative to positive the inequality reverses
n>2.99573/0.28768
n>10.4134
n>11
thanx freaked!! hehehe nice nickname!! ;D +1 rep ;D :D
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aight people anotha Qs ;D
its from nov 09 varient 1
find how many numbers between 5000 and 6000 can be formed from the digits 1,2,3,4,5,6 if repition of digits is allowed......i dnt get why is it 6 to da powr of 3? ??? ??? ??? >:(
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must start with 5 then there are three other digits from a choice of 5 so 5*4*3
this is in the ms. You are looking at the wrong answer
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must start with 5 then there are three other digits from a choice of 5 so 5*4*3
this is in the ms. You are looking at the wrong answer
oh sorry sir i meant for the second question.........if reptitions ARE allowed....thanks
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aight people anotha Qs ;D
its from nov 09 varient 1
find how many numbers between 5000 and 6000 can be formed from the digits 1,2,3,4,5,6 if repition of digits is allowed......i dnt get why is it 6 to da powr of 3? ??? ??? ??? >:(
Its a 4 digit num between 5000-6000
So da last 3 digits r changed wit those of da range..
Dat is each zero haz 6C1 different variables.
4 all 3 zero's 6C4*6C1*6C1=63
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Its a 4 digit num between 5000-6000
So da last 3 digits r changed wit those of da range..
Dat is each zero haz 6C1 different variables.
4 all 3 zero's 6C4*6C1*6C1=63
So immortal how did you find S1 past papers so far?
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So immortal how did you find S1 past papers so far?
y do u ask ???
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y do u ask ???
I dont know
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need help in June 09 number 1 , how to get Z....urgent
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When i get home
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need help in June 09 number 1 , how to get Z....urgent
Which paper is this frm??
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nvm abt last one....I would like to know how to solve question 5 in Nov.09 first varient......urgently
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I think this question waz already answered b4/
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5a)
i> no digits repeated range 5000-6000
so 5xxx r da possibilities, 1,2,3,4,5,6 5 cannot b used
remaining ones 5P3 as arrangement matters 4 numbers.
ii> repeated digits all num frm 1-6 cn b used
5xxx so 5(6c1)(6c1)(6c1)
so 6*6*6 num cn b formed dat is 63
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mmmmm...another question in Nov.09 first varient num.3..
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5b)
i> mor girls dan boys
so 3g 2b
4g 2b
5g 0b
8c3*6c2+8c4*6c2+8c5
=1316
ii>Total of 8+6=14 students 11 if cousins r ignored.
with cousins 11C2(wit 3 cousins already included)
without dem 11C5
so 55+462
=517
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mmmmm...another question in Nov.09 first varient num.3..
Will ans in 15min
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P(85 < x < 100)
= 0.5 – P [z < {(85 ? 100)/7}]
= 0.5 – P (z < – 2.143)
= 0.5 – (1 – ?(2.143))
= 0.9839 – 0.5
= 0.484
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why is it times 0.5..
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another question in Nov.09 2nd varient num.1.....how to get standard deviation ???
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why is it times 0.5..
because its
100-100
---------
7
which is zero
.The value of p corresponding to zero is 0.5
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part two of Question 3 is also very easy
its states that 33 percent of times are the longest and smallest
so value of p of the middle is 0.34.
1-0.34=0.67
which is +-0.67
(a-100)/7=0.67
will give u the maximum time
(a-100)/7=-0.67
will give you the least time
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i doubt anyone is awake
but if there is anyone on
answer the second variant of november 2009
question 4
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Can you post the paper? My file is corrupted.
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Can you post the paper? My file is corrupted.
Find how many different odd numbers greater then 500 can be made using some or all of the digits 1,3,5 and 6 with no digit being repeated
I did it but it took me half an hour to figure it out
But your output could proove to be valuable :D
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hey im stuck!!
n09 Q5 (a) (i) (ii) and b (ii)
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hey im stuck!!
n09 Q5 (a) (i) (ii) and b (ii)
Sorry m8, I do Edexcel.
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Hello , can someone plz help
June2005
Q7 (d)
P( sci | RH ) = P(sci 'n' RH) / P(RH) = (24÷110)/ (110÷148)
thats wat i did , but obviously its wrong
how do you find P(sci 'n' RH) ???
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Hello , can someone plz help
June2005
Q7 (d)
P( sci | RH ) = P(sci 'n' RH) / P(RH) = (24÷110)/ (110÷148)
thats wat i did , but obviously its wrong
how do you find P(sci 'n' RH) ???
On it now, will be finished soon iA.
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Hello , can someone plz help
June2005
Q7 (d)
P( sci | RH ) = P(sci 'n' RH) / P(RH) = (24÷110)/ (110÷148)
thats wat i did , but obviously its wrong
how do you find P(sci 'n' RH) ???
This is how you do it.
(c)You have to find the number of people that are right handed.
So, there are a total of right handers, and a total of students.
So,
(d) This is basically asking, out of 110 right handers, what is the probability that it will be a science student?
So, the number of science students that are right handed, are
Now,
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Oh! i was trying to use the formula
okay now it makes sense =')
thank you
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Oh! i was trying to use the formula
okay now it makes sense =')
thank you
Sometimes, you need to understand the whole concept, and understand the feel of the question before attempting it.
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Sometimes, you need to understand the whole concept, and understand the feel of the question before attempting it.
this also saves a lot of time..
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this also saves a lot of time..
Just doin ma job. 8)
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Find
E( x^2 + 4X - 3)
E(x) = 2.6
and the answer = 15.6
how did they get it ?????
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Find
E( x^2 + 4X - 3)
E(x) = 2.6
and the answer = 15.6
how did they get it ??
You must know the variance to answer this question. What is it?
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I have two questions...
From the January 2009 paper
question 3e) could you please explain fully how to do this..
and for 2b) why must we use the addition rule? When we didnt in 2a)
Thankyou SO much :)
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Find
E( x^2 + 4X - 3)
E(x) = 2.6
and the answer = 15.6
how did they get it ??
You must know the variance to answer this question. What is it?
Var 1.44
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Find
E( x^2 + 4X - 3)
E(x) = 2.6
and the answer = 15.6
how did they get it ??
You must know the variance to answer this question. What is it?
Var 1.44
V(X0=E(X^2)-(E(X))^2
1.44=E(X^2)-2.6^2
E(X^2)=1.44+6.76=8.18
E( x^2 + 4X - 3)=E(X^2)-4E(X)-3=8.18-4*2.6-3=8.18-1.4=-5.22
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June 2001
Question 1 a)
they said to find the mean and the standard deviation for the data given
in the markscheme they gave something different to what i solved
the mean i knew how to do but the standard deviation was the weird part
My solution : Standard Deviation = Square root of [(sum of x^2 divided by n)-(mean)^2]
my answer gives me Math Error
and in the markscheme the answer comes out 4 :/
plz someone help asap
thank you
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I dont have the cie 2001 jun paper. Can you upload it here?
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I dont have the cie 2001 jun paper. Can you upload it here?
not the cie the edexcel
6683 code
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sd=sqrt(sum x^2/n -mean^2)
=sqrt(46625/25-(1075/25)^2)=4
mistak in question sum x^2 should be 46625.
Dont blame me. I only stole them
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not the cie the edexcel
6683 code
Each of the 25 students on a computer course recorded the number of minutes x, to the nearest minute, spent surfing the internet during a given day. The results are summarised below.
,
(a) Find and for these data.
(3)
Two other students surfed the internet on the same day for 35 and 51 minutes respectively.
(b) Without further calculation, explain the effect on the mean of including these two students.
for part a
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Each of the 25 students on a computer course recorded the number of minutes x, to the nearest minute, spent surfing the internet during a given day. The results are summarised below.
,
(a) Find and for these data.
(3)
Two other students surfed the internet on the same day for 35 and 51 minutes respectively.
(b) Without further calculation, explain the effect on the mean of including these two students.
for part a
nvm nvm!! lol
thank you T.Q!!
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FROM NOW ON, ALL QUESTION PAPERS WILL BE ASSUMED TO BE EDEXCEL, UNLESS TOLD OTHERWISE.
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edexcel june 2009 question 3
i dont understand how they got the width
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Hello ppl i have som doubts plz can u try to solve them they are :
1) year 2002 june 6th question
2) year 2005 june 4th question
3) year 2007 june 2e) with explanation and reason
Plz reply soon ppl and best of luck for ur exam :D
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Hello ppl i have som doubts plz can u try to solve them they are :
1) year 2002 june 6th question
2) year 2005 june 4th question
3) year 2007 june 2e) with explanation and reason
Plz reply soon ppl and best of luck for ur exam :D
answering now.
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Thanks you Engraved for ur help cld u even help in normal distribution or if u have any good notes cld u plz attach it thanks a ton though :D may God bless u and ur exam goes well :)
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Hello ppl i have som doubts plz can u try to solve them they are :
1) year 2002 june 6th question
2) year 2005 june 4th question
3) year 2007 june 2e) with explanation and reason
Plz reply soon ppl and best of luck for ur exam :D
June 2007
The mean is the average of the 3 quartiles=45
Check the log table for the step above.
13.43
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Engraved y did u pick 54 or how did u knw u had to takee tht ?
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how do we calculate the width and the height of a rectangle representing a group class modal ???
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I just did CIE S1 today
so if you could just write down the whole question i will FOR SURE solve it
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edexcel june 2009 question 3
i dont understand how they got the width
u have to find out the scale factor
or solve it proportionally
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for histograms use
height_2/height_1=frequency density_2/frequency density_1
length of base_2/length of base_1=length of interval_2/length of interval_1
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astarmathsandphysics cld u plz solve these two question edexcel stats
1) year 2002 june 6th question
2) year 2005 june 4th question
thank u
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one mo.
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those two questions above
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see also
http://www.astarmathsandphysics.com/igcse_maths_notes/igcse_maths_notes_calculating_the_dimensions_of_histogram_bars.html
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Sorry for the late reply thanks a lot astar :)
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i dont get interpolation, can someone explain to me the exact steps of what to do
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I will do a page on interpolation for stats.
Here is a page also http://www.astarmathsandphysics.com/a_level_maths_notes/FP1/a_level_maths_notes_fp1_linear_interpolation.html
I will do the stats page tomorrow
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A discrete random variable is such that each of its values is assumed to be equally likely.
(a)Write down the name of the distribution that could be used to model this random variable.
(b)Give an example of such a distribution.
(c)Comment on the assumption that each value is equally likely.
(d)Suggest how you might refine the model in part (a).
Can someone please explain, im confused..?
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a)discrete uniform
b)SCORE ON A DICE
c)perfectly possible
d)throw the dice 1000000000000000000000 times and find from this
the probability of scoring each of 1,2,3,4,56,
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Hey long time no c all :)
So looks like AS isn't that easy after all. Year is gonna be full of doubts :D
Anyway,
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s09_qp_6.pdf (http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s09_qp_6.pdf)
Q6)ii) I drew the graph and decided to put the sign which means that there is skipped values on the y axis... Is this okay? like doesnt this change the shape of the first part of the graph?
Thanks
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hu
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7. A committee of 6 people, which must contain at least 4 men and at least 1 women, is to be chosen from 10 men and 9 women.
i) Find the number of possible committee that can be chosen. [3]
ii) Find the probability that one particular man, Albert and one particular woman, Tracey are both on the committee.[2]
iii) Find the number of possible committees that include either Albert or Tracey but not both. [3]
iv) The committee that is chosen of 4 men and 2 women. They queue up randomly in a line for refreshments. Find the probability that the women are not next to each other in the queue. [3]
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I am giving Exams for S1 this May/June. (2011)
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Can you help with this question, please?
7. A committee of 6 people, which must contain at least 4 men and at least 1 women, is to be chosen from 10 men and 9 women.
i) Find the number of possible committee that can be chosen. [3]
(6P1+8P1)*6!
ii) Find the probability that one particular man, Albert and one particular woman, Tracey are both on the committee.[2]
iii) Find the number of possible committees that include either Albert or Tracey but not both. [3]
iv) The committee that is chosen of 4 men and 2 women. They queue up randomly in a line for refreshments. Find the probability that the women are not next to each other in the queue. [3]
Let me work out the rest
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Oh
This is already done
sorry
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A dice is known to be biased in such a way that, when it is thrown, the probability of a 6 showing is . This biased dice and an ordinary fair dice are thrown. Find the probability that exactly one of the two dice shows a 6, given that at least one of them shows a 6.
Many thanks in advance; what does that last sentence even mean?!
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A dice is known to be biased in such a way that, when it is thrown, the probability of a 6 showing is . This biased dice and an ordinary fair dice are thrown. Find the probability that exactly one of the two dice shows a 6, given that at least one of them shows a 6.
Many thanks in advance; what does that last sentence even mean?!
You have to find the probability that; if together both the dices are thrown, at-least from both, 1 shows a six, and from both 1 has to show 6.
1 1 5
-- + --- = -----
4 6 12
Is this the answer ?
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You have to find the probability that; if together both the dices are thrown, at-least from both, 1 shows a six, and from both 1 has to show 6.
1 1 5
-- + --- = -----
4 6 12
Is this the answer ?
Nope........you did it wrong there dude.
The question asked at least a six form both dices but exactly one six should be obtained.
Hence probable results are :
1. 6 from biased die and any other number from unbiased die.
2. Any other number from unbiased die and 6 from unbiased die.
Probability : (1/4 x 5/6) + (3/4 x 1/6) = 1/3
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Nope........you did it wrong there dude.
The question asked at least a six form both dices but exactly one six should be obtained.
Hence probable results are :
1. 6 from biased die and any other number from unbiased die.
2. Any other number from unbiased die and 6 from unbiased die.
Probability : (1/4 x 5/6) + (3/4 x 1/6) = 1/3
Yeah !! :D
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Nope........you did it wrong there dude.
The question asked at least a six form both dices but exactly one six should be obtained.
Hence probable results are :
1. 6 from biased die and any other number from unbiased die.
2. Any other number from unbiased die and 6 from unbiased die.
Probability : (1/4 x 5/6) + (3/4 x 1/6) = 1/3
Hmm, apparently the answer is 8/9. Any ideas?
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got 8/9. faffing around with scanner
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here
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Instinct change ! No doubts here now, for anybody. :o
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A:one of the two dice shows a 6
B:at least one of them shows a 6
first find B
a tree diagram would help...........
1/4*1/6+1/4*5/6+3/4*1/6=3/8
P(B)=3/8
P(A)=1/4*5/6+3/4*1/6=1/3
P(A!B)=this means probabiltiy of A given that B
P(A)=1/3 / 3/8=8/9
hop u get it
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there are 6 boys and 3 women ..
how many possible ways are there if no 2 women are standing next to each other?
ans : 151200
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here
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here
thank you sir i got it now
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Hello :) im taking cie s1 and wondering if u know the way to find standard deviation on calculator , i use casio fx-991es and cant find any help on web :-\
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A coin is tossed 10 times
(a) How many different sequences of heads and tails are possible.
(b) How many different sequences containing six heads and four tails are possibles.
(c) What is the probability of getting six heads and four tails.
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Hello Smiley im taking cie s1 and wondering if u know the way to find standard deviation on calculator , i use casio fx-991es and cant find any help on web Undecided
From memory
Shift mode stat
choose 1 or 2 variable
input data
Choose shift 4
var
Then choose sigma_n or sigma_{n-1}
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A coin is tossed 10 times
(a) How many different sequences of heads and tails are possible.
(b) How many different sequences containing six heads and four tails are possibles.
(c) What is the probability of getting six heads and four tails.
a)2^10 =1024
b)10C4 or 10C4
c)10C4/2^10
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From memory
Shift mode stat
choose 1 or 2 variable
input data
Choose shift 4
var
Then choose sigma_n or sigma_{n-1}
Thank You ! :D
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A coin is tossed 10 times
(a) How many different sequences of heads and tails are possible.
(b) How many different sequences containing six heads and four tails are possibles.
(c) What is the probability of getting six heads and four tails.
a)2^10 =1024
b)10C4 or 10C4
c)10C4/2^10
thanxx
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can we solve basic binomial distribution questions on calculator ?! if yes then how ??
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can we solve basic binomial distribution questions on calculator ?! if yes then how ??
i think u want every thing in calculator
but for the factorial part
press shift and then press exactly below it
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i think u want every thing in calculator
but for the factorial part
press shift and then press exactly below it
lol ! yes ;)
my tcher sux big time...if we ppl follow what she explains,we'll definitely be ending up with Es and Us in stats...
so why not try to save some time and do what guarantees a full mark :)
and Thank You :)
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lol ! yes ;)
my tcher sux big time...if we ppl follow what she explains,we'll definitely be ending up with Es and Us in stats...
so why not try to save some time and do what guarantees a full mark :)
and Thank You :)
If you dont show the necessary workings you'll still land up with Es and Us. A correct answer doesnt guarantee full marks.
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If you dont show the necessary workings you'll still land up with Es and Us. A correct answer doesnt guarantee full marks.
the ones i asked for do. =)
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The letters of the word CONSTANTINOPLE are written on 14 cards, one on each card. The cards are shuffled and then arranged in a straight line.
How many arrangements are there where no two vowels are next to each other?
The answer is: 457228800
I'm not getting the right answer for some reason... :-\
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The letters of the word CONSTANTINOPLE are written on 14 cards, one on each card. The cards are shuffled and then arranged in a straight line.
How many arrangements are there where no two vowels are next to each other?
The answer is: 457228800
I'm not getting the right answer for some reason... :-\
9 consonants and 5 vowels.
Arrangement for consonants = 9!, thereby we have 10 spaces to fit in the vowels, and they can be arranged in any manner. So arrangement for these : 10C5 X 5
Multiplying all : 9! X 10C5 X 5
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What does " µ" symbol mean? I saw it in some questions..does it mean anything like sigma means standard deviation? are there any more weird symbols in S1 we should know about? What are the things given in the formula booklet for S1 that we shouldnt learn?
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What does " µ" symbol mean? I saw it in some questions..does it mean anything like sigma means standard deviation? are there any more weird symbols in S1 we should know about? What are the things given in the formula booklet for S1 that we shouldnt learn?
"µ" stands for the Greek letter, moew. Equivalent of the mean. You will use it in distributions most often.
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I have attached a question can someone try solving it?
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Can someone please solve for me Question 3 in May 2009 with steps?
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Can someone please solve for me Question 3 in May 2009 with steps?
(i)
P(0 buses) = (1-0.16)11
P(1 bus) = 11C1(0.16)(1-0.16)10
P(2 buses) = 11C2(0.16)2(1-0.16)9
where X is number of buses passed
so
P(X<3)= P(0 Busus)+P(1 Buses)+P(2 buses)
(ii)
USE normal distribution
mean = np
= 125(1-(0.16+0.2))
= 125 (0.64)
= 80
variance = npq
= 125 (0.64)(1-0.64)
= 28.8
let X be number of car passed
so
p(X>73)
= 1-P(X<=73)
=1 - ?((73-80)/root(28.8)
=1-?(-1.304)
=1-(1-?(1.304))
=1-1+?(1.304)
= ?(1.304)
use normal distribution tabel to get the value of ?(1.304)
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(i)
P(0 buses) = (1-0.16)11
P(1 bus) = 11C1(0.16)(1-0.16)10
P(2 buses) = 11C2(0.16)2(1-0.16)9
where X is number of buses passed
so
P(X<3)= P(0 Busus)+P(1 Buses)+P(2 buses)
(ii)
USE normal distribution
mean = np
= 125(1-(0.16+0.2))
= 125 (0.64)
= 80
variance = npq
= 125 (0.64)(1-0.64)
= 28.8
let X be number of car passed
so
p(X>73)
= 1-P(X<=73)
=1 - ?((73-80)/root(28.8)
=1-?(-1.304)
=1-(1-?(1.304))
=1-1+?(1.304)
= ?(1.304)
use normal distribution tabel to get the value of ?(1.304)
I think you just solved a wrong question?
My question was about histograms.
Here' the link to the paper:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20June/6683%20June%202009.pdf
I want Question 3 please! :)
& if you don't mind can you please explain to me Question 3e of January 2009 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20Jan/S1%202009-01.pdf
Thank you! :)
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ohhh!
ur edexcel studemt srry i thought u r a cie student
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I think you just solved a wrong question?
My question was about histograms.
Here' the link to the paper:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20June/6683%20June%202009.pdf
I want Question 3 please! :)
& if you don't mind can you please explain to me Question 3e of January 2009 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20Jan/S1%202009-01.pdf
Thank you! :)
widht would be::
3*2/6 = 1 cm
and height is :
10*9/15 = 6
so 6/1 = 6 cm
SIMPLE
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I think you just solved a wrong question?
My question was about histograms.
Here' the link to the paper:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20June/6683%20June%202009.pdf
I want Question 3 please! :)
& if you don't mind can you please explain to me Question 3e of January 2009 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20Jan/S1%202009-01.pdf
Thank you! :)
Var(X) = Sigma(xi2pi)-(mmean)2
= = Sigma(xi2pi)-(xipi)2
=[02(0.4)+12(0.3)+22(0.2)+32(0.1)]-[(0(0.4)+(1(0.3)+(2(0.2)+(3(0.1)/4]2
=[0(0.4)+1(0.3)+4(0.2)+9(0.1)]-[(0+0.3+0.4+0.3)]2
=[0+0.3+0.8+0.9]-(1)2
=2-1
=1
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widht would be::
3*2/6 = 1 cm
and height is :
10*9/15 = 6
so 6/1 = 6 cm
SIMPLE
Can you please explain to me what those numbers mean, what does the 3, 2, 6, 10, 9 and 15 stand for? :/
I'm sorry for the trouble, but this lesson has been annoying me since the IG's!
& again, you solved the wrong question, I wanted part e not d! ;)
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Okay, I've compiled my questions here in this post to make it easier! :)
1. Question 3 of June 2009 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20June/6683%20June%202009.pdf
2. Question 3e of January 2009 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20Jan/S1%202009-01.pdf
3. Question 7c of June 2008 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2008%20June/S1%202008-june-qp.pdf
4. Questions 6b and c of January 2008 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2008%20Jan/6683_01_que_20080115.pdf
With explanation please, I have the marking scheme, so I know the answers, but I can't understand the working/how to do it! :/
Thanks alot! :)
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Can you please explain to me what those numbers mean, what does the 3, 2, 6, 10, 9 and 15 stand for? :/
I'm sorry for the trouble, but this lesson has been annoying me since the IG's!
& again, you solved the wrong question, I wanted part e not d! ;)
Formula for
(i) frequency density (height) :
frequency
= ------------
class width
(ii) class width (interval) :
frequency
= ------------------
frequency density
Now for your question :
3 (a)
x = 10 - 15 (9.5 - 15.5); frequency = 15.
class width = (15.5 - 9.5) = 6. This is taken in histogram as 2 cm.
For 16 - 19, width is (19.5 - 16.5) = 3.
In the histogram, the width of the 16 - 18 class will be : (3 X 2)/6 = 1
(b)
Height (or frequency density) = (15 / 6) = 2.5. This is taken in histogram as 5 cm.
For 16 - 19, height is (9 / 3) = 3.
In the histogram, the height of the 16 - 18 class will be : (3 X 5)/2.5 = 6
Tell me if you still didn't get.
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Formula for
(i) frequency density (height) :
frequency
= ------------
class width
(ii) class width (interval) :
frequency
= ------------------
frequency density
Now for your question :
3 (a)
x = 10 - 15 (9.5 - 15.5); frequency = 15.
class width = (15.5 - 9.5) = 6. This is taken in histogram as 2 cm.
For 16 - 19, width is (19.5 - 16.5) = 3.
In the histogram, the width of the 16 - 18 class will be : (3 X 2)/6 = 1
(b)
Height (or frequency density) = (15 / 6) = 2.5. This is taken in histogram as 5 cm.
For 16 - 19, height is (9 / 3) = 3.
In the histogram, the height of the 16 - 18 class will be : (3 X 5)/2.5 = 6
Tell me if you still didn't get.
Thanks a lot! :)
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1. Question 3 of June 2009
Frequency density= Freq/ Class width
The freq. density will be the height of your histogram in case of unequal class widths.
x :10 – 15
Frequency= 15
Class width= 15 - 10 = 5
Frequency density= 15/5 = 3
Now, use proportions.
Class width, 5 : 2
Height, 3 : 5
For 16 – 18:
Class width=18 - 16 = 2
On histogram -> 2/5*2= 1/5
Frequency density=9/2 = 4.5
On histogram ->(5/3)*(9/2) = 15/2 = 7.5
I don't find the MS from your link... so, please check it.
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1. Question 3 of June 2009
Frequency density= Freq/ Class width
The freq. density will be the height of your histogram in case of unequal class widths.
x :10 – 15
Frequency= 15
Class width= 15 - 10 = 5
Frequency density= 15/5 = 3
Now, use proportions.
Class width, 5 : 2
Height, 3 : 5
For 16 – 18:
Class width=18 - 16 = 2
On histogram -> 2/5*2= 1/5
Frequency density=9/2 = 4.5
On histogram ->(5/3)*(9/2) = 15/2 = 7.5
I don't find the MS from your link... so, please check it.
I'm sorry, your answers are wrong, check Fidato's post above for the right answer.
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1. Question 3 of June 2009
Frequency density= Freq/ Class width
The freq. density will be the height of your histogram in case of unequal class widths.
x :10 – 15
Frequency= 15
Class width= 15 - 10 = 5
Frequency density= 15/5 = 3
Now, use proportions.
Class width, 5 : 2
Height, 3 : 5
For 16 – 18:
Class width=18 - 16 = 2
On histogram -> 2/5*2= 1/5
Frequency density=9/2 = 4.5
On histogram ->(5/3)*(9/2) = 15/2 = 7.5
I don't find the MS from your link... so, please check it.
Alpa, there you went wrong. Check the intervals before 16 - 18. They were 10 - 15, so in order to built a histogram, you need to amend the values to make them continuous (so you subtract 0.5 from the lower limit and add 0.5 to the upper limit).
Like take two class intervals, (10 - 15) and (16 - 18). There's gap of 1 between 15 and 16, so how can you make a histogram ?
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Doing the rest.
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Okay, I've compiled my questions here in this post to make it easier! :)
Here they follow.
1. Question 3 of June 2009 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20June/6683%20June%202009.pdf
Done.
2. Question 3e of January 2009 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20Jan/S1%202009-01.pdf
After 3 games Rohit has scored 6.
2 games are left and he has to get 4 points or more to win the prize.
List all the outcomes of the points he can get (see the image below).
You can see that there are 16 possible outcomes and six of which fetch him 4 points or more.
You know that :
P(0) = 0.4
P(1) = 0.3
P(2) = 0.2
P(3) = 0.1
So to win a price he should get points with the outcomes as follows (as circled in image).
{ (1,3) (2,2) (2,3) (3,1)(3,2) (3,3) }
Multiply and add the probabilities :
(0.3 X 0.1) + (0.2 X 0.2) + (0.2 X 0.1) + (0.1 X 0.3) + (0.1 X 0.2) + (0.1 X 0.1)
= 0.03 + 0.04 + 0.02 + 0.03 + 0.02 + 0.01
= 0.15
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3. Question 7c of June 2008 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2008%20June/S1%202008-june-qp.pdf
(7) (c)
For this you need to find out the probability that the bag weighs more than 53 kg.
(a)
More than 53, so (53 + 0.5 = 53.5)
53.5 - 50
Z = ------------
2
= si (1.75)
= 0.9599
Now, it's greater then 53, so :
1 - { si (1.75) }
= 1 - 0.959
= 0.0401
(c)
2 bags greater than 53 kg and 1 less than 53 kg .
Probability that any bag is greater then 53 kg = 0.0401
and
Probability that it is less then 53 kg = 0.959
So :
3C2 X (0.401)2 X (0.959)
= 0.004630586
= 0.0046
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4. Questions 6b and c of January 2008 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2008%20Jan/6683_01_que_20080115.pdf
(6) (b)
Weights between 190 and 210 are 60 %.
Now 60 % is equal to 0.6 of the probability right. So find out the value of si (0.6) which is equal to 0.7257 .
Now you know the formula :
X - mean
Z = -------------
standard deviation
Let standard deviation be s.
So :
210 - 200 190 - 200
------------ - ------------ = 0.7257
s s
10 - (-10)
----------- = 0.7257
s
20 = 0.7257s
s = 20/(0.7257)
standard deviation = 27.55959763 (or 27.56)
(c)
Use the same formula of normal distribution.
If you will use the original value of standard deviation (that is 27.55959763 and not 27.56), you will be benefited.
180 - 200
---------------
27.55959763
= si (-0.7257)
= 1 - { si (0.726) }
= 1 - 0.766
= 0.234
So 0.234 is the probability that the customer will complain.
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With explanation please, I have the marking scheme, so I know the answers, but I can't understand the working/how to do it! :/
Thanks alot! :)
Can you please provide all 4 marking schemes please ? Here itself.
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What Master_Key told is correct. Sorry it just got deleted, while I was playing with my slow internet around. :D
Firstly, don't forget to add (+0.5) if you are finding probability of a greater than number like in your 3rd question it was more than 53 kg, otherwise you may get a penalty (loss of marks) - I don't know whether the same is practised in ED-EXCEL or not.
Lastly, if you don't want to forget to keep in (+0.5), just apply the formula without adding (+0.5):
1 - { si (<=X) }
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I'm sorry, your answers are wrong, check Fidato's post above for the right answer.
Apologies. I wasn't in my right state of mind. -.-
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Here they follow.
Done.
After 3 games Rohit has scored 6.
2 games are left and he has to get 4 points or more to win the prize.
List all the outcomes of the points he can get (see the image below).
You can see that there are 16 possible outcomes and six of which fetch him 4 points or more.
You know that :
P(0) = 0.4
P(1) = 0.3
P(2) = 0.2
P(3) = 0.1
So to win a price he should get points with the outcomes as follows (as circled in image).
{ (1,3) (2,2) (2,3) (3,1)(3,2) (3,3) }
Multiply and add the probabilities :
(0.3 X 0.1) + (0.2 X 0.2) + (0.2 X 0.1) + (0.1 X 0.3) + (0.1 X 0.2) + (0.1 X 0.1)
= 0.03 + 0.04 + 0.02 + 0.03 + 0.02 + 0.01
= 0.15
This one's right and I understood it, thank you! :)
I'm sorry, the link for the marking scheme for this paper is not opening now! :/
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(7) (c)
For this you need to find out the probability that the bag weighs more than 53 kg.
(a)
More than 53, so (53 + 0.5 = 53.5)
53.5 - 50
Z = ------------
2
= si (1.75)
= 0.9599
Now, it's greater then 53, so :
1 - { si (1.75) }
= 1 - 0.959
= 0.0401
(c)
2 bags greater than 53 kg and 1 less than 53 kg .
Probability that any bag is greater then 53 kg = 0.0401
and
Probability that it is less then 53 kg = 0.959
So :
3C2 X (0.401)2 X (0.959)
= 0.004630586
= 0.0046
The answer is wrong, the mark scheme is here: http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2008%20June/S1%20june%2008%20ms.pdf
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(6) (b)
Weights between 190 and 210 are 60 %.
Now 60 % is equal to 0.6 of the probability right. So find out the value of si (0.6) which is equal to 0.7257 .
Now you know the formula :
X - mean
Z = -------------
standard deviation
Let standard deviation be s.
So :
210 - 200 190 - 200
------------ - ------------ = 0.7257
s s
10 - (-10)
----------- = 0.7257
s
20 = 0.7257s
s = 20/(0.7257)
standard deviation = 27.55959763 (or 27.56)
(c)
Use the same formula of normal distribution.
If you will use the original value of standard deviation (that is 27.55959763 and not 27.56), you will be benefited.
180 - 200
---------------
27.55959763
= si (-0.7257)
= 1 - { si (0.726) }
= 1 - 0.766
= 0.234
So 0.234 is the probability that the customer will complain.
The answer is also wrong, the mark scheme is here:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2008%20Jan/6683_01_rms_20080306.pdf
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Apologies. I wasn't in my right state of mind. -.-
No problem! :)
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The answer is wrong, the mark scheme is here: http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2008%20June/S1%20june%2008%20ms.pdf
In (a) there was a mistake so part (c) was influenced.
I took +0.5, so things happened.
You just find out the value of Z without adding (0.5) to 53, because at the end you have to subtract it from 1 anyhow.
I hope you understood and will do the (c) part yourself.
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The answer is also wrong, the mark scheme is here:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2008%20Jan/6683_01_rms_20080306.pdf
Oh ! I now realised that I had taken si (0.6) and not si-1 (0.6).
Anyways I think you got that (+-) method that they have used. Actually P(X<210) = 0.8 so they have taken si-1 ( 0.8 ) and nothing else.
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Hey
i just dont permutation and combinations.
any notes?
or what if i skip this question,I am fine with the rest of the question.
thanks :(
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Hey
i just dont permutation and combinations.
any notes?
or what if i skip this question,I am fine with the rest of the question.
thanks :(
but it is one of the scoring part of the paper
we just have to understand the condition of the question and do it
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Hey
i just dont permutation and combinations.
any notes?
or what if i skip this question,I am fine with the rest of the question.
thanks :(
see the video of combination and permutation in KHAN ACADEMEY
would help u
i and my frnds had the sme problem but is solved now
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i tried... :(
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i tried... :(
try try and you will succeed
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a fair dice is thrown once. what is the probability that the score is greatr than 3 or a prime number or both?
i added 3/6 + 3/6 + ( 3/6*3/6) , my ans being 5/4 wherews ans given is 5/6 =s
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a fair dice is thrown once. what is the probability that the score is greatr than 3 or a prime number or both?
i added 3/6 + 3/6 + ( 3/6*3/6) , my ans being 5/4 wherews ans given is 5/6 =s
(x>3)=4,5,6
p of (x>3)=3/6
(x a prime number)= 2,3,5,
p of (x a prime number)
So both =2,3,5 and 4,5,6
5 is repeated so (2,3,4,5,6)=p of both 5/6
the Question was to confuse you .
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rex has an old car. the probability that he is able to start the car on a rainy day is 0.2 and the probability that he is able to start it on a fine day is 0.6 . if the weather is either fine or rainy, and probability the weather is fine is p, the find in terms of p
1) the probability that he is unable to start the car on a day
2) the conditional probability that a day is fine and he is unable to start the car?
can someone solve this question please ???
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One mo
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prob solved in attachment
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prob solved in attachment
thanx Mr. Paul
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No probs
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Need help with this question
Part (e) to find median
I know we use interpolation but i am confused with the values to use for the upper/lower cumulative frequency..
thx
This is S1 Q5 from 2007 Jun
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Need help with this question
Part (e) to find median
I know we use interpolation but i am confused with the values to use for the upper/lower cumulative frequency..
thx
This is S1 Q5 from 2007 Jun
First of all make a cumulative frequency table.
you have a total of 100. Thus median should be the 50th entry.. so when you make the cf you will find the the 50th tab lies in 14-18 class and Cf is 26-50.. here it is not neccessary to do interpolation because 50 is the last one in the class so you can directly pick median of t is 18 seconds. but you can still do interpolation to check your answer.
(X - 14) / 18-14 = 50-26 / 50 -26
X-14 / 4 = 1
X-14 = 4
X = 14 + 4 = 18
In the same manner u find the quartiles... since the total frequency is 100 so lower quartile is 100/4 = 25 and upper is 100*3/4 = 75th entry
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I have a question regarding Normal Distribution..
I understand how to get and find probabilities but how do you find the mean and standard deviation when the lower quartile and upper quaritle are given? like in this question..
(http://)
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in normal distribution,
median =means
and so median= (LQ+UQ)/2
using mean value find s.d
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in normal distribution,
median =means
and so median= (LQ+UQ)/2
using mean value find s.d
BOLD - How can this be possible ?
Take upper quartile as 75 % (probability of 0.75) OR lower quartile as 25 % (probability of 0.25). Though I recommend you to take 0.75 because for LOWER QUARTILE (0.25) you have to do a bit of more subtraction which will be confusing.
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Can someone help me solve:
May 2010 Q7 Part (b)?
I'm kinda confused with Normal distribution.
Thanks. :)
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Can someone help me solve:
May 2010 Q7 Part (b)?
I'm kinda confused with Normal distribution.
Thanks. :)
Variant ?
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Variant ?
Variant 1
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Peter will continue UNTIL he has got an answer wrong. It will then not matter him getting answer by being helped.
Working in attachment :
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Can someone suggerst me really hard edexel S1 past papers for practice?
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Hey, I do Edexcel S1, please post all your doubts here!!!
hello
i just have one question
June 2008 Q7(c)
question about normal dist.
I dont get why they are multiplying by 3?
Thank you loads in advance.
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Guys, I need help in those theoretical questions that they ask, how do you 'apply' your knowledge of the subject there?
Like is there any rule sort of thing that we need to know, for example in mechanics by assuming that the string is inextensible, the acceleration is the same, so is there anything similar to that in S1?
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Guys try solving Solomon paper L and someone tell me why did they take the value of n as 121 to find the Quartiles and not 120 in Q7a.
http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FMaths%2FEdexcel%2FS1%2FSolomon/
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hello
i just have one question
June 2008 Q7(c)
question about normal dist.
I dont get why they are multiplying by 3?
Thank you loads in advance.
Because you are picking 3 bags at any order. so you can go like this. let the bag which is less than 53 be A and if its more than 53 be B.
They want 2 Bs and 1 A,
If they are chosen at random the arrangement can come
B B A
B A B
A B B
so there are 3 ways of getting 2 bags which weigh more than 53 and one bag which weighs less. so you multiply the probabilities and then add them or just find the probability once and multiply it by 3. Hope this helps.
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ok...i have my s1 exam coming(i mean we all do :P)...so i have a few doubts.......its from may/june 2007 qns 1....i dont get it at all
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ok...i have my s1 exam coming(i mean we all do :P)...so i have a few doubts.......its from may/june 2007 qns 1....i dont get it at all
For part a)
r=Sxy/?(SxxSyy) = -808.917/?(113573 x 8.657) = -0.816 (3sf)
and by the way, the formula is given in the formula booklet.
b) Since the product moment correlation coefficient is a negative value, then this means that as one value increases, the other value decreases, so in this case, the houses become cheaper as they become further away from the station.
c) The product moment correlation coefficient is not affected by coding i.e. if you add, subtract, multiply or divide the original data for coding, it'll not change, so it's value is the same.
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wait.....is this page for edexcel paper???....because im doing CIE!!!opsie :-[
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ok...i have my s1 exam coming(i mean we all do :P)...so i have a few doubts.......its from may/june 2007 qns 1....i dont get it at all
Let x = t - 35
Therefore, Mean = -15/12 = -1.25
However, this is the mean for (t-35)
The mean for t = -1.25 + 35 = 33.75 ~ 33.8
The s.d for x (i.e t-35) is the same as the s.d for t
So, S.d =
= = 2.3
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thanks alot!!!
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Guys try solving Solomon paper L and someone tell me why did they take the value of n as 121 to find the Quartiles and not 120 in Q7a.
http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FMaths%2FEdexcel%2FS1%2FSolomon/
& Solomon paper J, why are they taking the value of n as 56 instead of 55 in Q1a?
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wait.....is this page for edexcel paper???....because im doing CIE!!!opsie :-[
I thought you're Edexcel because the S1 paper is actually tomorrow.
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I thought you're Edexcel because the S1 paper is actually tomorrow.
no im CIE.......its ok.......BEST OF LUCK for ur s1 exam tomorrow!!!ALL of u ;)
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hi,
1 doubt...perhaps the last one
may/jun 2010 Q3 part e) ???
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okay I have something which I want to make clear of. When you code data.. Product moment correlation is not affected, Variance and standard deviation are only affected when you multiply and divide and Mean is affected by everything. Am I right?
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Can someone please explain Q7(a) in Jan 2011 S1 paper?
Thanks. :)
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I got it. LOL :P
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CIE
9709_s10_qp_62
Q 7 (ii).
I just want to know any short method, i have got the answer but with a way too long method.
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CIE
9709_s10_qp_62
Q 7 (ii).
I just want to know any short method, i have got the answer but with a way too long method.
as the markscheme suggests, its simple.
The no. of ways the pink card is not next to the green card = Total no. of ways - (the no. of ways the green card and the pink are together)
The no. of ways the pink and green are together...
There is a very neat trick to find this, it's in there in the CIE S1 book
U consider the 2 cards as a unit, and permute them with the other 7
Like this [pink, green] + 7 others... = total 8 (regarding the pink, green as one unit)
The permutation = 2*8!
Subtract this from 9! and u have ur answer
9!-(2*8!)
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as the markscheme suggests, its simple.
The no. of ways the pink card is not next to the green card = Total no. of ways - (the no. of ways the green card and the pink are together)
The no. of ways the pink and green are together...
There is a very neat trick to find this, it's in there in the CIE S1 book
U consider the 2 cards as a unit, and permute them with the other 7
Like this [pink, green] + 7 others... = total 8 (regarding the pink, green as one unit)
The permutation = 2*8!
Subtract this from 9! and u have ur answer
9!-(2*8!)
Thank you. So for combinations also I think this would be easier. +rep(I have to wait 114 minutes)
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Do we need to know the 52 cards pack!?
Do we need to know how to derive the formulas of vairance and standard deviations in sets of date or binomial distributions or why?!?!
Thanks...
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Do we need to know the 52 cards pack!?
Do we need to know how to derive the formulas of vairance and standard deviations in sets of date or binomial distributions or why?!?!
Thanks...
52 pack cards - yes. (No jokers ofcourse)
Derive the formulas for variance and s.d? Nah
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52 pack cards - yes. (No jokers ofcourse)
Derive the formulas for variance and s.d? Nah
can you tell me the 52 card pack?
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can you tell me the 52 card pack?
Erm, ok sure. 52 cards. 26 red and 26 black. four As, four 2s, four 3s, four 4s................four 10's, four jacks, four queens, four kings.
13 spades, 13 hearts, 13 diamonds and 13 clubs.
Find attached image. Source : wiki
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june 2008 paper 6 q5 part ii
when we get phi(t) = 0.7389
then how come the next step is
phi (t) = phi (0.64)
where did 0.64 come from? =S
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june 2008 paper 6 q5 part ii
when we get phi(t) = 0.7389
then how come the next step is
phi (t) = phi (0.64)
where did 0.64 come from? =S
I got that by the method given bellow
Take the medians of the time spent intervals
i.e. :- 0.3, 0.8, 1.55, 2.55, 3.8
now use the formula sigma fx
--------
sigma f
sigma fx = 3.3+12+27.9+76.5+79.8 = 199.5
sigma f = 11+15+18+30+21=95
:. 199.5/95=2.1
the mean is 2.1 hours
hope it helps
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june 2008 paper 6 q5 part ii
when we get phi(t) = 0.7389
then how come the next step is
phi (t) = phi (0.64)
where did 0.64 come from? =S
I guess the question/question paper you stated isn't the question you are confused about.
Anyway, i dont know the q, but i guess i understand ur doubt.
phi(t) = 0.7389
Meaning, for which value of 't' is the area 0.7389. For this u gotta look at the normal distribution table. You'll find it in the CIE S1 book
and the syl.
I've attached the part of the table where I found 0.7389. If you look closely, you'll see that cell (red) is in the 6th column (starting from column 'z') and the 7th row (starting from 0.0). 0.60 + 0.04 = 0.64,
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For something like (x>35) how do you know when to use x.5-meuow/s.d. how do you know its x.5 and not x, is there a pattern!! agghhh!! im panicking before my s1 exam!! :'( help someone!! like for 62/ON/10 qs 6 (iii) i had no clue which one to use!! :/!! ??? :(
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For something like (x>35) how do you know when to use x.5-meuow/s.d. how do you know its x.5 and not x, is there a pattern!! agghhh!! im panicking before my s1 exam!! :'( help someone!! like for 62/ON/10 qs 6 (iii) i had no clue which one to use!! :/!! ??? :(
Listen: You use the .5 thing (called Continuity correction) When you use Normal Approximation instead of binomiall.....For Example:
If its > then its 35.5
IF its < then its 34.5
If its > or equal its 34.5
If its < or equal its 35.5
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so for like 7(i) 63/ON/10 you wouldn't use it because it cant be done binomially?
thanks sooooo much you're a life saver!! - that clears that topic up for me! god bless u!!
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so for like 7(i) 63/ON/10 you wouldn't use it because it cant be done binomially?
thanks sooooo much you're a life saver!! - that clears that topic up for me! god bless u!!
In question 7 i) its already Normal distribution so you dont have to worry abt the C.C (0.5 thing)
I will give you a simple explanation...
If they ask about the prob. of 3 trials from 5 does this event X occur? You would use the binomial as 5C3 x ...x....
But if they ask you the prob of greater than 156 of event X occurs from 200... You could not possibly do 200C156 x...x... + 200C157 x...x... +.... Till 200C200
So thats why we use the continuity correction to approximate it as a Normal Distribution....So tht It would be much easier..So The Continuity correction here would be 15_._ You tell me :D
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156.5!!!!!!! ;D
thank you so much! good luck tomorow!! :)
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156.5!!!!!!! ;D
thank you so much! good luck tomorow!! :)
You are most welcome .. You tooo :D
By the way How are you in Permutations and Combinations??
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Good luck for 2morrow guyz.
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You are most welcome .. You tooo
By the way How are you in Permutations and Combinations??
Not too bad, but most of the times i just get lucky, everyone, i mean EVERYONE in our class can do them but are always unsure of the answer being right, i guess it's the same with everyone else from what i've heard
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Not too bad, but most of the times i just get lucky, everyone, i mean EVERYONE in our class can do them but are always unsure of the answer being right, i guess it's the same with everyone else from what i've heard
True, very true. Always unsure of the answer..
*sighs* , let's just hope tomorrow's paper is easy.
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Not too bad, but most of the times i just get lucky, everyone, i mean EVERYONE in our class can do them but are always unsure of the answer being right, i guess it's the same with everyone else from what i've heard
Well My problems are all In permutations and combinations as they started to get HArd from 2009-2010 :'(... I hope it comes easy tmrw... Best OF luck for all Of uss!! ;)
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June 2010 Variant 63 question 4 II)
Can some one plzzz explain the third way in the mark scheme???
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Im not sure but this is how i understood it, this is the first time im looking at it
SO the first three colas are next to each other, [CCC]
lets just let that be one big C, green tea = G, oranje juice = J
there are only three different places we can place the G:
1) G C G J J 2) G C J G J 3) G C J J G
hence the 3
note that the Gs can not be moved the only thing we can move is C J J hence 3! = 4x3/2
therefore 3x3!=18 or as the mark sceme 3x(4x3/2)
hope i helped ;), good luck everyone im out for the night!
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Im not sure but this is how i understood it, this is the first time im looking at it
SO the first three colas are next to each other, [CCC]
lets just let that be one big C, green tea = G, oranje juice = J
there are only three different places we can place the G:
1) G C G J J 2) G C J G J 3) G C J J G
hence the 3
note that the Gs can not be moved the only thing we can move is C J J hence 3! = 4x3/2
therefore 3x3!=18 or as the mark sceme 3x(4x3/2)
hope i helped ;), good luck everyone im out for the night!
By the way Remember! YOu cannOT do 3! only There is a repeated term which is the 2 JJ so It must be 3!/2....Couldnt the G be put 2nd and last? so It would have more than 3 laces to be moved to..Its infact 6! SO its actually 6x3!/2....But what I dont get is why they Did 4x3/2 ?!?!
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I guess the question/question paper you stated isn't the question you are confused about.
Anyway, i dont know the q, but i guess i understand ur doubt.
phi(t) = 0.7389
Meaning, for which value of 't' is the area 0.7389. For this u gotta look at the normal distribution table. You'll find it in the CIE S1 book
and the syl.
I've attached the part of the table where I found 0.7389. If you look closely, you'll see that cell (red) is in the 6th column (starting from column 'z') and the 7th row (starting from 0.0). 0.60 + 0.04 = 0.64,
ohhhhhh so DATS how u look at it...i ws lookin at it the other way round.....thankuu soooo veryyyy muchhh =D life saver i tel u :) all d bst wid ur papers bro !
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can someone pls explain qns 3 p62 oct/nov 2010 CIE!
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can someone pls explain qns 3 p62 oct/nov 2010 CIE!
See the attachment below.
Since this is a conditional probability you find out the probability of getting a 12.
If the spinner is thrown on a even side
then there are 4 ways 12 can come.
and the probability of a spinner landing on 12 would 2/5.
So here is Probability of getting a 12 if the spinner lands on the even side
2/5 * 4/36
similarly you do this for the spinner landing on the odd side
which would be 3/5 * 1/36
Now it says it lands on a even side so the fraction should look like this
2/5*4/36
---------
(2/5*4/36)+(3/5*1/36)
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thank you so much dumb economist!
+rep
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Ok...i'll post all my doubts here...pls help me;
1. may/june 2010 p61- qns 2 part2 p62-qns 5 part3and4 qns 7 part4and5 p63-qns6 part 2,3 and4
2.oct/nov 2010 -p63 qns 6 part3 and qns7 part2
3. oct/nov 2009 p61-qns 5(a) part1 and 2 (b) part2 and qns 6 part3
4.may/june 2007 qns 3(b) and 7 part3
5. oct/nov -qns 6 part3 and 7 part3
PLEASE HELP!!!
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thank you so much dumb economist!
+rep
I actually thought u wer dissing him until i read his name lol
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lol
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hez was the paper tady(cie
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it was a big upset for me :'(
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guys is S1 easier or M1 ???
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guys is S1 easier or M1 ???
S! anyday !! definitely WAY better dan m1..i mean I actualy ENJOY doin S1..... :O
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guys is S1 easier or M1 ???
if you do physics then M1 is Easy
if you do Commerce then S1 is easier
decide yourselves what do you do or what fell's easy
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guys is S1 easier or M1 ???
I did both this year!
M1 is good when you understand the concept and practice as much as possible, it's just the calculation part of Physics, so if you take Physics, even with just IGCSE knowledge, that would be fine for you.
S1 is easy to gain marks, and much easier to lose marks due to veryyyy silly mistakes that you may overlook when reading the question, noting down numbers and rounding them off to the correct number of significant figures, yet it's comparatively easier than M1, most of the formulas that you'll use are in the formula booklet and there isn't much of thinking that you'd need to do compared to M1, where you need common sense! :P
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if you do physics then M1 is Easy
if you do Commerce then S1 is easier
decide yourselves what do you do or what fell's easy
well i ve done igcse phy. only and in commerce ve doev acc and business ;D
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well i ve done igcse phy. only and in commerce ve doev acc and business ;D
then think by yourselves
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I did both this year!
M1 is good when you understand the concept and practice as much as possible, it's just the calculation part of Physics, so if you take Physics, even with just IGCSE knowledge, that would be fine for you.
S1 is easy to gain marks, and much easier to lose marks due to veryyyy silly mistakes that you may overlook when reading the question, noting down numbers and rounding them off to the correct number of significant figures, yet it's comparatively easier than M1, most of the formulas that you'll use are in the formula booklet and there isn't much of thinking that you'd need to do compared to M1, where you need common sense! :P
well basicly actually i 've done phy this year but igcse unfortiunately i cannot express foru how shocked i'm >:(with what i've done
so when i wanted to decide which paper to take i was like ::) as all of my friends tookM1 and i was like 'no one to move with me along this route ' :D anyway Thanks for ur advice ;)
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then think by yourselves
ofcourse i will do ;D
i just wanted to know abt the whole issue as to be able to decide Thanks
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ofcourse i will do ;D
i just wanted to know abt the whole issue as to be able to decide Thanks
hope this all helps
and my wishes that you take right step ahead
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well basicly actually i 've done phy this year but igcse unfortiunately i cannot express foru how shocked i'm >:(with what i've done
so when i wanted to decide which paper to take i was like ::) as all of my friends tookM1 and i was like 'no one to move with me along this route ' :D anyway Thanks for ur advice ;)
If you didn't like Physics IGCSE, then I'd advice you to go for S1.
The concepts you'd need from IGCSE for M1 are: speed, acceleration, distance, velocity, displacement, force, mass, weight, moments - go through the syllabus if you want, that's speaking of Edexcel's M1 of course.
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this question is of statistics S1
Find the total number of different 6-digits numbers that can be formed by using the digits 3, 4, 4, 5, 5, 6, 7 .
if one of the number is picked at random, find the probability that
(a) the number is odd,
(b) the number has two digits 5.
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this question is of statistics S1
Find the total number of different 6-digits numbers that can be formed by using the digits 3, 4, 4, 5, 5, 6, 7 .
if one of the digits is picked at random, find the probability that
(a) the number is odd,
(b) the number has two digits 5.
im not sure but is it like this
(a)6!/3!
(b)i didnt get the qns
i dont remember much but if u have the answer u can check with it
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Find the total number of different 6-digits numbers that can be formed by using the digits 3, 4, 4, 5, 5, 6, 7 .
if one of the number is picked at random, find the probability that
(a) the number is odd,
(b) the number has two digits 5.
Must have 2 4's, or 2 5's or both.
If 2 4's there are 6!/2! different numbers and same for two 5's.
If 2 4's and two 5's then there are 7!/2!2! ways for each choice of the other two numbers. There are three ways the other two can be chosen so altogether 2*6!/2! +3*6!/2! ways altogether.
a) if the number is odd then the last digit is 3, 5 or 7.
If the last digit is 3, then the other 5 numbers can be chosen from 4,4,5,5,6,7
Following the logic above there are 2*5!/2! +2*5!/2! ways of choosing these five. The same result is is 7 is the last number.
If 5 is the last number, and the other 5 numbers contain one 4, there are 5! ways and 5!/2! ways if two 4's.
Hence preb of being odd is (2(2*5!/2! +2*5!/2!)+5!+5!/2!)/2*6!/2! +3*6!/2!)
b)Two 5's so one 4. There are 6!/2! ways this can be down. Prob = 6!/2!/(Find the total number of different 6-digits numbers that can be formed by using the digits 3, 4, 4, 5, 5, 6, 7 .
if one of the number is picked at random, find the probability that
(a) the number is odd,
(b) the number has two digits 5.
Must have 2 4's, or 2 5's or both.
If 2 4's there are 6!/2! different numbers and same for two 5's.
If 2 4's and two 5's then there are 6!/2!2! ways for each choice of the other two numbers. There are three ways the other two can be chosen so altogether 2*6!/2! +3*6!/2! ways altogether.
a) if the number is odd then the last digit is 3, 5 or 7.
If the last digit is 3, then the other 5 numbers can be chosen from 4,4,5,5,6,7
Following the logic above there are 2*5!/2! +2*5!/2! ways of choosing these five. The same result is is 7 is the last number.
If 5 is the last number, and the other 5 numbers contain one 4, there are 5! ways and 5!/2! ways if two 4's.
Hence preb of being odd is (2(2*5!/2! +2*5!/2!)+5!+5!/2!)/2*6!/2! +3*6!/2!)
b)Two 5's so one 4. There are 6!/2! ways this can be done so 6!/2!/(2*6!/2! +3*6!/2!)
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Eight People sit in a minibus: four on the sunny side and four on the shady side. If two people want to sit on opposite sides to each other and another two people want to sit on the shady sides, in how many ways can this be done?
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here
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Thanks for clarifying this.. but you have written 6! instead of 4!
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you are right. Should be 4!
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Disco Lights are arranged in a vertical line. How many different arrangements can be made from two green, three blue and four red lights if at least eight lights are used?
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eight lights used
1 red three blue four red 1!*3!*4!
2 red two blue four red 2!*2!*4!
2 red three blue three red 2!*3!*3!
nine lights used 2!*3!*4!
add all these up
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Three letters are selected at random from the word SCHOOL. Find the probability that the selection
i. does not contain the letter O,
ii. contains both the letters O.
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i. Is it 4/6?
Doing probability maybe after 2 years. :)
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For both problems, answers are given as 1/5 i took out 2/7 for both.
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For both problems, answers are given as 1/5 i took out 2/7 for both.
Sorry maybe someone else would answer.
But why 2/7?
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No. of selections with no O's=4c3=4
No. of selections with one O= 4C2=6
No. of selections with two O's=4C1=4
Total Selections=6+4+4=14
P(Does not contain letter O)=4/14=2/7
P(Contains both the letters O)=4/14=2/7
This is how i did but the answers given were 1/5. I think my solution is right but want to confirm it.
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attached Q4 part (b0
and Q5 plz ??? :'( :'(
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Try this
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Try this
Thank you Sir .......... but I'm sorry if I disturbed you by asking you plz could you explain the book answer
attached .................what I don't understand is how he get the probability of sticking to the door choosen before and getting the correct door ??? ??? :-\ :-\ :-\ :-\
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There are three doors and initially the probability of picking the correct door is 1/3
After the host picks one of the two not chosen, it is as if the probability of the car being behind the door he has chosen is given to the one remaining unchosen door. so 1/3 +1/3=2/3
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Ann, Barry and Clare are three students taking a multiple choice examination paper. For each question a student has to select the correct answer from five that are offered. For Question 1, Ann has no idea of the correct answer, Barry correctly identifies one answer that is wrong and Clare correctly identifies two wrong answers. All three students decide to guess at random from the answers they think stand a chance of being correct. Calculate the probability that:
a. none of the three students chooses the correct answer.
b. Clare is the only one to choose the correct answer.
c. exactly one of the three students chooses the correct answer.
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here
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I have one confusion. For the question attached below, if we are said to draw a cumulative frequency curve, then can we use 100,150,200 on the x-axis of the curve or should we use 100.5,150.5,200.5 as we use the value of Upper Class Boundary while drawing C.F curve. I am really confused.
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I have one confusion. For the question attached below, if we are said to draw a cumulative frequency curve, then can we use 100,150,200 on the x-axis of the curve or should we use 100.5,150.5,200.5 as we use the value of Upper Class Boundary while drawing C.F curve. I am really confused.
You may use 100, 150, 200 etc. You'd use 100.5, 150.5, 200.5 when there wouldn't be an equal sign there. :)
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Cant see attachement but use 150 etc anyway
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@ astarmathsandphysics you can enlarge the attachment by clicking it. I am still in little confusion. I quote one sentence from my statistics books: "The cumulative frequencies are plotted against the upper class boundaries of the corresponding class." Our class boundary is written in the form x-y where x is inclusive but y is not. However, in the given question 150 200 are inclusive. So, to make it exclusive I add 0.5 and make it 150.5. For this same question mean has also been asked for 4 marks. and it has been done by using 0.5-100.5,100.5-150.5... in the mark scheme.By the way, the full question is from CIE 9709 May June 2011 Paper 61 question no. 6. I also read examiner report. In examiner report the statement was like this: "A mark was awarded for attempting to plot the cumulative frequencies against their respective upper class boundaries"
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@ astarmathsandphysics you can enlarge the attachment by clicking it. I am still in little confusion. I quote one sentence from my statistics books: "The cumulative frequencies are plotted against the upper class boundaries of the corresponding class." Our class boundary is written in the form x-y where x is inclusive but y is not. However, in the given question 150 200 are inclusive. So, to make it exclusive I add 0.5 and make it 150.5. For this same question mean has also been asked for 4 marks. and it has been done by using 0.5-100.5,100.5-150.5... in the mark scheme.By the way, the full question is from CIE 9709 May June 2011 Paper 61 question no. 6. I also read examiner report. In examiner report the statement was like this: "A mark was awarded for attempting to plot the cumulative frequencies against their respective upper class boundaries"
I did that paper some time back. And I got the same confusion as yours.
Yeah, now I get what you mean. You're right. Less or equal to 100 means it stops at 100. So, you cannot use 100 as the starting point for the other class. You have to add 0.5. Make it 100.5.
It's actually the difference between a discrete and continuous variable. The cumulative frequency uses a continuous variable, when you convert it to a simple frequency table, it becomes a discrete variable.
Hope you get me. :)
And thanks for asking that question. My confusion is cleared, albeit by myself.:P
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Hi! Doubt in a continuous random variable question.
Q) 5 from this paper attached!
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here
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1.
A machine is used to fill cans of soup with a nominal volume of 0.450 litres. Suppose that the machine delivers a quantity of Soup which is normally fistributed with mean x litres and standard deviation y litres. Given that x= 0.457 and y= 0.004, find the probability thata randomly chosen can will contain less than the nominal value.[Ans:0.0401]
It is required by law that no more than 1% of cans contain less than the nominal volume
Find.
i. the least value of x which will comply with the law if y=0.004 [Ans:0.459]
ii. the greatest value of y which will comply wit the law if x= 0.457 [Ans:0.003]
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Give me time to get to pc
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One more confusion:
Q. On a particular day, 50% of the employees in a large company had arrived at work by 8.30 am, and 10% had not arrived by 8.55 am.
a. Assuming a normal modal, find the standard deviation of arrival times, in minutes.[Ans:19.50]
b. It is given that only 5% of the employees had arrived by 8.05 am. Without further calculation, explain why this might suggest that normal model is not appropriate [Ans: not symmetrical ... but can you give me explanation..... what is means..]
c. Eighty employees are selected at random. Find the expectation of the number of these employees that arrived between 8.30 and 8.55 a.m.[Ans:32]
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1st questions above
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last question above
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I am taking CIE statistics 1 this year.
I have doubts relating to mean,median and mode. Although they are just the basics, i am still confused:
1) whenever we draw a histogram, do we always have to take 'frequency densities' on the y-axis? Or do we also take 'frequency' on the y-axis?
2) for grouped data, and the frequencies given, how do we find the median without drawing the cummulative frequency graph?
3) for grouped data, and frequencies given, how do we find the upper quartile and lower quartile?
4) what is the difference between variance and standard deviation?
Thanks.
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anyone who can answer my questions? Pls.
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You have to use frequency densities if the class intervals are not the same. Personally I think it is always better to use them cos you get used to them.
Variance = standard deviation squared.
For the median and quartile stuff see
http://www.astarmathsandphysics.com/a_level_maths_notes/S1/a_level_maths_notes_s1_quartiles_and_median_from_frequency_tables.html
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^^thankyou. now i can understand better and continue with my practice. :)
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no probs
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here's a question
Q) the mean snd standard deviation of the heights of 12 boys in a class are 148.8 and 5.4 cm respectively. A boy of height 153.4 cm joins the class. Find the mean and standard deviatiation of the 13 boys.
I found the mean but cant find SD.
____________________________
theres are formulas for finding median and quartiles in grouped data. But what if the data is not grouped and has large frequences and is summarised in a table.then how do we find the quartiles?
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see attachment
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here is my question plz explain his answer
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reasonable for the values given in the table as someone could easily practice 20 hours a week and professional musicians pactrice up to 60 or more. Not reasonable for 168 as only 168 hours in week.
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reasonable for the values given in the table as someone could easily practice 20 hours a week and professional musicians pactrice up to 60 or more. Not reasonable for 168 as only 168 hours in week.
Sir , I'm sorry but from where did you get these values ...is there anymore clarification ???
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Number of hourse x between 1 and 15 and number of hours y between 4 and 18 in the table in your attachment
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reasonable for the values given in the table as someone could easily practice 20 hours a week and professional musicians pactrice up to 60 or more. Not reasonable for 168 as only 168 hours in week.
which means that the graph will curve as it cease to be proportionalor form a linear relationship..........right ??? ::) ::) ::)
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Yes and it will definitetely curve as nutters practice 168 hours a week and make less than zero mistakes
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Yes and it will definitetely curve as nutters practice 168 hours a week and make less than zero mistakes
yes Thanks alot Sir ;D
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thankyou sir, i got how to do it.
But could you pls answer my other question?
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Is that the quantiles question?
http://www.astarmathsandphysics.com/a_level_maths_notes/S1/a_level_maths_notes_s1_quartiles_and_median_from_frequency_tables.html
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no sir. I am sorry but this isnt exactly my doubt. I wanted to know how we find the quartiles for large data with no classes.
For example:
Score : 0 1 2 3 4 5 6
frequency : 36 94 48 15 7 3 1
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For this one, there are 36+94+48+15+7+3+1=214
I would do this:214/4=53.5 so the lower quartile is 1
For the upper quartile find 3/4*215=160.5 so the upper quartile is 2
More complicated methods exist but for S1 do as above..
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thankyou,sir.oh, but my i already know about this method, but when solving some questions like this, i sometimes get the wrong answer.
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Give examples. This method is for Edexcel.
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A question:
An ordinary dice was thrown 50 times and the resulting scores were summarised in a frequency table. The mean score was calculated to be 3.42. It was later found that the frequencies 12 and 9,of two consecutive scores,had been swapped. What is the correct value of the mean?
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differene of12n+9(n-1)-12(n-1)-9n==12-9=3
3/50=0.06
3.42+0.06=3.48
or 3.42-0.06=3.36
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can somone help me with these two questions:
1-Three letters are selected at random from the word SCHOOL.Find the probablity that the selection:
a)does not contain the letter O
b)contains both letters O
2-In how many different ways can a football team of 11 players be photographed if there are to be two rows ,with 5 seated on the front row and 6 standing in the back row?In how many of these photographswould the captain be sitting in the centre of the front row?
thanx in adavnce
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1a.4/6*3/5=12/30=2/5
b.2/6*1/5=1/15
2.11! cos each player is an individual and they can be told apart.
The captain can sit on the front in one of five places and the others can arrange themselves in 10! ways so5*10!
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1a.4/6*3/5=12/30=2/5
b.2/6*1/5=1/15
2.11! cos each player is an individual and they can be told apart.
The captain can sit on the front in one of five places and the others can arrange themselves in 10! ways so5*10!
Sir why how did you find the first answer ??? why did you mulitply 4/6 * 3/5 ...???
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cos there are four non 'o' letters out of 6 so prob of not picking o is 4/6
when this letter is picked, there are three non 'o' letters out of five, so prob of picking non 'o' letter is 3/5
then 4/6*3/5
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theeducationchannel.info is working now. You can also ask for a 'video solution'
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cos there are four non 'o' letters out of 6 so prob of not picking o is 4/6
when this letter is picked, there are three non 'o' letters out of five, so prob of picking non 'o' letter is 3/5
then 4/6*3/5
but that's with assuming that the number is not replaced ......right ??? ok if that's the case then it might me 4/6*3/5*2/4 as he picked three numbers right ???
Thanks for the link I'll try it ;)
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exactly
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That's me in the videos
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hello
I would really appreciate it if someone explain the atttached question for me
Thanks in advance :)
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z=(23-25)/4=-.5
look this up. Since -.05<0, look up 0.5 and subtract from 1.
1-0.6915=0.3085
Since this is for each of the three bikes the answer is 0.3085^3
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z=(23-25)/4=-.5
look this up. Since -.05<0, look up 0.5 and subtract from 1.
1-0.6915=0.3085
Since this is for each of the three bikes the answer is 0.3085^3
Sir I did the same but the answer was that one
attached
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They are wrong. Phone them and tell them.
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They are wrong. Phone them and tell them.
Actually if I do have their number I would do immediately ;)
here is other question which I want to be sure about it
why the Q3 is not 34 only
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They have done it the way I would do it.
Note that if you have 3 number 1,5,7 the middle is the second, the middle of the first and third one (1+3)/2=2
There are 30 numbers so 3/4(1+30)=23.25
34 is the 23rd one but I want the 23.25th one so go 1/4 of the way from 34 to 36 which is 1/2 then add it to 34.
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They have done it the way I would do it.
Note that if you have 3 number 1,5,7 the middle is the second, the middle of the first and third one (1+3)/2=2
There are 30 numbers so 3/4(1+30)=23.25
34 is the 23rd one but I want the 23.25th one so go 1/4 of the way from 34 to 36 which is 1/2 then add it to 34.
then how the Q1 was 19 ???
Sir is that method above known ??? I mean in my S1 book they never taught us about this method just find 75% of the no. then if it is not a whole no. then round up and find the corresponding term
:-X
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I know. I don't like the book method - it is technically incorrect. My method is the statistically correct method.
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I know. I don't like the book method - it is technically incorrect. My method is the statistically correct method.
ok can plz Sir give me the equations for :
Q1
Q2
Q3 by your method
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There is no equation but actually a method. I treat a list as continuous and use
1/4 (n+1) for lower quartile
1/2 (n+1) for median
3/4 (n+1) for upper quartile.
If any of these result in a decimal I interpolate as I did for your question.
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There is no equation but actually a method. I treat a list as continuous and use
1/4 (n+1) for lower quartile
1/2 (n+1) for median
3/4 (n+1) for upper quartile.
If any of these result in a decimal I interpolate as I did for your question.
Thanks alot Sir
Q1 -the above method can be used for Edexcel too ???
Q2 in the attached pic. it's wrong I'm pretty sure but want to know if you agree or not ???
Thanks in advance :D
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negative skew actually since longer tale on left than right.
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negative skew actually since longer tale on left than right.
hahaha.....Thanks alot
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Sir plz explain that question :o
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because you can have two vowels in three different ways
VVC
VCV
CVV
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because you can have two vowels in three different ways
VVC
VCV
CVV
ohhhhhhh.....it's that simple ??? ......
Thanks alot Sir ;)
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No probs
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Sir , I don't understand in (f) in the attached question
Could you please explain it by your own way as the mark scheme is confusing enough for me ??? ???
ADDED DOUBT
part (c) the 2nd line plz in answer
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Positive skew. The skew is always in the direction of the tail, where there are less readings.
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Positive skew. The skew is always in the direction of the tail, where there are less readings.
ahaaaa...so I can't get it from the box plot but what I shloud think about is the attached one right
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2 question to answer plz ???
regarding the first question what I don't understand is why he didn't use ''square root 10'' as a standard variation ??? what's your opinion
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Your skew diagram is exactly right
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Normal distribution is written N(mean, standard deviation^2) so standard deviation = sigma=10
As for the second I will post an attachment
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here
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here
Thanks alot Sir
+rep
also plz see my answer to this one which one is right ???
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You are using sqrt(10) as the standard deviation but it is 10.
They have the correct zanswer
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You are using sqrt(10) as the standard deviation but it is 10.
They have the correct zanswer
Sir Thanks ...........can you plz look here
I'm sorry........I know I might be asking too much about that question but I really want to understand it :-[ :-[ :-[ :-[ :-[ :-[
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It says sigma^2=10^2
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here we go exam after tom...hoping for a good..soon reply :-[ :-\
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has any one done s1 edexcel june 2011 q8 (g) and June 2004 S1 edexcel q5 (C)?
Please help!
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has any one done s1 edexcel june 2011 q8 (g) and June 2004 S1 edexcel q5 (C)?
Please help!
I think it would be best of you provided a link to the paper or posted the whole question since many of us do not have the paper at hand :(
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http://www.xtremepapers.com/Edexcel/Advanced%20Level/Mathematics/2004%20June/S1Jun04Q.pdf
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june 2011.
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Miss relina will do your q when I get home
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Just 4 u miss relina
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Just 4 u miss relina
OMG!!!!!!!!!!! :o :o :o :o...I don't know how can I thank you :-[ :-[
I can not find the words to say that you are MY LIFE SAVER..........PERFECT 8)
I 've posted these questions on TSR but people there are kind of lazy ::) and annoying :-\
but you really helped me throughout my long journey ..alllll credit goes to you ;D :-*
Thanks alot Sir ;) ;) ;) ;) :) :) :)
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No probs. Pass the goodwill around
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Q) I have 7 fruit bars to last a week. 2 are apricot, 3 are fig and 2 are peach. I select one bar each day. In how many different orders can i eat the bars?
If i select a fruit bar at random each day,what is the probability that i eat the 2 apricot ones on consecutive days?
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7/(2!3!2!)
6/(7/(2!3!2!))
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7/(2!3!2!)
6/(7/(2!3!2!))
thankyou for answering the question but the book answers given in the book are, 210 and 2/7.
And i got a couple of more doubts:
Q) 6 people are going to travel in a six-seater minibus but only 3 of them can drive. In how many diffrent ways can they seat themselves?
Q) 10 people travel in 2 cars, a saloon and a mini. If the saloon has seats for six and the mini has seats for 4, find the number of different ways the party can travel, assuming that the order of seating in each car does not matter and all the people can drive.
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should be 7!/(2!3!2!) and 6/(7!(2!3!2!))
1. first pick the driver - choice of three and the remaining 5 can sit in 5! ways so 3*5!
2.10!/(4!6!)
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thankyou sir, got it.
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good go see the tests here on my website.
Putting up two good tests every day now
http://www.astarmathsandphysics.com/test_centre/
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good go see the tests here on my website.
Putting up two good tests every day now
http://www.astarmathsandphysics.com/test_centre/
found just what i was looking for. The tests are perfect for my revision. Thankyou sir.
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Thanks. Any tests in particular you need ask me.
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Thanks. Any tests in particular you need ask me.
well, i find permutations and combinations questions a little confusing so thats what i need the most practise for.
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I have done 3 tests and will put them uptonight
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I have done 3 tests and will put them uptonight
nice tests. Thankyou sir.
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they are up now. try them. more on way
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Hi
I can't figure out how to answer this question:
On a particular farm, 60% of all eggs laid by hens are classified 'large'. Everyday, the farmer selects the same number of eggs at random and sends them as a batch to the market. Given that the standard deviation of the number of eggs in a batch is 12, find
(a) the number of eggs in a batch,
(b) the mean number of large eggs in a batch.
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12=sqrt(np(1-p))
144=n*0.6*0.4 so n=600
Mean=np=600*0.6=360
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I don't believe it was that easy. Thanks!
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See my S1 videos here
http://theeducationchannel.info/videos/18/A-Level-Maths-S1/most_recent/all_time/
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A commuter, who caught the same rain to work everyday, kept a recod over six months and found that the train was late on 49 days out of a total of 126 working days. Use a binomial probability model to estimate the probability that, in five day working week, the train will be late exactly 3 times. State briefly, giving reasons, whether a binomial model is likely to be a satisfactory or unsatisfactory mathematical model in this situation.
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p=49/126=7/18
P(Late 3 times)=5C3*(7/18)^3*(11/18)^2=0.2196
not suitable cos of strikes etc (in UK at least)
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Can anyone explain me about permutation? With few examples?
Thanks,
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I can explain 2000 times today
See the first two videos here
http://theeducationchannel.info/videos/18/A-Level-Maths-S1/most_recent/all_time/
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A box contains 300 discs of different colours. There are 100 pink discs, 100 blue discs and 100 orange
discs. The discs of each colour are numbered from 0 to 99. Five discs are selected at random, one at
a time, with replacement. Find
(ii) the probability that exactly 2 discs with numbers ending in a 6 are selected, [3]
The marking Scheme says:
P(2 end in 6) = (1/10)2 × (9/10)3 × 5C2 =0.0729
My doubt is..
How on earth is prob success is 1/10???!
for a single disc to be numbered 6.. i am getting 1/100
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Read the question properly brother. It doesn't say "Numbered 6", it says "ending in a 6".
6, 16, 26, 36, 46, 56, 66, 76, 87, 96. 10 numbers from 100. 10/100 = 1/10
I think you can figure out the rest :)
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whoooaaa! jeez, thanks alot bro! :D
that paper trolled the hell outta me xD
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:)
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HI,
Can anyone please tell me hoe to do S1 january 2012 question 7 part C .. tried alot but cant understand ???
thanks :)
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HI,
Can anyone please tell me hoe to do S1 january 2012 question 7 part C .. tried alot but cant understand ???
thanks :)
Link to the paper?