IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: I Jimmy I on May 23, 2011, 07:17:31 pm
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Hello everyone!
Im having an AS reset this year and frankly forgot some basics.....
My doubt is Question 5(ii) of Oct/Nov 2006
Given,
Mean(meou)=?
Variance=root(21)
This is how I tried solving it,
1.P(X>10)=0.7389
2.1-phi( (10-meou) / root21)=0.7389
3.-phi(10-meou/root21)=0.7389-1
4.phi(10-meou/root21)=0.2611
After step4 I get confused on how to find z of 0.2611, whether to do phy-1(0.2611) and make it -ve or what!
Someone plz solve this and then ill ask my doubt more clearly
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My doubt was if given to find the phy-1 of any probability below 0.5 then I should do 1-(x) and add a -ve sign infront of that value or not...?
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My doubt was if given to find the phy-1 of any probability below 0.5 then I should do 1-(x) and add a -ve sign infront of that value or not...?
Correct.
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My doubt was if given to find the phy-1 of any probability below 0.5 then I should do 1-(x) and add a -ve sign infront of that value or not...?
Yes add a negative value whenever you do this. This was the only doubt ? So the same you apply to the question in the paper you have posted.
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Ye that was my only doubt :D
Other working is easy :D Just wanted to understand the concept! Anyways tyvm for confirming :D
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Ye that was my only doubt :D
Other working is easy :D Just wanted to understand the concept! Anyways tyvm for confirming :D
Otherwise, you know what to do when value of z is positive or negative right ? Or make two equations and find mean and standard deviation.
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I didnt quite understand what you meant. Could u give a question as an example plz? :)
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I didnt quite understand what you meant. Could u give a question as an example plz? :)
Question 3 :
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s10_qp_61.pdf
Just try these types. They cover almost whole normal distribution and will make the topic easier for you.
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Good news, I solved it with ease but that was definitely a first in using simultaneous equations in statistics or atleast the questions I covered.
Again TYVM for bringing this up :)
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I got this other doubt in question 7(iii) of Oct/Nov 2006
The question is, Using a suitable approximation, calculate the probability that the number of small bands in the office pack is between 88 and 97 inclusive!
My doubt is how to write the inequality...? Whether to write it as 88<X<97 OR 88<=X<=97!
The reason Im asking is because this affects the c.c(continuity correction) while solving the question and therefore get a wrong answer :(
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It says "inclusive" right. So greater than and equal to
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So if the word inclusive wasn't there then I would go with 88<X<97 only?:o
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And could someone confirm the following by simply writing "confirmed"
If given in B.D (Binomial Distribution) and changed to N.D(Normal Distribution) then use of C.C is necessary so.....
If it was 88<X<97 then it would become 88.5<X<96.5
However if it was 88<=X<=97 then it would become 87.5<=X<=97.5
Could some1 plz confirm whether the above is either correct or wrong! TYVM!
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And could someone confirm the following by simply writing "confirmed"
If given in B.D (Binomial Distribution) and changed to N.D(Normal Distribution) then use of C.C is necessary so.....
If it was 88<X<97 then it would become 88.5<X<96.5
However if it was 88<=X<=97 then it would become 87.5<=X<=97.5
Could some1 plz confirm whether the above is either correct or wrong! TYVM!
Yup thts correct..
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Tyvm :D
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Tyvm :D
YAMW :P
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To my knowledge, using the normal distribution to calculate the probability of a random variable X > q (where q is an integer) is the same as
the probability of x >=q. So by mistake, if u make an error in the sign, your final answer won't change. However, you may lose marks for
not using the correct sign if the question clearly stated the conditions.
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P(X>10) = P(V>10.5) when using C.C
P(X>=10) = P(V>=9.5) when using C.C
So it does matter a LOT :)
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P(X>10) = P(V>10.5) when using C.C
P(X>=10) = P(V>=9.5) when using C.C
So it does matter a LOT :)
Yes, for normal distribution model, it does matter agreed. :)
What's C.C?
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Tyvm :D
what is this Tyvm????
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what is this Tyvm????
Thank You Very Much :o
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Thank You Very Much :o
ohh tyvm
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ohh tyvm
Lol your welcome ;)
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Yes, for normal distribution model, it does matter agreed. :)
What's C.C?
Continuity Correction When aproximating a B.D as a Normal D....
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Yes, for normal distribution model, it does matter agreed. :)
What's C.C?
Continuity Correction. Its used when Binomial distributions are converted to Normal distributions when np and nq>5
So u either add or subtract 0.5 to the given P(X>n)
P.S "n" in the P(X>n) is any constant :)
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Ok guys, I have another doubt. :(
Question 5(iii) and (iv) of M/J 2010 ----2nd variant-----
Given, P(Q)= 1/24
P(R)=1/9
Question is,
(iii)Are events Q and R exclusive? Justify your answer.
(iv)Are events Q and R independent? Justify your answer.
If my memory serves me well then exclusive is (A.B)=0 and independent is (A.B) NOT= 0
Marking Scheme says that Q and R are exclusive but ( 1/24*1/9) isnt 0 so HOW...?
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Ok guys, I have another doubt. :(
Question 5(iii) and (iv) of M/J 2010 ----2nd variant-----
Given, P(Q)= 1/24
P(R)=1/9
Question is,
(iii)Are events Q and R exclusive? Justify your answer.
(iv)Are events Q and R independent? Justify your answer.
If my memory serves me well then exclusive is (A.B)=0 and independent is (A.B) NOT= 0
Marking Scheme says that Q and R are exclusive but ( 1/24*1/9) isnt 0 so HOW...?
5(iii)
if P(R and Q)=0
then it is Exclusive
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Ok guys, I have another doubt. :(
Question 5(iii) and (iv) of M/J 2010 ----2nd variant-----
Given, P(Q)= 1/24
P(R)=1/9
Question is,
(iii)Are events Q and R exclusive? Justify your answer.
(iv)Are events Q and R independent? Justify your answer.
If my memory serves me well then exclusive is (A.B)=0 and independent is (A.B) NOT= 0
Marking Scheme says that Q and R are exclusive but ( 1/24*1/9) isnt 0 so HOW...?
5 (iv)
if P(R and Q)=P(R) * P(Q)
then it is Independent
But if it is not equal to it then it is not independent
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Continuity Correction. Its used when Binomial distributions are converted to Normal distributions when np and nq>5
So u either add or subtract 0.5 to the given P(X>n)
P.S "n" in the P(X>n) is any constant :)
Ah ok, thanks
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5(iii)
if P(R and Q)=0
then it is Exclusive
thats what I thought but P(R)*P(Q) is not 0 even tho the Marking scheme said the opposite...
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Ermm guys! I got another doubt.....
Question 6(i) of MJ 2010, 2nd varient.
Plz reply ASAP
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x = 0,1,2
0 -> all ducks -> 5/7*4/6*3/5*2/4 = 1/7
1-> any place out of the 4 can be a geese - > 4C1 * 5/7*4/6*3/5*2/4= 4/7
2-> 4C2*5/7*4/6*2/5*1/4 = 2/7