IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: sweetie on May 25, 2010, 04:15:31 pm
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hey ppl,
can sum1 plzzz xplain q1 of paper6 ( statistics) second variant?????/
Thank You
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which year?
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First of all, identify the possible outcomes (i.e. possible sums of the numbers on the dice). These will be 16 in number ('4 sides of dice x 2' as there are 2 dice). If you want (though it's not needed), a possibility diagram can be drawn.
1|2 3 4 5
2|3 4 5 6
3|4 5 6 7
4|5 6 7 8
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1 2 3 4
Hence, the probability of scoring a sum more then or equal to 7 is 3/16. As you can see, it is a binomial distribution, because the probability 'p' (probability of success, as in this case the probability of scoring a sum more then or equal to 7 is 3/16) remains constant with each throw i.e. the probability is independent. The experiment is also being repeated several times. For this,
E(X)=np i.e. Expected Value=(Number of trials)(Probability of success)
therefore, for this distribution E(X)=(200)(3/16)
=37.5
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I guess u xplained the wrong que., but anywayz thanx :)
its no09 q1 2nd variant
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Okay. :)
If you notice the box-and-whisker plot is normally distributed i.e. the mid-point of the interquartile range is the median, and the distribution extends to equal length on either side. Hence, for part (a), the mid-point is the mean, just like in the normal distribution curve, the line of symmetry passes through the mean. i.e. (63+39)/2 = 51.
In the second part, the examiner wants the standard deviation. You have the mean, two values of x and z which can be found by keeping the fact that P(x<=63)=P(x>=39)=0.75. Remember, the quartiles divide the distribution into four parts, such that the Q1 is the lower 1/4th of the distribution, while the Q3 is the 1 - upper 3/4th of the distribution. Meaning thereby, that the probability of x being more than or equal to the Q1 is 0.75, and less than or equal to Q3 is also 0.75. So, z = +0.674 or -0.674. Now just insert these values into the formula:
z = (x - mu)/sigma
Remember, z = -0.674 for x=63, and vice versa. You can use either of the two sets of values.
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Thank You sooooooooo much
+rep
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in nov 09 paper 62.. how do u do q 4???
please explain with detail..
thanks in advance :D :D
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in nov 09 paper 62.. how do u do q 4???
please explain with detail..
thanks in advance :D :D
Look greater then 500 means
it can only be >500
>600
>1000
>3000
>5000
>6000
because thats all numbers we got 1.3.5 and 6
There can four ways to make odd numbers greater then 500
513
531
561
563
similarly greater then 600
613
631
653
635
651
615
which means 6 ways
greater then 1000
1653
1365
1563
1635
= 4 ways
greater then 3000
3165
3561
3651
3615
=four ways
greater then 5000
5163
5361
5631
5613
four ways
greater then 6000
6135
6513
6531
6351
6153
6315
=six ways
4+6+4+4+4+6=28 ways
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Thanks+rep
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You welcome!
Any more questions (past years) i will be glad to help(i hope so)
Good luck!
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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)
varient?
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q3 part 2 nov 2009 variant 1
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varient?
1
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varient?
To ifraha
question 3 part 2 is easy
0.33=short
0.33=long
0.34=standard
this means that area of Z of the maximum time of standard
and x<z=0.34+0.33=0.67
the corresponding value of 0.67 will be z =0.44
a-100
-----=0.44
7
a will be 103.08
the minimum would be
a-100
------=-0.44
7
which will 96.1
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hey im stuck!!
n09 Q5 (a) (i) (ii) and b (ii)
5000 till 6000
so five is fixed
5xxx
5P3=60
part 2)
5xxx
any number could come
so 5C5(6C1)(6C1)(6c1)
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Thanks
could u please answer part b(ii) of the same question
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thnku so muich..:)
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Thanks
could u please answer part b(ii) of the same question
6 boys 8 girls
5 pupils
either they are in the team or all are not in the team
11C2(because only two places will be left if they all are in the team and there will be 11 people altogether remaining if 3 boys are already in the team) +11C5( 3 boys cannot enter so 11 people remaining for 5 places)
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wow u r so smart
u r a life saviour!! r u gonna do the exam in the mornin or r u a teacher?
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wow u r so smart
u r a life saviour!! r u gonna do the exam in the mornin or r u a teacher?
yeah i am going to do
wish me luck
i am not a genius By the way.i just worked really hard for statistics which is basically my weakest subject
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huhuh!!!
u r definitely kidding!!!!!!!!!!!
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huhuh!!!
u r definitely kidding!!!!!!!!!!!
wallah
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oh dear the percentile is goin go high!!!
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r u in pakistan??
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neways n09 q1 part 2
variant 2 please
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neways n09 q1 part 2
variant 2 please
Sorry man i went to sleep
3 hours to go for CIE exam.
I dont have the question paper for this but i think this is the one where you have to find standard deviation sa7?
x<39=0.25 (as this is the lower quartile)
and then
(39-51)sigma=0.25
1-0.25=0.75
which if you look at the table at the bottom on the normal distribution graph states the corresponding value to be 0.674
which will be -0.674 because its on the negative side
-12/sigma=-0.674
sigma=
-12/-0.674=sigma
sigma=17.8