Official Edexcel S3 thread Wednesday 20 May 2015 Morning [6691]
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Original post by nayilgervinho
For Solomon Paper D, q2b why is the answer: 19/20
19/20 = 0.95
0.95 of the students will be expected to have a confidence interval which includes the mean simply because its a 95% confidence interval
Why do you multiply your probability by 2 at the end of part a.
5. An organic farm produces eggs which it sells through a local shop. The weight of the eggs produced on the farm are normally distributed with a mean of 55 grams and a standard deviation of 3.9 grams. (a) Find the probability that two of the farm’s eggs chosen at random differ in weight by more than 4 grams. (5 marks) The farm sells boxes of six eggs selected at random. The weight of the boxes used are normally distributed with a mean of 28 grams and a standard deviation of 1.2 grams. (b) Find the probability that a randomly chosen box with six eggs in weighs less than 350 grams.
5. An organic farm produces eggs which it sells through a local shop. The weight of the eggs produced on the farm are normally distributed with a mean of 55 grams and a standard deviation of 3.9 grams. (a) Find the probability that two of the farm’s eggs chosen at random differ in weight by more than 4 grams. (5 marks) The farm sells boxes of six eggs selected at random. The weight of the boxes used are normally distributed with a mean of 28 grams and a standard deviation of 1.2 grams. (b) Find the probability that a randomly chosen box with six eggs in weighs less than 350 grams.
Original post by nayilgervinho
Why do you multiply your probability by 2 at the end of part a.
5. An organic farm produces eggs which it sells through a local shop. The weight of the eggs produced on the farm are normally distributed with a mean of 55 grams and a standard deviation of 3.9 grams. (a) Find the probability that two of the farm’s eggs chosen at random differ in weight by more than 4 grams. (5 marks) The farm sells boxes of six eggs selected at random. The weight of the boxes used are normally distributed with a mean of 28 grams and a standard deviation of 1.2 grams. (b) Find the probability that a randomly chosen box with six eggs in weighs less than 350 grams.
5. An organic farm produces eggs which it sells through a local shop. The weight of the eggs produced on the farm are normally distributed with a mean of 55 grams and a standard deviation of 3.9 grams. (a) Find the probability that two of the farm’s eggs chosen at random differ in weight by more than 4 grams. (5 marks) The farm sells boxes of six eggs selected at random. The weight of the boxes used are normally distributed with a mean of 28 grams and a standard deviation of 1.2 grams. (b) Find the probability that a randomly chosen box with six eggs in weighs less than 350 grams.
X=weight of egg X1 - X2 = A (make this a single normal distribution)
p( |X1-X2| > 4 )
means 2 * P( A > 4 )
as the difference can be greater than 4 or less than -4
P (A < -4 ) = 1 - phi (4-mew/sigma) = 1 - Y
P (A > 4 ) = 1 - phi (A < 4 - mew/sigma)
add the 2 together, 1- Y + 1 -Y
= 2- 2Y
factor out the 2
2 ( 1 - Y )
(edited 8 years ago)
Also, my predictions for wednesday
Q1. something to do with sampling; probably Quota (hasn't come up yet)
Q2. Spearman's rank question
Q3. Mean testing
Q4. Contingency tables
Q5. Difference of mean
Q6. Testing a distribution, probably normal or uniform distribution (hasn't come up yet)
Q7. Combining 2 normal distributions to form a single distribution
Q8. Bias question
what do you think?
Q1. something to do with sampling; probably Quota (hasn't come up yet)
Q2. Spearman's rank question
Q3. Mean testing
Q4. Contingency tables
Q5. Difference of mean
Q6. Testing a distribution, probably normal or uniform distribution (hasn't come up yet)
Q7. Combining 2 normal distributions to form a single distribution
Q8. Bias question
what do you think?
Original post by tazza ma razza
Also, my predictions for wednesday
Q1. something to do with sampling; probably Quota (hasn't come up yet)
Q2. Spearman's rank question
Q3. Mean testing
Q4. Contingency tables
Q5. Difference of mean
Q6. Testing a distribution, probably normal or uniform distribution (hasn't come up yet)
Q7. Combining 2 normal distributions to form a single distribution
Q8. Bias question
what do you think?
Q1. something to do with sampling; probably Quota (hasn't come up yet)
Q2. Spearman's rank question
Q3. Mean testing
Q4. Contingency tables
Q5. Difference of mean
Q6. Testing a distribution, probably normal or uniform distribution (hasn't come up yet)
Q7. Combining 2 normal distributions to form a single distribution
Q8. Bias question
what do you think?
I think you're right about uniform distribution, except i think itll be continuous. also for bias, they've only ever asked about the mean, so we don't need to know for variance right?
also just wanted to ask, if it's (for e.g)
P(|z|<3) is that the same thing at P(-3<z<3) or is it better to do 2 x P(z<3) (at if the answer is > 1 you minus 1)
is this right??
(edited 8 years ago)
Original post by mmms95
I think you're right about uniform distribution, except i think itll be continuous. also for bias, they've only ever asked about the mean, so we don't need to know for variance right?
also just wanted to ask, if it's (for e.g)
P(|z|<3) is that the same thing at P(-3<z<3) or is it better to do 2 x P(z<3) (at if the answer is > 1 you minus 1)
is this right??
also just wanted to ask, if it's (for e.g)
P(|z|<3) is that the same thing at P(-3<z<3) or is it better to do 2 x P(z<3) (at if the answer is > 1 you minus 1)
is this right??
Variance isn't that hard to learn basically just squaring 1/n as your coeff of sigma, n lots of sigma and then cancel the nsigma^2/n^2 -> sigma^2/n
proof done!
so like a rectangular one?
i would work it out the long way (more room for mistakes due to rounding / calculator error but arguably more room for method marks) then check at the end - you finish in like 45 - 50 mins anyway so you have bags of time!
but regarding your issue, yeah, i would believe so - check page 4 in your s3 textbook to double check!
(edited 8 years ago)
For 6 b ii I can't quite explain it.6. A survey found that of the 320 people questioned who had passed their driving test aged under twenty-five, 104 had been involved in an accident in the two years following their test. Of the 80 people in the survey who were aged twenty-five or over when they passed their test, 16 had been involved in an accident in the following two years. (a) Draw up a contingency table showing this information. (2 marks) It is desired to test whether the proportion of drivers having accidents within two years of passing their test is different for those who were aged under twenty-five at the time of passing their test than for those aged twenty-five or over. (b) (i) Stating your hypotheses clearly, carry out the test at the 5% level of significance. (ii) Explain clearly why there is only one degree of freedom. (11 marks)
A quick question:
what would you do if there is tied ranks? (2 marks) how would you answer that?
On the textbook, it said use spearman rank since it is good approximation anyway, but i have a feeling that we need to use pmcc instead? in this case, do you use pmcc on the ranks or the original values of x and y
what would you do if there is tied ranks? (2 marks) how would you answer that?
On the textbook, it said use spearman rank since it is good approximation anyway, but i have a feeling that we need to use pmcc instead? in this case, do you use pmcc on the ranks or the original values of x and y
Original post by mmms95
Yes I know the significance of the modulus :P anyways, I got it! (drew a diagram to understand) but thanks anyways 
Yeah, the diagram does help. If it p(|W| > 2) say, that = 2p(W >2). But if it was p(|W| < 2) that's equal to p(-2 < W < 2)
Original post by DCMed96
A quick question:
what would you do if there is tied ranks? (2 marks) how would you answer that?
On the textbook, it said use spearman rank since it is good approximation anyway, but i have a feeling that we need to use pmcc instead? in this case, do you use pmcc on the ranks or the original values of x and y
what would you do if there is tied ranks? (2 marks) how would you answer that?
On the textbook, it said use spearman rank since it is good approximation anyway, but i have a feeling that we need to use pmcc instead? in this case, do you use pmcc on the ranks or the original values of x and y
There was a previous question in the last couple of years where you have to find the SRCC by taking intervals of the tied ranks and then using and explaining why PMCC is better.
please can someone explain june 13 7b to me I dont understand at all!
here is the question and markscheme:
here is the question and markscheme:
Attachment not found
Original post by mmms95
I think you're right about uniform distribution, except i think itll be continuous. also for bias, they've only ever asked about the mean, so we don't need to know for variance right?
also just wanted to ask, if it's (for e.g)
P(|z|<3) is that the same thing at P(-3<z<3) or is it better to do 2 x P(z<3) (at if the answer is > 1 you minus 1)
is this right??
also just wanted to ask, if it's (for e.g)
P(|z|<3) is that the same thing at P(-3<z<3) or is it better to do 2 x P(z<3) (at if the answer is > 1 you minus 1)
is this right??
P(|z|<3) = P(-3<z<3) always, by definition.
It is only 2 x P(z<3) if the distribution is symmetric about 0. Thus it is true for the standard normal, but won't be for a uniform distribution between 0 and 1 say.
JUNE 13 PART B ANSWER: The question wants you to combine both sets of data together to work out the variance. Firstly, what you need to do is work out Σx for the one which has 32 samples by doing: 32 x 4.55 = 145.6 , then you add 145.6 to the total of the 8 samples from part a (33.29) which gives 178.89. Then what you want to do the same to find the total for Σx^2. once you've done that you can work out the unbiased estimate for the variance. Once you have that, use the standard error of the mean formula to work out the standard error of the mean. Hope this helps!
Original post by roscoe xbo
JUNE 13 PART B ANSWER: The question wants you to combine both sets of data together to work out the variance. Firstly, what you need to do is work out Σx for the one which has 32 samples by doing: 32 x 4.55 = 145.6 , then you add 145.6 to the total of the 8 samples from part a (33.29) which gives 178.89. Then what you want to do the same to find the total for Σx^2. once you've done that you can work out the unbiased estimate for the variance. Once you have that, use the standard error of the mean formula to work out the standard error of the mean. Hope this helps!
Could you kindly explain how to calculate a combined variance?
Original post by roscoe xbo
JUNE 13 PART B ANSWER: The question wants you to combine both sets of data together to work out the variance. Firstly, what you need to do is work out Σx for the one which has 32 samples by doing: 32 x 4.55 = 145.6 , then you add 145.6 to the total of the 8 samples from part a (33.29) which gives 178.89. Then what you want to do the same to find the total for Σx^2. once you've done that you can work out the unbiased estimate for the variance. Once you have that, use the standard error of the mean formula to work out the standard error of the mean. Hope this helps!
thank you so much!
I'm not sure I completely understand how to find Σx^2, in the mark scheme they do this:
why do I get why they do 4.552 x32 but why do they also do 32x0.25??
Original post by pol234
thank you so much!
I'm not sure I completely understand how to find Σx^2, in the mark scheme they do this:
why do I get why they do 4.552 x32 but why do they also do 32x0.25??
I'm not sure I completely understand how to find Σx^2, in the mark scheme they do this:
why do I get why they do 4.552 x32 but why do they also do 32x0.25??
they do the 31 x 0.25 because of the formula of the unbiased estimator of the vairance: σ^2 = (nS^2) / n-1 . They re-arranged it to get 31 x 0.25( this is the unbiased estimator for the variance).
Again hope this helps
Original post by V0ldemort17
Could you kindly explain how to calculate a combined variance?
You use the unbiased estimator of the variance formula to work out the combined variance. However, the hardest part about this question is working out what Σx^2 is of the combined sample is. once you have that it is easy
To work out the total sample value of Σx^2, you have to work backwards using the 0.25 you got given in the question. Using the unbiased estimator formula you can work out what the biased estimator of the variance is (from the sample of size 32). if you do: 0.25 = (32 x S^2) / 31 and re-arrange it, you get S^2 as 31/128. then use that number to work out what Σx^2 is for the sample of size 32. You do this by using the 'normal' variance formula: (Σx^2 / n) - (4.55)^2 = 31/128. Re-arrange that and you get Σx^2 = 670.23. Then you just add 141.4035 from the sample size of 8 to get the total Σx^2!
Sorry this is a bit long winded but this is a tough question.
Hope this helps
Original post by roscoe xbo
You use the unbiased estimator of the variance formula to work out the combined variance. However, the hardest part about this question is working out what Σx^2 is of the combined sample is. once you have that it is easy
To work out the total sample value of Σx^2, you have to work backwards using the 0.25 you got given in the question. Using the unbiased estimator formula you can work out what the biased estimator of the variance is (from the sample of size 32). if you do: 0.25 = (32 x S^2) / 31 and re-arrange it, you get S^2 as 31/128. then use that number to work out what Σx^2 is for the sample of size 32. You do this by using the 'normal' variance formula: (Σx^2 / n) - (4.55)^2 = 31/128. Re-arrange that and you get Σx^2 = 670.23. Then you just add 141.4035 from the sample size of 8 to get the total Σx^2!
Sorry this is a bit long winded but this is a tough question.
Hope this helps
To work out the total sample value of Σx^2, you have to work backwards using the 0.25 you got given in the question. Using the unbiased estimator formula you can work out what the biased estimator of the variance is (from the sample of size 32). if you do: 0.25 = (32 x S^2) / 31 and re-arrange it, you get S^2 as 31/128. then use that number to work out what Σx^2 is for the sample of size 32. You do this by using the 'normal' variance formula: (Σx^2 / n) - (4.55)^2 = 31/128. Re-arrange that and you get Σx^2 = 670.23. Then you just add 141.4035 from the sample size of 8 to get the total Σx^2!
Sorry this is a bit long winded but this is a tough question.
Hope this helps
Thanks
Original post by roscoe xbo
they do the 31 x 0.25 because of the formula of the unbiased estimator of the vairance: σ^2 = (nS^2) / n-1 . They re-arranged it to get 31 x 0.25( this is the unbiased estimator for the variance).
Again hope this helps
Again hope this helps
thanks for helping me!
(edited 8 years ago)
Original post by pol234
thanks for helping me!
Its ok
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