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# Help on AQA statistics please :) watch

1. I've got myself stuck on a couple of questions on the January 2006 AQA statistics paper so if anyone could explain the working to me that would be really great. I've got the mark scheme but I can't work out how they get from the question to their answer.

Questions:

4)b)ii) & 4)b)iii)

2. (Original post by sammyria)
I've got myself stuck on a couple of questions on the January 2006 AQA statistics paper so if anyone could explain the working to me that would be really great. I've got the mark scheme but I can't work out how they get from the question to their answer.

Questions:

4)b)ii) & 4)b)iii)

and all of question 7

For bii) I would put the data into your calculator , using mid interval approximations. For the standard deviation do the same.
3. For biii) What you need to do is convert 1.5 minutes to how many standard deviations it is away from the mean, then use the tables in the formula booklet to work out how much percent is below/above that.
Sorry about the separate reply I can't see the picture so I have to keep posting and going back
4. (Original post by qwerty54321)
For bii) I would put the data into your calculator , using mid interval approximations. For the standard deviation do the same.
That works for finding the mean of y bar as the result in part 4a) for the mean of the population gives the same answer as the mean of the sample of 36 in part b), the answer being approx 80.25.

But I don't understand for the standard error of y bar. 4a) gives an answer of approx 31.134 for the standard deviation of the population, and in the mark scheme for part b) it gives the answer for the s.d of the sample y bar as 31.134/ root 36. They've divided the population standard deviation by the root of the number of customers in the sample and I just didn't understand why they did this.

Maybe that is just how you find it out but I haven't been taught stats for over a year now and my textbook isn't being very helpful so I didn't want to just assume that's how it's done.

Edit: Ahh I think you use the central limit theorem by which if a sample size n is taken from a distribution the means will be equal but the s.d will be the s.d of the original/root n
If you could confirm this I'd be gratefull.
5. (Original post by sammyria)
That works for finding the mean of y bar as the result in part 4a) for the mean of the population gives the same answer as the mean of the sample of 36 in part b), the answer being approx 80.25.

But I don't understand for the standard error of y bar. 4a) gives an answer of approx 31.134 for the standard deviation of the population, and in the mark scheme for part b) it gives the answer for the s.d of the sample y bar as 31.134/ root 36. They've divided the population standard deviation by the root of the number of customers in the sample and I just didn't understand why they did this.

Maybe that is just how you find it out but I haven't been taught stats for over a year now and my textbook isn't being very helpful so I didn't want to just assume that's how it's done.

Edit: Ahh I think you use the central limit theorem by which if a sample size n is taken from a distribution the means will be equal but the s.d will be the s.d of the original/root n
If you could confirm this I'd be gratefull.
Yeah to find the standard deviation of a sample you do population s.d / sqrt of number of data in the sample.
And CLT says that if you have a large enough sample (n>30) then the data in the sample shows normal dist. Helpful?
6. (Original post by qwerty54321)
Yeah to find the standard deviation of a sample you do population s.d / sqrt of number of data in the sample.
And CLT says that if you have a large enough sample (n>30) then the data in the sample shows normal dist. Helpful?
Yes thanks. I'll be back if I get stuck again.

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