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URGENT Help with A level maths

QP: https://filestore.aqa.org.uk/resources/mathematics/AQA-73573-SQP.PDF

MS: https://filestore.aqa.org.uk/resources/mathematics/AQA-73573-SMS.PDF

Question 14ai, i keep getting that P(X < 127) = 1.516 x 10^-10, what am i doing wrong?
Reply 1
Original post by efsfsf\\eeeeeee
QP: https://filestore.aqa.org.uk/resources/mathematics/AQA-73573-SQP.PDF
MS: https://filestore.aqa.org.uk/resources/mathematics/AQA-73573-SMS.PDF
Question 14ai, i keep getting that P(X < 127) = 1.516 x 10^-10, what am i doing wrong?

> 127?

But the cumulative probability of being > 6.3 standard deviations greater than the mean is tiny ~ 10^(-10) so your ans sounds right (is it single or double tailed? not that it really matters). The numbers here, for whatever reason, are chosen to be well into the tail of a normal distribution.
(edited 1 month ago)
Original post by mqb2766
> 127?
But the cumulative probability of being > 6.3 standard deviations greater than the mean is tiny ~ 10^(-10) so your ans sounds right (is it single or double tailed? not that it really matters). The numbers here, for whatever reason, are chosen to be well into the tail of a normal distribution.

If it is >127 it doesnt change the answer, but if you are saying >127 i agree with that but the mark scheme says <127.

I used normal CD on my calc, lower: -999, upper: 127, mean (u): 123, s.d: 70 / root(12144)

the regular casio a level calc has this, do you know why i didnt get the right answer?
Reply 3
Original post by efsfsf\\eeeeeee
If it is >127 it doesnt change the answer, but if you are saying >127 i agree with that but the mark scheme says <127.
I used normal CD on my calc, lower: -999, upper: 127, mean (u): 123, s.d: 70 / root(12144)
the regular casio a level calc has this, do you know why i didnt get the right answer?

Tbh, I just saw the first one (didnt scroll down) but
https://www.wolframalpha.com/input/?i=Integrate%5BPDF%5BNormalDistribution%5B%5B%2F%2Fquantity%3A123%2F%2F%5D%2C%5B%2F%2Fquantity%3A0.635%2F%2F%5D%5D%2Cx%5D%2C%5B%2F%2Fquantity%3A127%2F%2F%5D%2C%5B%2F%2Fquantity%3A150%2F%2F%5D%5D
gives 1.5*10^(-10) for P(X>127) which is pretty much what your calculator is giving (though use lower as 127 and upper as 150 say) and agree with the previous back of the envelope calc.

The mark scheme should have P(X>127) and as they half the the 5% to 0.025 they should be testing 1.5*10^(-10) against 0.025 (or double both values).
(edited 1 month ago)
Original post by mqb2766
Tbh, I just saw the first one (didnt scroll down) but
https://www.wolframalpha.com/input/?i=Integrate%5BPDF%5BNormalDistribution%5B%5B%2F%2Fquantity%3A123%2F%2F%5D%2C%5B%2F%2Fquantity%3A0.635%2F%2F%5D%5D%2Cx%5D%2C%5B%2F%2Fquantity%3A127%2F%2F%5D%2C%5B%2F%2Fquantity%3A150%2F%2F%5D%5D
gives 1.5*10^(-10) for P(X>127) which is pretty much what your calculator is giving (though use lower as 127 and upper as 150 say) and agree with the previous back of the envelope calc.
The mark scheme should have P(X>127) and as they half the the 5% to 0.025 they should be testing 1.5*10^(-10) against 0.025 (or double both values).

ah so mark scheme was wrong the whole time, thank you for your help.
Reply 5
Original post by efsfsf\\eeeeeee
ah so mark scheme was wrong the whole time, thank you for your help.

Would appear so. While the numbers in this case are extreme (sample mean "miles" in the confidence interval and nowhere near the boundary), Im surprised at them.

Note, Id get in the habit of sketching a bell curve and mark the mean and std dev on, so you get the appropriate confidence interval(s).
(edited 1 month ago)
Original post by mqb2766
Would appear so. While the numbers in this case are extreme (sample mean "miles" in the confidence interval and nowhere near the boundary), Im surprised at them.
Note, Id get in the habit of sketching a bell curve and mark the mean and std dev on, so you get the appropriate confidence interval(s).

Do you mean using Z values? As i prefer to do it this way, would it be wrong in any type of question to do it as i did?
Reply 7
Original post by efsfsf\\eeeeeee
Do you mean using Z values? As i prefer to do it this way, would it be wrong in any type of question to do it as i did?

Whether its z values or probabilities, a sketch would indicate what youre calculating. So if you had
https://onlinestatbook.com/2/calculators/normal_dist.html
and put in
mean 123
std dev 0.635
above 124.3
Youd get the right shaded tail and hopefully youd realise it corresponded to P(X>124.3). Obviously here 124.3 is 127, but thats so far in the tail you cant see it.
Original post by mqb2766
Whether its z values or probabilities, a sketch would indicate what youre calculating. So if you had
https://onlinestatbook.com/2/calculators/normal_dist.html
and put in
mean 123
std dev 0.635
above 124.3
Youd get the right shaded tail and hopefully youd realise it corresponded to P(X>124.3). Obviously here 124.3 is 127, but thats so far in the tail you cant see it.

Links not working?
Reply 9
Original post by efsfsf\\eeeeeee
Links not working?

It was 1/2 hr ago, but
https://homepage.divms.uiowa.edu/~mbognar/applets/normal.html
is similar.

yh i see it now, so the conclusion is basically overwhelming evidence that the mean has changed?

ie almost everyone from the 12144 sample had their mean expenditure changed?
Original post by efsfsf\\eeeeeee
yh i see it now, so the conclusion is basically overwhelming evidence that the mean has changed?
ie almost everyone from the 12144 sample had their mean expenditure changed?

Yes about the overwhelming evidence that the sample mean has changed. Even if you dont like doing it that way, its usually worth estimating (or calculating) the rough z score as this tells you the difference in the means as a multiple of the standard deviations. You normally expect a 95% interval to be mean+/-2 std devs (confidence interval boundary) and 99.7% to be mean+/-3 std devs (confidence interval boundary) and you could mark these values on a sketch. Here 4/0.63~6.3 which is well into the tail of the distribution as the pdf goes down like e^(-#^2), so very quickly.
(edited 1 month ago)

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