The Student Room Group

HomeWork is way too hard this week...another q

Scroll to see replies

Original post by Fyer1234
Thanks

but I do t have th z-score because I don't know what value or how to find the value I need to minus the mean with...?


Oh, okay. So this question is doing it the other way round. You look in the table to see which z score corresponds to the top 10%. Then you know z, the mean and the standard deviation so you can use z=xμσz=\frac{x-\mu}{\sigma} to find the score.
Reply 21
Original post by brittanna
Oh, okay. So this question is doing it the other way round. You look in the table to see which z score corresponds to the top 10%. Then you know z, the mean and the standard deviation so you can use z=xμσz=\frac{x-\mu}{\sigma} to find the score.


Thanks

lol I am so lost, been at this for 7 hours and f..I..g hate it.

im looking at the normal distribution table and I can't tell what I'm looking for. I still don't see how the zscore helps though...

Surely the cut-off score is 577 and the percentage of applicants who have test score within two sd of the mean is 95%

i don't see why these two answers are wrong...
Original post by Fyer1234
Thanks

lol I am so lost, been at this for 7 hours and f..I..g hate it.

im looking at the normal distribution table and I can't tell what I'm looking for. I still don't see how the zscore helps though...

Surely the cut-off score is 577 and the percentage of applicants who have test score within two sd of the mean is 95%

i don't see why these two answers are wrong...


On page 19 of this document: https://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/N38210A-GCE-Mathematical-Formulae-Statistical-Tables.pdf
.

You can see that the z score corresponding to being in the top 10% is 1.2816. You can now put this into the standardisation formula.
Reply 23
Original post by brittanna
On page 19 of this document: https://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/N38210A-GCE-Mathematical-Formulae-Statistical-Tables.pdf
.

You can see that the z score corresponding to being in the top 10% is 1.2816. You can now put this into the standardisation formula.


Thanks for your help mate

so, 1.2816 - 500 /60 = -8.316

gosh that looks so wrong. If it is right, I still can't see how that fits into the question....?
Original post by Fyer1234
Thanks for your help mate

so, 1.2816 - 500 /60 = -8.316

gosh that looks so wrong. If it is right, I still can't see how that fits into the question....?


Remember 1.2816 is the z score, so 1.2816=x500601.2816 = \frac{x - 500}{60}.

Where x is the score needed to be in the top 10%.
Reply 25
Original post by brittanna
Remember 1.2816 is the z score, so 1.2816=x500601.2816 = \frac{x - 500}{60}.

Where x is the score needed to be in the top 10%.


thanks again

so this is what I've got -rounded up if that's correct

577-500/60 = z score...?

so if 77 is correct, 577 is the cut-off score which means anything above 577 is in the top 10%

I hope I'm getting closer lol
Original post by Fyer1234
thanks again

so this is what I've got -rounded up if that's correct

577-500/60 = z score...?

so if 77 is correct, 577 is the cut-off score which means anything above 577 is in the top 10%

I hope I'm getting closer lol


I think that's correct, yeah :biggrin:.
Reply 27
Original post by brittanna
I think that's correct, yeah :biggrin:.



Ok awesome! Hopefully someone can verify this

thanks for your patience today, really really appreciated
Yes basically you are rearranging the formula to get x.
Reply 29
Ah ha ok thanks

just one last thing on this bloomin question, is 95% correct for the 2nd part of it? I'm guessing it's not quite as simple as that lol

thanks everyone for your patience
Reply 30
Original post by Fyer1234
Ah ha ok thanks

just one last thing on this bloomin question, is 95% correct for the 2nd part of it? I'm guessing it's not quite as simple as that lol

thanks everyone for your patience


Did you end up finding the answer eventually?
what was it ?
Original post by Fyer1234
Ah ha ok thanks

just one last thing on this bloomin question, is 95% correct for the 2nd part of it? I'm guessing it's not quite as simple as that lol

thanks everyone for your patience

I think it is that simple.
For a normal distribution, 95% of all data lies within 2 standard deviations away from the mean.

Quick Reply

Latest