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Mathematics 7 Online
OpenStudy (anonymous):

I could really use some help, please.. Nicole missed the lesson on normal distribution and needs to do her homework. Explain to Nicole how to use the mean and standard deviation of a normal distribution to determine the top 5% of the population.

OpenStudy (amistre64):

did you miss the lesson as well?

OpenStudy (anonymous):

Pretty much xD

OpenStudy (anonymous):

I'd really appreciate the help if you could.

OpenStudy (amistre64):

this question is vague, but it assumes that a person knows how to find a zscore.

OpenStudy (anonymous):

I realize that it's vague, thank my school for that. -.-

OpenStudy (amistre64):

otherwise, you have to integrate a complicated setup to make any useful determinantions with JUST a mean and sd

OpenStudy (anonymous):

I'm just no really sure on what they want me to say. Should I just write something like you just said?

OpenStudy (amistre64):

well, ive covered this question in depth twice, and theres no pretty answer unless we know what a zscore is since:\[z=\frac{x-mean}{sd}\] to find x, is just algebra \[mean+z(sd)=x\] but as you can see, a knowledge of the zscore is needed in order to determine the results

OpenStudy (anonymous):

Okay, I appreciate it! thanks for the help!

OpenStudy (amistre64):

good luck :)

OpenStudy (anonymous):

Thank you!

OpenStudy (anonymous):

@amistre64, so the answer is "we can't find the answer unless they give us the score"?

OpenStudy (amistre64):

if we take the question verbatim, then that is the way i see it, the only way to use ONLY the mean and sd to find the results would be overtly complicated for a beginners class. what they most likely think is that a zscore is already known how to determine and that the mean and sd are then applied into the situation to determine the needed solution

OpenStudy (anonymous):

ok thanks!

OpenStudy (amistre64):

IF a zscore is already been established (its just a table lookup for the most part) then we can use the mean and sd in the formula:\[mean+z(sd)=x\] which is most likely what they expect for an answer

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