Normal distribution problem

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  • #1
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Homework Statement


A student sits a test and is told that the marks follow a normal distribution with mean 100. The student receives a mark of 124 and is told that he is at the 68th percentile. Calculate the variance of the distribution.

Homework Equations


https://lh3.googleusercontent.com/proxy/XW4iFBVEU-T7v_I-U5zV6mo7CvCYxVfS6kbEiAuBBsiEcJWb-P7AHpF6ciRBv4skZW5pqDilEHh9F5zdeCfu0N0ztdfDS_lm-A=w685-h215-nc

The Attempt at a Solution


I replaced 0.68 for f (x) and 100 for μ and 124 for x. I tried to rearange to solve for σ but was unsuccessful only getting as far as,
0.553 = -288/σ^2 - lnσ
I think my approach might be completely wrong, or is there a way of doing this with the aid of a graphic calculator?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


A student sits a test and is told that the marks follow a normal distribution with mean 100. The student receives a mark of 124 and is told that he is at the 68th percentile. Calculate the variance of the distribution.

Homework Equations


https://lh3.googleusercontent.com/proxy/XW4iFBVEU-T7v_I-U5zV6mo7CvCYxVfS6kbEiAuBBsiEcJWb-P7AHpF6ciRBv4skZW5pqDilEHh9F5zdeCfu0N0ztdfDS_lm-A=w685-h215-nc

The Attempt at a Solution


I replaced 0.68 for f (x) and 100 for μ and 124 for x. I tried to rearange to solve for σ but was unsuccessful only getting as far as,
0.553 = -288/σ^2 - lnσ
I think my approach might be completely wrong, or is there a way of doing this with the aid of a graphic calculator?

Your ##f##-formula is incorrect: is should be
[tex] f(x) = \frac{1}{\sqrt{2 \pi} \delta} e^{- \displaystyle \frac{(x-\mu)^2}{2 \delta^2}} [/tex]
Anyway, that is not relevant for this problem: a percentile is a point on the cumulative distribution, not on the probability density function.

There is no closed-form formula for the cumulative distribution, so you need to use tables of the cumulative, or use appropriate software. Some scientific calculators even have a "normal cumulative" button.
 
Last edited:
  • #3
SteamKing
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Homework Statement


A student sits a test and is told that the marks follow a normal distribution with mean 100. The student receives a mark of 124 and is told that he is at the 68th percentile. Calculate the variance of the distribution.

Homework Equations


https://lh3.googleusercontent.com/proxy/XW4iFBVEU-T7v_I-U5zV6mo7CvCYxVfS6kbEiAuBBsiEcJWb-P7AHpF6ciRBv4skZW5pqDilEHh9F5zdeCfu0N0ztdfDS_lm-A=w685-h215-nc

The Attempt at a Solution


I replaced 0.68 for f (x) and 100 for μ and 124 for x. I tried to rearange to solve for σ but was unsuccessful only getting as far as,
0.553 = -288/σ^2 - lnσ
I think my approach might be completely wrong, or is there a way of doing this with the aid of a graphic calculator?

Your formula shows the PDF for the normal distribution with mean μ and SD δ. The percentile of the student's score represents the area under the PDF distribution curve from -∞ to x. For example, the 50th percentile means that x = μ. Like Ray Vickson posted, you'll have to use standard normal tables to find out the SD.
 
  • #4
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The percentile of the student's score represents the area under the PDF distributio

∫ f (x)dx=0.68

Where f (x) is the PDF

Will this work as a definite integral from 0 to 124 and I try to solve again for σ ?
Well this seemed like it would work until I actually tried to integrate it:)
 
Last edited:
  • #5
SteamKing
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∫ f (x)dx=0.68

Where f (x) is the PDF

Will this work as a definite integral from 0 to 124 and I try to solve again for σ ?
No, that's not how the standard normal tables are constructed.

The standard normal distribution has μ = 0 and δ = 1.0. In order to accommodate normal distributions with other values of μ and δ, a transformation formula is used:

##Z = \frac{x-μ}{δ}##

The total area under the standard normal distribution curve is

##\int_{-∞}^{+∞} f(Z)dZ = 1.0##,

but since the curve is symmetrical about μ = 0

##\int_{-∞}^0 f(Z)dZ = 0.5##

Knowing the value of the cumulative distribution, you want to find Z which corresponds to that amount from a standard normal table, like this one:

https://www.mathsisfun.com/data/standard-normal-distribution-table.html

Once you find the correct Z value, then using the student's score and the mean of the distribution, you should be able to work back and solve for the S.D., δ.
 
  • #6
134
12
So I set it up so that
P (X≤124) = 0.68
and
P (Z≤(124-100)/σ)
Using invNorm on my calculator for the standard (z) distribution I found
(124-100)/σ≈0.468
σ≈51.3
Variance = 10051
Nice :)

Thank you
 
  • #7
SteamKing
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So I set it up so that
P (X≤124) = 0.68
and
P (Z≤(124-100)/σ)
Using invNorm on my calculator for the standard (z) distribution I found
(124-100)/σ≈0.468
σ≈51.3
Variance = 10051
Nice :)

Thank you
Everything looks OK, except the variance should be equal to σ2. IDK how you got 10051.
 
  • #8
134
12
Everything looks OK, except the variance should be equal to σ2. IDK how you got 10051.
Your right, I must have put the wrong number in the calculator.
 

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