# Normal distribution problem

## 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.

## 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?

Ray Vickson
Homework Helper
Dearly Missed

## 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.

## 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
$$f(x) = \frac{1}{\sqrt{2 \pi} \delta} e^{- \displaystyle \frac{(x-\mu)^2}{2 \delta^2}}$$
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:
NihalRi
SteamKing
Staff Emeritus
Homework Helper

## 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.

## 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.

NihalRi
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:
SteamKing
Staff Emeritus
Homework Helper
∫ 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.

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

SteamKing
Staff Emeritus
Homework Helper
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.

NihalRi
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.