# Normal distribution problem

1. Apr 4, 2016

### NihalRi

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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?

2. Apr 4, 2016

### Ray Vickson

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: Apr 4, 2016
3. Apr 4, 2016

### SteamKing

Staff Emeritus
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. Apr 4, 2016

### NihalRi

∫ 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: Apr 4, 2016
5. Apr 4, 2016

### SteamKing

Staff Emeritus
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. Apr 4, 2016

### NihalRi

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. Apr 4, 2016

### SteamKing

Staff Emeritus
Everything looks OK, except the variance should be equal to σ2. IDK how you got 10051.

8. Apr 4, 2016

### NihalRi

Your right, I must have put the wrong number in the calculator.