# Homework Help: Normal distribution head-scratcher

1. Feb 17, 2010

### Mglafas

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

A study investigates the performance of students in an exam, which is known to follow a normal distribution, with a mean of 63 and a standard deviation of 21.

What is the probability that the difference between two randomly selected
students is more than 5 marks?

2. Relevant equations

P(|x1-x2|>5)

Any ideas? Many thanks :)

2. Feb 17, 2010

### korican04

Try using

Y=X1-X2 as a new random variable with mean 0 and standard deviation of sqrt(2)*21

P(Y>5)
Z=(Y-mean)/std

P(Z>.052) = 1-P(Z<.052)

so now you can find the values in a z-table.

I haven't thought of the other case in which X2>X1, but this is at least how you should think about the question.

With such a high deviation intuitively you should be getting a high probability.

Last edited: Feb 17, 2010
3. Feb 21, 2010

### vintwc

How did you get 0.052? Mean=0, SD= sqrt(42)?

4. Feb 21, 2010

### korican04

For normal distributions you can use the properties if X1 and X2 are two independent random variables then Y=aX1+bX2 is a random variable with normal distribution and mean a*u1+b*u2 and variance a^2*sigma1^2+b^2*sigma2^2

So in our case Y=X1-X2, mean = 1*63+-1*63=0
variance = 1^2*(21^2)+(-1)^2*(21^2)=21^2+21^2=2*21^2
std=sqrt(2)*21

P(Y>5)
You can't look this up. so you look up Z which is (Y-mean)/std.