# Find the Variance of this distribution

1. Aug 10, 2010

### thereddevils

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

The number of hours spent in a library per week by arts and science students in a college is normally distributed with mean 12 hours and standard deviation 5 hours for arts students and mean 15 hours and standard deviation 4 hours for science students. A random sample of four arts students and six science students is chosen. Assuming that X is the mean number if hours spent by these 10 students in a week. Calculate Var(X)

2. Relevant equations

3. The attempt at a solution

Let X_A and X_S be the random variable of number of hours spent by arts and science students respectively.

X=(4X_A+6X_S)/(10)

Var(X)=Var((4X_A+6X_S)/(10))

=1/100 (16 Var(X_A)+36 Var(X_s))

=1/100 (16 x 25 + 36 x 16)

=9.76

Am i correct?

2. Aug 10, 2010

### hgfalling

Re: variance

Does this pass the smell test? You are indicating that the standard deviation of the mean of the ten students is like 3.1 hrs.

But suppose we just took ten arts students (who have the higher standard deviation). You should know that the deviation of the group will be $\frac{5}{\sqrt{10}} \approx 1.6$. So something is wrong!

What happened is that you got confused between two things:

Var(X + X) = 2 Var(X) (the sum of two random variables X)
Var(2X) = 4 Var(X) (a random variable X multiplied by 2)

When we pick four arts students, that's not one arts student multiplied by 4. Hence we don't get 16*Var(X_A), etc.