1. ## Mean and Variance

The daily distance,X,covered by a bread seller in a housing estate from Monday to Friday has mean 12 km and standard deviation 0.5 km.For Saturday and Sunday,the daily distance,Y,covered by the bread seller has mean 10 km and standard deviation 0.4 km.

Find mean and variance of the average daily distance,S,covered by the bread seller in 7 days from Monday to Sunday and state whatever assumptions you have made.

I've calculated the mean of S to be 11.43 km from :

S = [5X+2Y]/7
E(S) = (5/7)E(X) + (2/7)E(Y)
= (5/7)(12) + (2/7)(10)
= 11.43 km

Var(S) is given as

Var(S) = (5/49)(0.5^2) + (2/49)(0.4^2)
=0.032 km^2

Shouldn't Var(S) = (25/49)(0.5^2) + (4/49)(0.4^2)
= 0.141 km^2 ???

In this case,we assume that the distance travelled by the bread seller on a certain day as independent of the distance travelled on other days isn't it?

Thanks a lot for your help.

2. Okay, so exactly where did you need that assumption of independence?

If they were not independent, what would you get for the variance?

3. Originally Posted by TKHunny
Okay, so exactly where did you need that assumption of independence?

If they were not independent, what would you get for the variance?
I have no idea.

4. When you added the variances, and did not include a covariance term, this is where you assumed they were independent.