# Computing covariance between normal and uniform distributions [closed]

I think that this question is related: Covariance of a compound distribution, but it's too abstract for me (I believe that what I have to do must be simpler).

Here's the given: machines $A$ and $B$ produce hoses, the distribution of hose length in $A$ is $X_A \sim N(\mu = 100, \sigma = 5)$, the distribution of $X_B \sim U(85, 115)$. Hoses are accepted when their length is in $[90, 110]$ length range. We are also given that together $A$ and $B$ produce 90% of acceptable hoses.

# Need to find

The variance of overly long hoses per 100 units.

# What I found so far

My understanding of the problem tells me that $X_A$ and $X_B$ are linearly dependent, $A$ must produce 0.751395448691 parts of a unit, while $B$ produces the remaining 0.25-ish parts. So, I thought that I need to use the

$$\sigma^2_{A+B} = \sigma^2_A + \sigma^2_B + 2\rho\sigma_A\sigma_B$$

formula. $\sigma_A$ is a given. $\sigma_B$ is easy to find, I'm stuck at finding $\rho$. I know that it should be along the lines of

$$\rho = \frac{E(X_AX_B)}{E(X_A)E(X_B)}$$

Where, again, $E(X_A)$ and $E(X_B)$ are easy to find, but I have no idea how to find $E(X_AX_B)$.

## closed as off-topic by kjetil b halvorsen, Michael Chernick, Peter Flom♦May 16 at 12:07

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• You are asked to reason about the sum of independent (not "linearly independent") variables $X_A$ and $X_B$. That immediately implies their covariance is zero. – whuber Jun 18 '15 at 14:19
• Thanks @whuber, but what makes you believe that the variables are independent? If knowing $X_A$ I immediately know $X_B$, doesn't it imply they are dependent? Also I can find the value of $X_B$ using a linear transformation on $X_A$, wouldn't this mean that the dependence is linear? How do you rule that they are (in)dependent in this case? – wvxvw Jun 18 '15 at 19:21
• If you don't assume the variables are independent, you cannot solve this problem and you should complain to the teacher or textbook writer. – whuber Jun 18 '15 at 19:23