1. ## Linear Function of Random Variables

I've got this problem here, but I'm a little concerned about my answer, not sure where I'm going wrong.

Making handcrafted pottery generally takes two major steps: wheel throwing and firing. The time of wheel throwing and the time of firing are normally distributed random variables with means of 40 min and 60 min and standard deviations of 2 min and 3 min, respectively
Let $\displaystyle X_1$ be the time throwing the wheels.
$\displaystyle \mu_X_1 = 40 min$
$\displaystyle \sigma_X_1 = 2 min$

Let $\displaystyle X_2$ be the time firing.
$\displaystyle \mu_X_2 = 60 min$
$\displaystyle \sigma_X_2 = 3 min$

$\displaystyle Y=X_1 + X_2$

$\displaystyle E(Y)=E(X_1)+E(X_2)=40min + 60min = 100min$
$\displaystyle V(Y)=\sigma_X_1 ^2 + \sigma_X_2 ^2 = 2^2 + 3^2 = 13min^2$

$\displaystyle \therfore P(Y<=85)=P(z< \frac{85-100}{sqrt(13)})=0.0000159$

This answer seems extremely low, considering 85 is just outside the first standard deviation. Not sure where I messed up here.

2. Originally Posted by Kasper
I've got this problem here, but I'm a little concerned about my answer, not sure where I'm going wrong.

Let $\displaystyle X_1$ be the time throwing the wheels.
$\displaystyle \mu_X_1 = 40 min$
$\displaystyle \sigma_X_1 = 2 min$

Let $\displaystyle X_2$ be the time firing.
$\displaystyle \mu_X_2 = 60 min$
$\displaystyle \sigma_X_2 = 3 min$

$\displaystyle Y=X_1 + X_2$

$\displaystyle E(Y)=E(X_1)+E(X_2)=40min + 60min = 100min$
$\displaystyle V(Y)=\sigma_X_1 ^2 + \sigma_X_2 ^2 = 2^2 + 3^2 = 13min^2$

$\displaystyle \therfore P(Y<=85)=P(z< \frac{85-100}{sqrt(13)})=0.0000159$

This answer seems extremely low, considering 85 is just outside the first standard deviation. Not sure where I messed up here.