Bivariate Normal

**Kasper** · October 30th 2010, 10:44 AM

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

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

$Y=X_1 + X_2$

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

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

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

**mr fantastic** · October 30th 2010, 01:56 PM

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

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

$Y=X_1 + X_2$

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

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

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

You haven't posted the whole question but, reading between the lines, your answer is correct.

**Kasper** · October 30th 2010, 03:09 PM

Originally Posted by mr fantastic

You haven't posted the whole question but, reading between the lines, your answer is correct.

Whoops had it posted, must have accidentally deleted it when I edited it. Thanks!

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