1. ## job probability.....problem

John has two jobs. For daytime work at a jewelry store he is paid $200 per month, plus a commission. His monthly commission is normally distributed with mean$600 and standard deviation $40. At night he works as a waiter, for which his monthly income is normally distributed with mean$100 and standard deviation $30. John's income levels from these two sources are independent of each other. Referring to Table 6-4, for a given month, what is the probability that John's total income from these two jobs is less than$825?

The toal mean is 900 and total Std is 50

My question is how did they come up with the total mean and std?

2. Originally Posted by Blevil
John has two jobs. For daytime work at a jewelry store he is paid $200 per month, plus a commission. His monthly commission is normally distributed with mean$600 and standard deviation $40. At night he works as a waiter, for which his monthly income is normally distributed with mean$100 and standard deviation $30. John's income levels from these two sources are independent of each other. Referring to Table 6-4, for a given month, what is the probability that John's total income from these two jobs is less than$825?

The toal mean is 900 and total Std is 50

My question is how did they come up with the total mean and std?
Let X be the random variable 'monthly commission from day job (dollars)'.
X ~ Normal $\displaystyle (\mu_X = 600, \, \sigma_X = 40)$.

Let Y be the random variable 'monthly income from night job (dollars)'.
Y ~ Normal $\displaystyle (\mu_Y = 100, \, \sigma_Y = 30)$.

Let U = X + Y.

It's well know that when X and Y are independent normal variates, then U ~ Normal $\displaystyle (\mu = \mu_X + \mu_Y, \, \sigma^2 = \sigma^2_X + \sigma^2_Y)$ (so I disagree with "total mean is 900").

So U ~ Normal $\displaystyle (\mu = 700, \, \sigma = .....)$ and your job is to calculate Pr(U < 625).