# Variance of normally distributed RV

1. May 7, 2013

### Gauss M.D.

1. The problem statement, all variables and given/known data

The time it takes for a nurse to look up a patient journal is uniformly distributed between three and seven minutes. One morning there's 96 journal orders for the nurse to take care of.

Calculate the probability that she will get them all done during an eight hour work day, with no more than 15 minutes overtime. Assume normal distribution.

2. Relevant equations

Let X = number of minutes to look up one journal

3. The attempt at a solution

First, the probability function f(x) = 0.25 for 3 < x < 7

E(X) = int(0.25x dx) = 5

V(X) = E((X-5)^2) =

$\int 1/4(x-5)^2 dx = 1/12(x-5)^3 _{(3 to 7)} = 4/3$

σx = 2/√3

Let Y = 96X.

E(Y) = 96E(X) = 480

This is where I'm not sure:

σy = 96σx = 111

Y = N(480,111)

P(Y < 8*60+15) = P(Y < 495)

We are 15/111 = 0.136 standard deviations above expectation.

P(Y < 495) = P(Z < 0.136) = roughly 0.55

The answer says P(Y < 495) = P(Z < 1.36) so I'm probably missing something dumb? Re-done my calcs four times now :(

2. May 7, 2013

### Gauss M.D.

Oh... σy = sqrt(96)σx

Nevermind then.