The mean of a sample of size N from a normal distribution with mean and sd s

is normally distributed with m and the sd (standard error) s/sqrt(N).

Here the sample size is 6 (one pack of 6), and m=32 hrs, and

s=theta/sqrt(6). Therefore the z-score corresponding to a mean life of 30 hrs

is z=-2*sqrt(6)/theta.

We are told to find theta so that p(mean of pack<30hrs)=0.01. This is equivalent to asking that for a standard normal RV that:

P(z<-2*sqrt(6)/theta)=0.01.

Now looking up the critical value that corresponds to a P of 0.01 we find

it occurs at z~=-2.33, so:

2*sqrt(6)/theta = 2.33,

or:

theta = 2*sqrt(6)/2.33

RonL.