# Sampling question-number one- have no idea

• Nov 7th 2006, 04:32 PM
kingkaisai2
Sampling question-number one- have no idea
The life of powerlong batteries, sold in packs of 6, may be assumed to have a normal distribution with mean 32 hours and standard deviation of theta hours. Find the value of theta so that for one box in 100 (on average) the mean life of the batteries is less than 30 hours.
• Nov 8th 2006, 07:43 AM
CaptainBlack
Quote:

Originally Posted by kingkaisai2
The life of powerlong batteries, sold in packs of 6, may be assumed to have a normal distribution with mean 32 hours and standard deviation of theta hours. Find the value of theta so that for one box in 100 (on average) the mean life of the batteries is less than 30 hours.

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.