Question

The lifetime of a certain type of battery is normally distributed with mean value 11 hours and standard deviation 1 hour. There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages?

Answer 1

The lifetime value needed is 11.8225 hours.

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by

`Z = (X - \mu)/(\sigma)`

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The lifetime of a certain type of battery is normally distributed with mean value 11 hours and standard deviation 1 hour. This means that
`\mu = 11, \sigma = 1`.

What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages?

This is the value of THE MEAN SAMPLE X when Z has a pvalue of 0.95. That is between Z = 1.64 and Z = 1.65. So we use
`Z = 1.645`

Since we need the mean sample, we need to find the standard deviation of the sample, that is:

`s = (\sigma)/(√(4)) = 0.5`

So:

`Z = (X - \mu)/(s)`

`1.645 = (X - 11)/(0.5)`

`X - 11 = 0.5*1.645`

`X = 11.8225`

The lifetime value needed is 11.8225 hours.