Question

The batteries produced in a manufacturing plant have a mean time to failure of 27 months with a standard deviation of 2 months. A simple random sample of 100 batteries is taken and it is recorded how long it takes for each battery to fail. What is the level L so there is only a 4% chance that the sample mean time to failure falls above L? Round your answer to two decimal places.

Answer 1

There is only a 4% chance that the sample mean time to failure falls above 27.35 months.

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

For this problem, we have that:

The batteries produced in a manufacturing plant have a mean time to failure of 27 months with a standard deviation of 2 months, so
`\mu = 27, \sigma = 2`.

We have a sample of 100 students, so we need to find the standard deviation of the sample, to use in the place of
`\sigma` in the z score formula.

`s = (2)/(√(100)) = (2)/(10) = 0.2`

What is the level L so there is only a 4% chance that the sample mean time to failure falls above L?

This level is the value of X when Z has a pvalue of 0.96.

`Z = 1.75` has a pvalue of 0.96. So we have to find X when
`Z = 1.75`.

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

`1.75 = (X - 27)/(0.2)`

`X = 27.35`

There is only a 4% chance that the sample mean time to failure falls above 27.35 months.