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

Answer:

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

Step-by-step 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

$Z = \frac{X - \mu}{\sigma}$

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.

$s = \frac{2}{\sqrt{100}} = \frac{2}{10} = 0.2$

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 $Z = 1.75$.

$Z = \frac{X - \mu}{s}$

$1.75 = \frac{X - 27}{0.2}$

$X = 27.35$

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