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

Answer:

The lifetime value needed is 11.8225 hours.

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, 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:

$s = \frac{\sigma}{\sqrt{4}} = 0.5$

So:

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

$1.645 = \frac{X - 11}{0.5}$

$X - 11 = 0.5*1.645$

$X = 11.8225$

The lifetime value needed is 11.8225 hours.