Question

A large manufacturing plant uses lightbulbs with lifetimes that are normally distributed with a mean of 1200 hours and a standard deviation of 60 hours. To minimize the number of bulbs that burn out during operating hours, all bulbs are replaced at once. How often should the bulbs be replaced so that no more than 1% burn out between replacement periods? (Round your answer to one decimal place.)

Answer 1

Answer:

The bulbs should be replaced each 1060.5 days.

Step-by-step explanation:

Problems of normally distributed samples are 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:

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

The Z-score measures how many standard deviations the measure is from the mean. 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 get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 1200, \sigma = 60$

How often should the bulbs be replaced so that no more than 1% burn out between replacement periods?

This is the first percentile, that is, the value of X when Z has a pvalue of 0.01. So X when Z = -2.325.

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

$-2.325 = \frac{X - 1200}{60}$

$X - 1200 = -2.325*60$

$X = 1060.5$

The bulbs should be replaced each 1060.5 days.