Question

Suppose a manufacturer of light bulbs produces a 75-watt bulb that burns a mean of 7500 hours before it burns out. It has a standard deviation of 220 hours. What percentage of their bulbs would they have to replace under warranty if they offered a warranty of 7200 hours?

Answer 1

8.69% of their bulbs would they have to replace under warranty if they offered a warranty of 7200 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)`

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 = 7500, \sigma = 220`

What percentage of their bulbs would they have to replace under warranty if they offered a warranty of 7200 hours?

This is the probability that X is lower or equal than 7200 hours. So this is the pvalue of Z when X = 7200.

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

`Z = (7200 - 7500)/(220)`

`Z = -1.36`

`Z = -1.36` has a pvalue of 0.0869

So 8.69% of their bulbs would they have to replace under warranty if they offered a warranty of 7200 hours.