Question

The life of light bulbs is distributed normally. The standard deviation of the lifetime is 25 hours and the mean lifetime of a bulb is 590 hours. Find the probability of a bulb lasting for at most 637 hours. Round your answer to four decimal places.

Answer 1

0.9699 = 96.99% probability of a bulb lasting for at most 637 hours.

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 = (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 = 590, \sigma = 25`

Find the probability of a bulb lasting for at most 637 hours.

This is the pvalue of Z when X = 637. So

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

`Z = (637 - 590)/(25)`

`Z = 1.88`

`Z = 1.88` has a pvalue of 0.9699

0.9699 = 96.99% probability of a bulb lasting for at most 637 hours.