Question

The life of light bulbs is distributed normally. The standard deviation of the lifetime is 15 hours and the mean lifetime of a bulb is 520 hours. Find the probability of a bulb lasting for at most 528 hours. Round your answer to four decimal places.

Answer 1

0.7031 = 70.31% probability of a bulb lasting for at most 528 hours.

Explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

The standard deviation of the lifetime is 15 hours and the mean lifetime of a bulb is 520 hours.

This means that
`\sigma = 15, \mu = 520`

Find the probability of a bulb lasting for at most 528 hours.

This is the p-value of Z when X = 528. So

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

`Z = (528 - 520)/(15)`

`Z = 0.533`

`Z = 0.533` has a p-value of 0.7031

0.7031 = 70.31% probability of a bulb lasting for at most 528 hours.