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 590 hours. Find the probability of a bulb lasting for at most 605 hours. Round your answer to four decimal places.

Answer 1

Answer:

0.8413 = 84.13% probability of a bulb lasting for at most 605 hours.

Step-by-step 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:

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

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 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 590 hours.

This means that $\sigma = 15, \mu = 590$

Find the probability of a bulb lasting for at most 605 hours.

This is the pvalue of Z when X = 605. So

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

$Z = \frac{605 - 590}{15}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

0.8413 = 84.13% probability of a bulb lasting for at most 605 hours.