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

Answer:

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

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:

$\mu = 590, \sigma = 25$

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

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

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

$Z = \frac{637 - 590}{25}$

$Z = 1.88$

$Z = 1.88$ has a pvalue of 0.9699

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