Question

The life of light bulbs is distributed normally. The standard deviation of the lifetime is 2525 hours and the mean lifetime of a bulb is 590590 hours. Find the probability of a bulb lasting for at most 622622 hours. Round your answer to four decimal places.

Answer 1

Answer:

0.8997 = 89.97% probability of a bulb lasting for at most 622 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.

Mean of 590 hours, standard deviation of 25 hours.

This means that $\mu = 590, \sigma = 25$

Find the probability of a bulb lasting for at most 622 hours.

This is the p-value of Z when X = 622.

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

$Z = \frac{622 - 590}{25}$

$Z = 1.28$

$Z = 1.28$ has a p-value of 0.8997.

0.8997 = 89.97% probability of a bulb lasting for at most 622 hours.