Question

Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a standard deviation of 1900 hours and a mean life span of 19,000 hours. If a monitor is selected at random, find the probability that the life span of the monitor will be more than 15,579 hours. Round your answer to four decimal places.

Answer 1

Answer:

0.9641 = 96.41% probability that the life span of the monitor will be more than 15,579 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:

$\mu = 19000, \sigma = 1900$

Find the probability that the life span of the monitor will be more than 15,579 hours.

This is 1 subtracted by the pvalue of Z when X = 15579. So

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

$Z = \frac{15579 - 19000}{1900}$

$Z = -1.80$

$Z = -1.80$ has a pvalue of 0.0359

1 - 0.0359 = 0.9641

0.9641 = 96.41% probability that the life span of the monitor will be more than 15,579 hours.