Question

Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a variance of 4,000,000 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 20,179 hours. Round your answer to four decimal places.

Answer 1

Using the normal distribution, it is found that there is a 0.2776 = 27.76% probability that the life span of the monitor will be more than 20,179 hours.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 19000, \sigma = \sqrt{4000000} = 2000$

The probability that the life span of the monitor will be more than 20,179 hours is one subtracted by the p-value of Z when X = 20179, hence:

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

$Z = \frac{20179 - 19000}{2000}$

Z = 0.59.

Z = 0.59 has a p-value of 0.7224.

1 - 0.7224 = 0.2776.

0.2776 = 27.76% probability that the life span of the monitor will be more than 20,179 hours.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1