Question

Suppose that replacement times for washing machines are normally distributed with a mean of 9.4 years and a standard deviation of 2 years. Find the replacement time that separates the top 18% from the bottom 82%.

Answer 1

Using the normal distribution, it is found that the replacement time that separates the top 18% from the bottom 82% is of 11.23 years.

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.

In this problem, the mean and the standard deviation are, respectively, given by $\mu = 9.4, \sigma = 2$.

The desired value is the 82nd percentile, which is X when Z = 0.915, hence:

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

$0.915 = \frac{X - 9.4}{2}$

X - 9.4 = 0.915(2)

X = 11.23

The replacement time that separates the top 18% from the bottom 82% is of 11.23 years.

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