Question

a company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and other features. the distribution of the number of months in service for the fleet of cars is bell - shaped and has a mean of 62 months and a standard deviation of 7 months. using the 68 - 95 - 99.7 (empirical) rule, what is the approximate percentage of cars that remain in service between 76 and 83 months? the approximate percentage of cars that remain in service between 76 and 83 months is % . do not enter the percent symbol.) question help: video

Explanation:

Step1: Calculate the number of standard deviations from the mean

The mean \(\mu = 62\) and the standard deviation \(\sigma=7\).
For \(x = 76\): \(z_1=\frac{76 - 62}{7}=\frac{14}{7} = 2\)
For \(x = 83\): \(z_2=\frac{83 - 62}{7}=\frac{21}{7}=3\)

Step2: Use the Empirical Rule

The Empirical Rule states that about \(95\%\) of the data lies within \(z=\pm2\) (\(\mu\pm2\sigma\)) and about \(99.7\%\) of the data lies within \(z = \pm3\) (\(\mu\pm3\sigma\)).
The percentage of data within \(z=- 3\) to \(z = 3\) is \(99.7\%\), and the percentage of data within \(z=-2\) to \(z = 2\) is \(95\%\).
The percentage of data between \(z = 2\) and \(z=3\) is \(\frac{99.7\%-95\%}{2}\)

$$ \frac{99.7 - 95}{2}=\frac{4.7}{2}=2.35 $$

Answer:

\(2.35\)