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 64 months and a standard deviation of 7 months. Using the empirical (68-95-99.7) rule, what is the approximate percentage of cars that remain in service between 43 and 50 months?

Answer 1

Answer:

$z =\frac{50-64}{7}= -2$

$z =\frac{43-64}{7}= -3$

We know that within two deviations from the mean we have 95% of the data from the empirical rule so then below 2 deviation from the mean we have (100-95)/2 % =2.5%. And within 3 deviations from the mean we have 99.7% of the data so then below 3 deviations from the mean we have (100-99.7)/2% =0.15%

And then the final answer for this case would be:

$2.5 -0.15 = 2.35\%$

Step-by-step explanation:

For this case we have the following parameters from the variable number of motnhs in service for the fleet of cars

$\mu = 64, \sigma =7$

For this case we want to find the percentage of values between :

$P(43< X< 50)$

And we can use the z score formula given by:

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

In order to calculate how many deviation we are within from the mean. Using this formula for the limits we got:

$z =\frac{50-64}{7}= -2$

$z =\frac{43-64}{7}= -3$

We know that within two deviations from the mean we have 95% of the data from the empirical rule so then below 2 deviation from the mean we have (100-95)/2 % =2.5%. And within 3 deviations from the mean we have 99.7% of the data so then below 3 deviations from the mean we have (100-99.7)/2% =0.15%

And then the final answer for this case would be:

$2.5 -0.15 = 2.35\%$