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 60 months and a standard deviation of 10 months. Using the 68-95-99.7 rule, what is the approximate percentage of cars that remain in service between 30 and 40 months?

The 68-95-99.7 rule, also known as the empirical rule, is used to describe the distribution of data in a normal distribution. According to this rule:

- Approximately 68% of the data falls within one standard deviation of the mean.
- Approximately 95% of the data falls within two standard deviations of the mean.
- Approximately 99.7% of the data falls within three standard deviations of the mean.

Given:
- Mean (\(\mu\)) = 60 months
- Standard deviation (\(\sigma\)) = 10 months

We need to find the approximate percentage of cars that remain in service between 30 and 40 months.

First, we convert the months into standard deviations from the mean:

1. For 30 months:
   \[
   Z = \frac{X - \mu}{\sigma} = \frac{30 - 60}{10} = -3
   \]
   
2. For 40 months:
   \[
   Z = \frac{X - \mu}{\sigma} = \frac{40 - 60}{10} = -2
   \]

So, we are looking for the percentage of cars that fall between \(Z = -3\) and \(Z = -2\).

Using the 68-95-99.7 rule:
- 68% of the data falls within \( \pm 1 \sigma\) (between 50 and 70 months).
- 95% of the data falls within \( \pm 2 \sigma\) (between 40 and 80 months).
- 99.7% of the data falls within \( \pm 3 \sigma\) (between 30 and 90 months).

Since the distribution is symmetric, the percentage of data that lies beyond \(Z = -2\) but within \(Z = -3\) is half of the remaining percentage beyond \(Z = -2\). From the empirical rule:

- Between \(Z = -2\) and \(Z = +2\), we have 95%.
- Beyond \(Z = -2\), we have \(100\% - 95\% = 5\%\).

Since the distribution is symmetric, 2.5% lies beyond \(Z = +2\) and 2.5% lies below \(Z = -2\).

Further, the percentage beyond \(Z = -3\) is:
- Beyond \(Z = -3\), we have \(100\% - 99.7\% = 0.3\%\).

Since the distribution is symmetric, 0.15% lies beyond \(Z = +3\) and 0.15% lies below \(Z = -3\).

Thus, the percentage of cars that lie between \(Z = -3\) and \(Z = -2\) is:
- Percentage between \(Z = -3\) and \(Z = -2\) = \(2.5\% - 0.15\% = 2.35\%\).

Therefore, approximately 2.35% of the cars remain in service between 30 and 40 months.

Do you want more details or have any questions?

Here are 8 related questions to expand your understanding:

1. What is the probability of a car remaining in service between 50 and 70 months?
2. How would the percentage change if the mean was different but the standard deviation remained the same?
3. What is the Z-score for a car that has been in service for

Learn how to calculate the approximate percentage of cars remaining in service between 30 and 40 months using the 68-95-99.7 rule for normal distribution. Understand the concepts of mean, standard deviation, Z-score, and empirical rule applied to fleet management policies.

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