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 53 months and a standard deviation 55 months. Using the Empirical Rule rule, what is the approximate percentage of cars that remain in service between 38 and 43 months? Do not enter the percent symbol. ans =

The mean number of months in service is 53 and the standard deviation is 55. We want the percentage of cars with service times between 38 and 43 months. Step 1. Convert the endpoints (38 and 43 months) to z‐scores:   For 38 months:     z₁ = (38 − 53) / 55 = (−15) / 55 ≈ −0.273   For 43 months:     z₂ = (43 − 53) / 55 = (−10) / 55 ≈ −0.182 Step 2. Although the Empirical Rule is best used for intervals of 1, 2, or 3 standard deviations (giving 68%, 95%, or 99.7%, respectively), here the range from 38 to 43 months is very small compared to one standard deviation. In such cases, we can approximate the probability by finding the area under the normal curve between these two z‐scores. Since the entire range from –0.273 to –0.182 is only about 0.091 standard deviations wide, and the normal density near these z‐values is roughly 0.39 (the height doesn’t change much over this narrow interval), the probability is approximately:   Probability ≈ (width in z) × (density)     ≈ 0.091 × 0.39 ≈ 0.0355 Step 3. Convert this to a percentage:   0.0355 ≈ 3.55% Thus, using the Empirical Rule reasoning (and recognizing the limitation for small intervals), about 3.6 percent of the cars are in service between 38 and 43 months. Since the answer is to be given without the percent symbol, the final answer is: ans = 3.6