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 of 4 months. Using the \( 68-95-99.7 \) rule, what is the approximate percentage of cars that remain in service between 41 and 49 months? Do not enter the percent symbol. ans =

We are given that the number of months in service is approximately normally distributed with a mean of 53 months and a standard deviation of 4 months. We want to find the approximate percentage of cars that remain in service between 41 and 49 months. Step 1. Convert the endpoints to z‑scores.  For 41 months:   z = (41 – 53) / 4 = –12/4 = –3  For 49 months:   z = (49 – 53) / 4 = –4/4 = –1 Step 2. Interpret these z‑scores using the 68-95-99.7 rule. The rule tells us that approximately:  • 68% of the data lie within ±1 standard deviation (i.e. between z = –1 and 1).  • 99.7% lie within ±3 standard deviations (i.e. between z = –3 and 3). Since roughly 68% are within ±1 standard deviation, about 32% (100% – 68%) lie outside that range. Because the normal distribution is symmetric, about 16% lie below z = –1. Similarly, approximately 0.15% lie below z = –3 (because 99.7% are within ±3, leaving about 0.3% in both tails, or 0.15% in each tail). Step 3. Find the approximate percentage between z = –3 and z = –1. Thus, the percentage of cars with z-scores between –3 and –1 is approximately:  16% – 0.15% ≈ 15.85% Rounding this value gives approximately 16%. So, the approximate percentage of cars that remain in service between 41 and 49 months is 16.