2. The stopping distance of a certain car traveling at 60 miles per hour follows an approximately normal distribution with a mean of 130 feet and a standard deviation of 5 feet. Approximately what percent of the time does the car stop at a distance of between 120 feet and 140 feet?

Step 1 ：Calculate the z-scores for 120 and 140 feet: \(z_1 = \frac{120 - 130}{5}\) and \(z_2 = \frac{140 - 130}{5}\)

Step 2 ：Find the corresponding probability for each z-score using the standard normal table or calculator: \(P(z_1)\) and \(P(z_2)\)

Step 3 ：Subtract the probabilities to find the percentage of time the car stops between 120 and 140 feet: \(P(120 \le x \le 140) = P(z_2) - P(z_1)\)