Problem

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?

Answer

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Answer

Subtract the probabilities to find the percentage of time the car stops between 120 and 140 feet: P(120≤x≤140)=P(z2)−P(z1)

Steps

Step 1 ：Calculate the z-scores for 120 and 140 feet: z1=120−1305 and z2=140−1305

Step 2 ：Find the corresponding probability for each z-score using the standard normal table or calculator: P(z1) and P(z2)

Step 3 ：Subtract the probabilities to find the percentage of time the car stops between 120 and 140 feet: P(120≤x≤140)=P(z2)−P(z1)

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