Question

The distribution of passenger vehicle speeds traveling on a residential street is nearly normal with a mean of 31.1 miles/hour and a standard deviation of 3.6 miles/hour. What percent of passenger vehicles travel slower than 35 miles/hour?

a) 69.85%
b) 81.11%
c) 91.23%
d) 94.77%

Answer 1

The percent of passenger vehicles that travel slower than 35 miles/hour is approximately 85.9%.

Step-by-step explanation:

To find the percent of passenger vehicles that travel slower than 35 miles/hour, we need to calculate the z-score for 35 miles/hour using the mean and standard deviation. The formula for calculating the z-score is z = (x - mean) / standard deviation. Plugging in the values, we get z = (35 - 31.1) / 3.6 = 1.083. To find the percent of vehicles that travel slower than 35 miles/hour, we need to find the area under the normal distribution curve to the left of the z-score of 1.083. Using a standard normal distribution table, we can find that the area to the left of 1.083 is approximately 0.859, or 85.9%. Therefore, the percent of passenger vehicles that travel slower than 35 miles/hour is approximately 85.9%.