Questions: Use the Empirical Rule. The mean speed of a sample of vehicles along a stretch of highway is 64 miles per hour with a standard deviation of 3 miles per hour. Estimate the percent of vehicles whose speeds are between 61 and 67 miles per hour. (Assume the data set has a normal distribution.) Approximately % of vehicles travel between 61 miles per hour and 67 miles per hour.

Solution Steps

To estimate the percent of vehicles whose speeds are between \( 61 \) mph and \( 67 \) mph, we first calculate the Z-scores for these values using the formula:

\[ Z = \frac{X - \mu}{\sigma} \]

where:

• \( X \) is the value,
• \( \mu = 64 \) mph (mean speed),
• \( \sigma = 3 \) mph (standard deviation).

Calculating the Z-score for the lower bound \( 61 \) mph:

\[ Z_{start} = \frac{61 - 64}{3} = \frac{-3}{3} = -1.0 \]

Calculating the Z-score for the upper bound \( 67 \) mph:

\[ Z_{end} = \frac{67 - 64}{3} = \frac{3}{3} = 1.0 \]

Next, we use the Z-scores to find the probability that a vehicle's speed falls between these two values. This is done using the cumulative distribution function \( \Phi \):

\[ P(61 < X < 67) = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(1.0) - \Phi(-1.0) \]

From standard normal distribution tables or calculators, we find:

\[ \Phi(1.0) \approx 0.8413 \quad \text{and} \quad \Phi(-1.0) \approx 0.1587 \]

Thus, the probability is:

\[ P(61 < X < 67) = 0.8413 - 0.1587 = 0.6826 \]

To express this probability as a percentage, we multiply by \( 100 \):

\[ P(61 < X < 67) \times 100 \approx 68.26\% \]

Final Answer