Questions: A random sample of vehicle mileage expectancies has a sample mean of x̄=169,200 miles and sample standard deviation of s=19,400 miles. Use the Empirical Rule to estimate the percentage of vehicle mileage expectancies that are more than 188,600 miles. Round your answer to the nearest whole number (percent).

Solution Steps

To determine how many standard deviations the value \( 188600 \) miles is from the mean \( \bar{x} = 169200 \) miles, we calculate the z-score using the formula:

\[ z = \frac{x - \mu}{\sigma} \]

Substituting the values:

\[ z = \frac{188600 - 169200}{19400} = \frac{14400}{19400} \approx 0.7412 \]

Using the z-score, we find the cumulative probability \( P(Z \leq z) \):

\[ P(Z \leq 0.7412) \approx 0.8413 \]

This value represents the proportion of vehicle mileage expectancies that are less than \( 188600 \) miles.

To find the percentage of vehicle mileage expectancies that are more than \( 188600 \) miles, we subtract the cumulative probability from \( 1 \):

\[ P(Z > 0.7412) = 1 - P(Z \leq 0.7412) \approx 1 - 0.8413 = 0.1587 \]

Converting this to a percentage:

\[ \text{Percentage beyond} = 0.1587 \times 100 \approx 15.87\% \]

Rounding to the nearest whole number gives us \( 16\% \).

Final Answer