Question

When Camila goes bowling, her scores are normally distributed with a mean of 150 and a standard deviation of 12. What percentage of the games that Camila bowls does she score less than 137, to the nearest tenth?

Answer 1

To find the percentage of the games that Camila bowls where she scores less than 137, we can use the properties of the normal distribution. We'll first convert Camila’s score of 137 into a z-score, which tells us how many standard deviations away from the mean this score is.

The formula for calculating the z-score is:

\[
z = \frac{(X - \mu)}{\sigma}
\]

where:
- \(X\) is the score we are examining (137),
- \(\mu\) is the mean (150),
- \(\sigma\) is the standard deviation (12).

Plugging in the values:

\[
z = \frac{(137 - 150)}{12} = \frac{-13}{12} \approx -1.0833
\]

Next, we will look up this z-score in the standard normal distribution table, or use a calculator or software that provides cumulative probabilities for the standard normal distribution.

Looking up \(z = -1.08\) (or using a more precise calculator for \(z = -1.0833\)), we find that the cumulative probability \(P(Z < -1.08)\) is approximately 0.1401.

This means that about 14.01% of the time, Camila scores less than 137. Rounding this to the nearest tenth of a percent, we find:

\[
\text{Percentage} \approx 14.0\%
\]

Thus, the final answer is:

**14.0%** of the games that Camila bowls, she scores less than 137.