Question

Anna and Barbara are two friends who love to bowl against each other. Both bowlers’ scores are approximately Normally distributed. Anna has a mean score of 184 points, with a standard deviation of 15 points, while Barbara has a mean score of 191 points, with a standard deviation of 18 points. If the higher score wins, and they bowl independently of each other, what is the probability that Anna will beat Barbara in their next match?

A) 0.0218
B) 0.383
C) 0.416
D) 0.495
E) 0.613

Answer 1

The correct option is 0.383.

To find the probability that Anna will beat Barbara in their next match, we need to compare their scores statistically. Since the scores are normally distributed, we can use the z-score formula to standardize the scores and then compare them.

The z-score is calculated as
`\( Z = ((X - \mu))/(\sigma) \), where \( X \)` is the individual score,
`\( \mu \)` is the mean, and
`\( \sigma \)` is the standard deviation.

For Anna:

`\[ Z_{\text{Anna}} = \frac{(X_{\text{Anna}} - \mu_{\text{Anna}})}{\sigma_{\text{Anna}}} = \frac{(X_{\text{Anna}} - 184)}{15} \]`

For Barbara:

`\[ Z_{\text{Barbara}} = \frac{(X_{\text{Barbara}} - \mu_{\text{Barbara}})}{\sigma_{\text{Barbara}}} = \frac{(X_{\text{Barbara}} - 191)}{18} \]`

Now, we need to find the probability that Anna's score is higher than Barbara's. This can be calculated using the standard normal distribution table or a calculator. The difference in the z-scores is:

`\[ Z_{\text{diff}} = Z_{\text{Anna}} - Z_{\text{Barbara}} \]`

By looking up this z-score difference in a standard normal distribution table, we find the probability that Anna's score is higher than Barbara's. From calculations, the answer is approximately 0.383.