Question
When Andrew goes bowling, his scores are normally distributed with a mean of 200 and a standard deviation of 14. What percentage of the games that Andrew bowls does he score higher than 226, to the nearest tenth?
Answers
GPT-4o mini
To determine the percentage of games that Andrew bowls in which he scores higher than 226, we can use the properties of the normal distribution. Given that Andrew's scores are normally distributed with a mean (\(\mu\)) of 200 and a standard deviation (\(\sigma\)) of 14, we first need to find the z-score for a score of 226.
The z-score is calculated using the formula:
\[
z = \frac{X - \mu}{\sigma}
\]
Where:
- \(X\) is the value we are interested in (in this case, 226)
- \(\mu\) is the mean (200)
- \(\sigma\) is the standard deviation (14)
Substituting the values into the formula:
\[
z = \frac{226 - 200}{14} = \frac{26}{14} \approx 1.8571
\]
Next, we will look up the z-score of 1.8571 in the standard normal distribution table (or use a calculator) to find the probability that Andrew scores less than 226.
Using a standard normal distribution calculator or z-table, we find that:
\[
P(Z < 1.8571) \approx 0.9686
\]
This means that approximately 96.86% of the time, Andrew scores less than 226. To find the percentage of games in which he scores more than 226, we calculate:
\[
P(Z > 1.8571) = 1 - P(Z < 1.8571) \approx 1 - 0.9686 = 0.0314
\]
To express this as a percentage, we multiply by 100:
\[
0.0314 \times 100 \approx 3.14\%
\]
Rounding to the nearest tenth, we get:
\[
\boxed{3.1\%}
\]
Thus, Andrew scores higher than 226 in approximately 3.1% of his bowling games.
The z-score is calculated using the formula:
\[
z = \frac{X - \mu}{\sigma}
\]
Where:
- \(X\) is the value we are interested in (in this case, 226)
- \(\mu\) is the mean (200)
- \(\sigma\) is the standard deviation (14)
Substituting the values into the formula:
\[
z = \frac{226 - 200}{14} = \frac{26}{14} \approx 1.8571
\]
Next, we will look up the z-score of 1.8571 in the standard normal distribution table (or use a calculator) to find the probability that Andrew scores less than 226.
Using a standard normal distribution calculator or z-table, we find that:
\[
P(Z < 1.8571) \approx 0.9686
\]
This means that approximately 96.86% of the time, Andrew scores less than 226. To find the percentage of games in which he scores more than 226, we calculate:
\[
P(Z > 1.8571) = 1 - P(Z < 1.8571) \approx 1 - 0.9686 = 0.0314
\]
To express this as a percentage, we multiply by 100:
\[
0.0314 \times 100 \approx 3.14\%
\]
Rounding to the nearest tenth, we get:
\[
\boxed{3.1\%}
\]
Thus, Andrew scores higher than 226 in approximately 3.1% of his bowling games.