When Andrefw 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?
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attempt 1 out of 2

Step 1 ：Given that Andrew's bowling scores are normally distributed with a mean of 200 and a standard deviation of 14, we need to find the percentage of games where he scores higher than 226.

Step 2 ：We start by calculating the z-score, which is a measure of how many standard deviations an element is from the mean. The formula for the z-score is \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the score, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 3 ：Substituting the given values into the formula, we get \(Z = \frac{226 - 200}{14} = 1.8571428571428572\).

Step 4 ：Next, we use a z-table or a standard normal distribution to find the percentage of games where Andrew scores higher than 226. The z-score of 1.8571428571428572 corresponds to a percentage of 3.2.

Step 5 ：Therefore, the percentage of games that Andrew bowls and scores higher than 226 is approximately \(\boxed{3.2\%}\).