Susan has been on a bowling team for 14 years. After examining all of her scores over that period of time, she finds that they follow a normal distribution. Her average score is 225, with a standard deviation of 13. If during a typical week Susan bowls 16 games, what is the probability that her average score is more than 230? (A) 0.0620(B) 0.3520(C) 0.6480(D) 0.9382

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex]

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Her average score is 225, with a standard deviation of 13. This means that [tex]\mu = 225, \sigma = 13[/tex].

If during a typical week Susan bowls 16 games, what is the probability that her average score is more than 230?

This is 1 subtracted by the pvalue of Z when [tex]X = 230[/tex].

By the Central Limit Theorem, we have [tex]s = \frac{\sigma}{\sqrt{n}} = \frac{13}{4} = 3.25[/tex]

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{230 - 225}{3.25}[/tex]

[tex]Z = 1.54[/tex]

[tex]Z = 1.54[/tex] has a pvalue of 0.9382. This means that there is a 1-0.9382 = 0.0618 = 6.18% probability that her average score is more than 230.