Question

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

Answer 1

It would be nice if you spaced the question out properly

Answer 2

There is a 6.18% probability that her average score is more than 230.

Explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean
`\mu` and standard deviation
`\sigma`, a large sample size can be approximated to a normal distribution with mean
`\mu` and standard deviation
`(\sigma)/(√(n))`

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

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

`Z = (X - \mu)/(\sigma)`

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
`\mu = 225, \sigma = 13`.

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
`X = 230`.

By the Central Limit Theorem, we have
`s = (\sigma)/(√(n)) = (13)/(4) = 3.25`

`Z = (X - \mu)/(s)`

`Z = (230 - 225)/(3.25)`

`Z = 1.54`

`Z = 1.54` 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.