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 for the week is between 220 and 228?

Score of Susan follow a Normal Distribution and we have the population standard deviation, as this standard deviation is of her scores of previous 14 years.

In a given week, Susan bowls 16 games. This means, our sample size is 16. So,

n = 16

We have to find the probability that her average score of the week is between 220 and 228. Since the distribution is normal and value of population standard deviation is known, we will use the concept of z-score and z-distribution to find the desired probability.

First we will convert the given numbers to their equivalent z-scores. The formula to calculate the z-scores is:

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

x = 220 converted to z-score will be:

[tex]z=\frac{220-225}{\frac{13}{\sqrt{16}}}=-1.54[/tex]

x = 228 converted to z-score will be:

[tex]z=\frac{228-225}{\frac{13}{\sqrt{16}}}=0.92[/tex]

So, probability that Susan's score is between 220 and 228 is equivalent to probability of z score being in between - 1.54 and 0.92

i.e.

P (220 < X < 228) = P( -1.54 < z < 0.92)

From the z-table we can find the following values:

P( -1.54 < z < 0.92) = P(x < 0.92) - P(x<-1.54)

P( -1.54 < z < 0.92) = 0.8212 - 0.0618

P( -1.54 < z < 0.92) = 0.7594

Since, P (220 < X < 228) is equivalent to P( -1.54 < z < 0.92), we can conclude that the probability that Susan's average score for the week is between 220 and 228 is 0.7594.