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. What is the probability that in a one-game playoff, her score is more than 227?

We are given that 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.

Let, X = scores over that period of time

X ~ N([tex]\mu = 225, \sigma = 13^{2}[/tex])

The z score probability distribution is given by;

Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = average score

[tex]\sigma[/tex] = standard deviation

So, probability that in a one-game playoff, her score is more than 227 is given by = P(X > 227)

P(X > 227) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{227-225}{13}[/tex] ) = P(Z > 0.15) = 1 - P(Z [tex]\leq[/tex] 0.15)

= 1 - 0.55962 = 0.4404

Therefore, probability that in a one-game playoff, her score is more than 227 is 0.4404.