Question

6.9.) Major League Baseball teams have become concerned about the length of games. During a recent season, games averaged 2 hours and 52 minutes (172 minutes) to complete. Assume the length of games follows the normal distribution with a standard deviation of 16 minutes. a.) What is the probability that a randomly selected game will be completed in 1.) 200 minutes or less

Answer 1

Probability that a randomly selected game will be completed in 200 minutes or less is 0.95994.

Explanation:

We are given that during a recent season, games averaged 2 hours and 52 minutes (172 minutes) to complete. Assume the length of games follows the normal distribution with a standard deviation of 16 minutes.

Let X = length of games

So, X ~ N(
`\mu = 172, \sigma^(2) = 16^(2))`

The z score probability distribution is given by;

Z =
`(X-\mu)/(\sigma)` ~ N(0,1)

where,
`\mu` = average time = 172 minutes

`\sigma` = standard deviation = 16 minutes

So, Probability that a randomly selected game will be completed in 200 minutes or less is given by = P(X
`\leq` 200 min)

P(X
`\leq` 200) = P(
`(X-\mu)/(\sigma)`
`\leq`
`(200-172)/(16)` ) = P(Z
`\leq` 1.75)

= 0.95994

Hence, Probability that a randomly selected game will be completed in 200 minutes or less is 0.95994.

Answer 2

95.99% probability that a randomly selected game will be completed in 200 minutes or less.

Explanation:

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:

`\mu = 172, \sigma = 16`

What is the probability that a randomly selected game will be completed in 200 minutes or less

This is the pvalue of Z when X = 200. So

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

`Z = (200 - 172)/(16)`

`Z = 1.75`

`Z = 1.75` has a pvalue of 0.9599

95.99% probability that a randomly selected game will be completed in 200 minutes or less.