## 6.9.) Major League Baseball teams have become concerned about the length of games. During a recent season, games averaged 2 hours and 52 min

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

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2021-09-16T14:59:53+00:00
2021-09-16T14:59:53+00:00 2 Answers
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## Answers ( )

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

Step-by-step explanation:Problems of normally distributed samples can be solved using the z-score formula.In a set with mean and standard deviation , the zscore of a measure X is given by:

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:What is the probability that a randomly selected game will be completed in 200 minutes or lessThis is the pvalue of Z when X = 200. So

has a pvalue of 0.9599

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

Answer:Probability that a randomly selected game will be completed in 200 minutes or less is 0.95994.Step-by-step 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 gamesSo, X ~ N(

The z score probability distribution is given by;Z = ~ N(0,1)

where, = average time = 172 minutes

= standard deviation = 16 minutes

So, Probability that a randomly selected game will be completed in 200 minutes or less is given by = P(X200 min)P(X200) = P( ) = P(Z 1.75)

= 0.95994

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