Respuesta :
Using the normal approximation to the binomial distribution, by the Central Limit Theorem, it is found that there is a 0.102 = 10.2% probability that, in a simple random sample of 95 games, 11 or more of them end in extra innings.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It 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, which is the percentile of X.
- The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. By the Central Limit Theorem, it can be approximated to the normal distribution with [tex]\mu = np, \sigma = \sqrt{np(1-p)}[/tex].
In this problem:
- The percentage of games that go into extra innings is 7.6%, hence p = 0.076.
- His teams played 95 games, hence n = 95.
The mean and the standard deviation are given by:
[tex]\mu = np = 95(0.076) = 7.22[/tex]
[tex]\sigma = \sqrt{np(1-p)} = \sqrt{95(0.076)(0.924)} = 2.5829[/tex]
Using continuity correction, as the normal distribution is continuous and the binomial is discrete, the probability that, in a simple random sample of 95 games, 11 or more of them end in extra innings is P(X > 11 - 0.5) = P(X > 10.5), which is 1 subtracted by the p-value of Z when X = 10.5.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{10.5 - 7.22}{2.5829}[/tex]
[tex]Z = 1.27[/tex]
[tex]Z = 1.27[/tex] has a p-value of 0.898.
1 - 0.898 = 0.102
0.102 = 10.2% probability that, in a simple random sample of 95 games, 11 or more of them end in extra innings.
More can be learned about the normal approximation to the binomial distribution at https://brainly.com/question/14424710
