Respuesta :
Using the normal probability distribution and the central limit theorem, it is found that there is a 0.0207 = 2.07% probability that the sample proportion of baseball players will be greater than 63%.
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.
- By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex], as long as [tex]np \geq 10[/tex] and [tex]n(1 - p) \geq 10[/tex].
In this problem, we have that:
- 59% of the students in a university are baseball players, hence p = 0.59.
- A sample of 632 students is selected, hence n = 632.
The mean and the standard error are given by:
[tex]\mu = p = 0.59[/tex]
[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.59(0.41)}{632}} = 0.0196[/tex]
The probability that the sample proportion of baseball players will be greater than 63% is one subtracted by the p-value of Z when X = 0.63, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.63 - 0.59}{0.0196}[/tex]
[tex]Z = 2.04[/tex]
[tex]Z = 2.04[/tex] has a p-value of 0.9793.
1 - 0.9793 = 0.0207.
0.0207 = 2.07% probability that the sample proportion of baseball players will be greater than 63%.
To learn more about the normal probability distribution and the central limit theorem, you can check https://brainly.com/question/24663213
