Respuesta :
Using the normal distribution and the central limit theorem, it is found that there is a 0.2119 = 21.19% probability that more than 77% of the sample have a driver's license.
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:
- 75% of high school seniors have a driver's license, hence p = 0.75.
- A sample of 300 is taken, hence n = 300.
The conditions are verified because:
np = 300 x 0.75 = 225 > 10.
n(1-p) = 300 x 0.25 = 75 > 10.
The mean and the standard error are given by:
[tex]\mu = p = 0.75[/tex]
[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.75(0.25)}{300}} = 0.025[/tex]
The probability that more than 77% of the sample have a driver's license is one subtracted by the p-value of Z when X = 0.77, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{0.77 - 0.75}{0.025}[/tex]
Z = 0.8
Z = 0.8 has a p-value of 0.7881.
1 - 0.7881 = 0.2119.
0.2119 = 21.19% probability that more than 77% of the sample have a driver's license.
To learn more about the normal distribution and the central limit theorem, you can check https://brainly.com/question/24663213
