Respuesta :
Answer: 0.0885
Step-by-step explanation:
Given : According to a recent Catalyst Census, 16% of executive officers were women with companies that have company headquarters in the Midwest.
i.e. p= 0.16
Sample size : n= 154
Now, the probability that more than 20% of this sample is comprised of female employees is given by :-
[tex]P(p>0.20)=P(z>\dfrac{0.20-0.16}{\sqrt{\dfrac{0.16(1-0.16)}{154}}})[/tex]
[∵ [tex]z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]]
[tex]=P(z>1.35)\approx1-P(z\leq1.35)=1-0.9115=0.0885[/tex] [Using the standard z-value table]
Hence, the required probability = 0.0885
Using the normal probability distribution and the central limit theorem, it is found that there is a 0.0869 = 8.69% probability that more than 20% of this sample is comprised of female employees.
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:
- 16% of executive officers were women with companies that have company headquarters in the Midwest, hence p = 0.16.
- A random sample of 154 executive officers from these companies was selected, hence n = 154.
The mean and the standard error are given by:
[tex]\mu = p = 0.16[/tex]
[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.16(0.84)}{154}} = 0.0295[/tex]
The probability that more than 20% of this sample is comprised of female employees is 1 subtracted by the p-value of Z when X = 0.2, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.2 - 0.16}{0.0295}[/tex]
[tex]Z = 1.36[/tex]
[tex]Z = 1.36[/tex] has a p-value of 0.9131.
1 - 0.9131 = 0.0869.
0.0869 = 8.69% probability that more than 20% of this sample is comprised of female employees.
To learn more about the normal probability distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213
