Respuesta :
Answer:
(a) 0.40
(b) 0.049
(c) [tex]\bar p\sim N(0.40,0.049^{2})[/tex]
(d) Explained below
Step-by-step explanation:
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
 [tex]E(\bar p)=p[/tex]
The standard deviation of this sampling distribution of sample proportion is:
 [tex]SE(\bar p)=\sqrt{\frac{p(1-p)}{n}}[/tex]
Given:
n = 100
p = 0.40
As n = 100 > 30 the Central limit theorem is applicable.
(a)
Compute the expected value of [tex]\bar p[/tex] as follows:
[tex]E(\bar p)=p=0.40[/tex]
The expected value of [tex]\bar p[/tex] is 0.40.
(b)
Compute the standard error of [tex]\bar p[/tex] as follows:
[tex]SE(\bar p)=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.40(1-0.40)}{100}}=0.049[/tex]
The standard error of [tex]\bar p[/tex] is 0.049.
(c)
The sampling distribution of [tex]\bar p[/tex] is:
[tex]\bar p\sim N(0.40,0.049^{2})[/tex]
(d)
The sampling distribution of p show that as the sample size is increasing the distribution is approximated by the normal distribution.
