The student newspaper at a large university asks an SRS of 250 undergraduates, "Do you favor eliminating the carnival from the term-end celebration?" All in all, 150 of the 250 are in favor. Suppose that (unknown to you) 55\% of all undergraduates favor eliminating the carnival. If you took a very large number of SRSs of size n=250 n=250 from this population, the sampling distribution of the sample proportion ${\mathrm{p}}^{\wedge }\hat{p}$would be

a. exactly Normal with mean 0.55 and standard deviation 0.03.

b. approximately Normal with mean 0.55 and standard deviation 0.03.

c. exactly Normal with mean 0.60 and standard deviation 0.03.

d. approximately Normal with mean 0.60 and standard deviation 0.03.

e. heavily skewed with mean 0.55 and standard deviation 0.03.

Step by step solution

01

Given Information

Given,

$$\begin{gathered} p=55\%=0.55 \\ n=250 \end{gathered}$$

Formula used:

${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$

02

Explanation for correct option

If the sample fraction is close to normal, $np\ge 10$and $n(1-p)\ge 10$

$$\begin{gathered} n{p}_0=250(0.55)=137.5\ge 10 \\ n\left(1-p_0\right)=250(1-0.55)=112.5\ge 10 \\ {\mu }_{\hat{p}}=p=0.55 \\ {\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ =\sqrt{\frac{0.55(1-0.55)}{250}} \\ =0.03 \end{gathered}$$

Hence, the correct option is (b)

03

Explanation for incorrect option

a. exactly Normal with mean 0.55 and standard deviation 0.03 is not the answer.

c. exactly Normal with mean 0.60 and standard deviation 0.03 is not the answer.

d. approximately Normal with mean 0.60 and standard deviation 0.03 is not the answer.

e. heavily skewed with mean 0.55 and standard deviation 0.03 is not the answer.

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