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 $n=250$from this population, the sampling distribution of the sample proportion $\stackrel{\wedge }{P}$would be

(a) exactly Normal with mean $0.55$and standard deviation of $0.03$

(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 that Population proportion$\left(p\right)=0.55$

Sample size$\left(n\right)=250$

02

Explanation

Here,

$$\begin{gathered} np=250(0.55)=137.5>10 \\ n(1-p)=250(1-0.55)=112.5>10 \end{gathered}$$

Now the mean and standard deviation can be calculated as:

$$\begin{gathered} {\mu }_{\hat{p}}=p \\ =0.55 \\ {\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ =\sqrt{\frac{0.55(1-0.55)}{250}} \\ =0.03 \end{gathered}$$

So the answer is(b)

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