Suppose that $Y$ has a Binomial distribution with $n=80$ and $p=0.67$ . Let $\widehat{p}=Y / 80$ . The standard deviation of the distribution of $\hat{p}$ is ____. $\newline$ (A) $\frac{17.688}{80}$ $\newline$ (B) $0$ . $67$ $\newline$ (C) $53$ . $6$ $\newline$ (D) $\sqrt{\frac{0.22 \overline{11}}{80}}$

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Q. Suppose that

Y

has a Binomial distribution with

n=80

and

p=0.67

. Let

\widehat{p}=Y / 80

. The standard deviation of the distribution of

\hat{p}

is ____.

\newline

(A)

\frac{17.688}{80}

\newline

(B)

0

67

\newline

(C)

53

6

\newline

(D)

\sqrt{\frac{0.22 \overline{11}}{80}}

Given formula for standard deviation: We know that $\hat{p}$ is the sample proportion, which is the estimator of the population proportion $p$ in a binomial distribution. The standard deviation of the sample proportion $\hat{p}$ is given by the formula: $\newline$ $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$ $\newline$ where $p$ is the probability of success, $1-p$ is the probability of failure, and $n$ is the sample size.
Substitute values into formula: Given that $n=80$ and $p=0.67$ , we can substitute these values into the formula to calculate the standard deviation of $\hat{p}$ : $\newline$ $\sigma_{\hat{p}} = \sqrt{\frac{0.67(1-0.67)}{80}}$
Calculate $1$ -p: First, calculate $1-p$ : $\newline$ $1-0.67 = 0.33$
Multiply p by $1$ -p: Now, multiply $p$ by $1-p$ : $\newline$ $0.67 \times 0.33 = 0.2211$
Divide product by sample size: Next, divide this product by the sample size $n$ : $\newline$ $\frac{0.2211}{80} = 0.00276375$
Find square root of result: Finally, take the square root of the result to find the standard deviation of $\hat{p}$ : $\newline$ $\sigma_{\hat{p}} = \sqrt{0.00276375} \approx 0.05257$
Check closest option: The final answer needs to be in the form of one of the given options. The calculated standard deviation is approximately $0$ . $05257$ , which is not exactly in the form of any of the options provided. However, we can recognize that the square root of a fraction is involved, and the closest option that represents this is option (D), which is in the form of a square root of a fraction over $80$ . Let's express our calculated standard deviation in a similar form to see if it matches: $\newline$ $\sigma_{\hat{p}} = \sqrt{\frac{0.2211}{80}}$
Confirm equivalence with $2$ / $9$ : We can see that the fraction inside the square root is slightly different from the one in option (D). The fraction in option (D) is $\frac{0.22\overline{11}}{80}$ , which is a repeating decimal equivalent to $\frac{2}{9}$ . Let's check if our calculated fraction is equivalent to $\frac{2}{9}$ : $\newline$ $0.2211 \approx \frac{2}{9}$
Conclusion: To confirm, we can divide $2$ by $9$ : $\newline$ $\frac{2}{9} = 0.2222\overline{2}$ $\newline$ This is very close to our calculated value of $0$ . $2211$ , and the slight difference is likely due to rounding during the calculation. Therefore, we can conclude that the standard deviation of $\hat{p}$ is indeed represented by option (D): $\newline$ $\sigma_{\hat{p}} = \sqrt{\frac{0.22\overline{11}}{80}}$