QuestionAccording to a survey in a country, $39 \%$ of adults do not own a credit card. Suppose a simple random sample of 400 adults is obtained. Comple $\qquad$ B. Not normal because $\mathrm{n} \leq 0.05 \mathrm{~N}$ and $\mathrm{np}(1-\mathrm{p})<10$ C. Not normal because $n \leq 0.05 \mathrm{~N}$ and $\mathrm{np}(1-\mathrm{p}) \geq 10$ D. Approximately normal because $n \leq 0.05 \mathrm{~N}$ and $\mathrm{np}(1-\mathrm{p}) \geq 10$
Determine the mean of $\tilde{n}$ e sampling distribution of $\hat{p}$. $\mu_{\hat{p}}=.39$ (Round to two decimal places as needed.) Determine the standard deviation of the sampling distribution of $\hat{p}$. $\sigma_{\hat{p}}=.024$ (Round to three decimal places as needed.) (b) What is the probability that in a random sample of 400 adults, more than $42 \%$ do not own a credit card?
The probability is .1056 . (Round to four decimal places as needed.) Interpret this probability. If 100 different random samples of 400 adults were obtained, one would expect 11 to result in more than $42 \%$ not owning a credit card. (Round to the nearest integer as needed.) (c) What is the probability that in a random sample of 400 adults, between $34 \%$ and $42 \%$ do not own a credit card?
The probability is $\square$ (Round to four decimal places as needed.)