Points: 1 of 4
Save
According to a survey in a country, $25 \%$ of adults do not own a credit card. Suppose a simple random sample of 200 adults is obtained. Complete parts (a) through (d) below.
(a) Describe the sampling distribution of $\hat{p}$, the sample proportion of adults who do not own a credit card. Choose the phrase that best describes the shape of the sampling distribution of $\hat{p}$ below.
A. Not normal because $n \leq 0.05 \mathrm{~N}$ and $\mathrm{np}(1-\mathrm{p})< 10$
B. Approximately normal because $n \leq 0.05 \mathrm{~N}$ and $n p(1-p)< 10$
C. Approximately normal because $n \leq 0.05 \mathrm{~N}$ and $n p(1-p) \geq 10$
D. Not normal because $n \leq 0.05 \mathrm{~N}$ and $n p(1-p) \geq 10$
Determine the mean of the sampling distribution of $\hat{p}$.
$\mu_{\hat{p}}=0.25$ (Round to two decimal places as needed.)
Determine the standard deviation of the sampling distribution of $\hat{p}$.
$\sigma_{\hat{p}}=0.031$ (Round to three decimal places as needed.)
(b) What is the probability that in a random sample of 200 adults, more than $28 \%$ do not own a credit card?
The probability is $\square$.
(Round to four decimal places as needed.)
Answer
Thus, the probability that in a random sample of 200 adults, more than 28% do not own a credit card is approximately \(\boxed{0.1636}\).
Step 1 ：Given that the sample size (n) is 200 and the probability of success (p), which is the probability of an adult not owning a credit card, is 0.25.
Step 2 ：Calculate the mean of the binomial distribution using the formula \(np\), which gives \(200 \times 0.25 = 50.0\).
Step 3 ：Calculate the standard deviation of the binomial distribution using the formula \(\sqrt{np(1-p)}\), which gives \(\sqrt{200 \times 0.25 \times (1-0.25)} = 6.123724356957945\).
Step 4 ：Standardize the value 0.28 to find the corresponding z-score using the formula \(\frac{0.28 - p}{\sqrt{\frac{p(1-p)}{n}}}\), which gives \(\frac{0.28 - 0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}} = 0.9797958971132724\).
Step 5 ：Find the probability that the z-score is greater than this value, which is approximately 0.1636.
Step 6 ：Thus, the probability that in a random sample of 200 adults, more than 28% do not own a credit card is approximately \(\boxed{0.1636}\).
