Question

Suppose a simple random sample of size n = 150 is obtained from a population whose size is N= 30,000 and whose population proportion with a specified characteristic is p=0.6. wick here to viewithe standarcinommaldistributionlabelnagel Click here to view the standard normal distribution table (page 2). (a) Describe the sampling distribution of p. Choose the phrase that best describes the shape of the sampling distribution below. O A. Approximately normal because ns 0.05N and np(1 - p) 10 O B. Not normal because ns0.05N and np(1-p) < 10. O C. Approximately normal because ns 0.05N and np(1-p) < 10. O D. Not normal because n30.05N and np(1-p) 2 10. Determine the mean of the sampling distribution of p. HA p (Round to one decimal place as needed.) Determine the standard deviation of the sampling distribution of p. on = (Round to six decimal places as needed) (b) What is the probability of obtaining x = 99 or more individuals with the characteristic? That is, what is P(p=0.66)? PO 20.66)= 1 (Round to four decimal places as needed.) (c) What is the probability of obtaining x = 84 or fewer individuals with the characteristic? That is what is P(P <0.56)? PỘs 0.56)= (Round to four decimal places as needed.)

Answer 1

(a) **Sampling Distribution Description:**

- The correct choice is **C. Approximately normal because \(n \cdot p \cdot (1 - p) < 10\) and \(n \cdot 0.05 \cdot N\).**

- This condition ensures that the sampling distribution of the sample proportion is approximately normal due to the central limit theorem.

- The mean (\(\mu_p\)) of the sampling distribution of \(p\) is given by \(p\), which is the population proportion.

- The standard deviation (\(\sigma_p\)) of the sampling distribution of \(p\) is given by \(\sqrt{\frac{p \cdot (1 - p)}{n}}\).

(b) **Probability of Obtaining \(x \geq 99\) individuals with the characteristic (P(p \geq 0.66)):**

- The mean of the sampling distribution (\(\mu_p\)) is \(p = 0.6\).

- The standard deviation (\(\sigma_p\)) is calculated using the formula mentioned in part (a).

- Use the standard normal distribution table to find \(P(p \geq 0.66)\).

(c) **Probability of Obtaining \(x \leq 84\) individuals with the characteristic (P(p \leq 0.56)):**

- Use the standard normal distribution table to find \(P(p \leq 0.56)\).

Please provide the values obtained for \(\mu_p\) and \(\sigma_p\) so that I can assist you further with parts (b) and (c).

Answer 2

The sampling distribution of the proportion is approximately normal when certain conditions are met. The mean of the sampling distribution is equal to the population proportion. The standard deviation of the sampling distribution can be calculated using a formula.

Step-by-step explanation:

The sampling distribution of a proportion can be approximated by a normal distribution when the sample size is large and certain conditions are met. In this case, the conditions are np > 5 and nq > 5, where np is the number of successes and nq is the number of failures in the sample. The shape of the sampling distribution of the proportion is approximately normal when these conditions are satisfied.

The mean of the sampling distribution of the proportion is equal to the population proportion p. In this case, the population proportion is 0.6, so the mean of the sampling distribution is also 0.6.

The standard deviation of the sampling distribution of the proportion can be calculated using the formula σ = sqrt((p(1-p))/n), where p is the population proportion and n is the sample size. Plugging in the values, we get σ = sqrt((0.6(1-0.6))/150) = 0.0422 (rounded to six decimal places).

To find the probability of obtaining x = 99 or more individuals with the characteristic, we need to find P(p≥0.66). We can use the standard normal distribution table to find the z-score corresponding to the proportion 0.66, and then use the table to find the probability of obtaining a z-score equal to or greater than that value.

Similarly, to find the probability of obtaining x = 84 or fewer individuals with the characteristic, we need to find P(p≤0.56). We can again use the standard normal distribution table to find the z-score corresponding to the proportion 0.56, and then use the table to find the probability of obtaining a z-score equal to or less than that value.