Suppose a simple random sample of size n=150 is obtained from a population whose size is N=20,000 and whose population proportion with a specified characteristic is p=0.8. Please answer questions (1) through (5) below.

(1) describe the sampling ditriribution of p^ (choose the corrct phrase that best decribes the shape of the sampling below)

a) Not normal because n <, 0.05N and np (1-p) <10.

b) Approximateley normal because n <, 0.05N and np (1-p) >10.

c) Approximateley normal because n <, 0.05N and np (1-p) <10.

d) Not normal because n <, 0.05N and np (1-p) >10.

2) Determine the mean of the sampling distribution of p^.

p^ = ______ (round to one decimal place as needed)

3) Determine the standard deviation of the sampling distribution of p^.

p^ = ______ (roaund six decimal places as needed)

4) What is the probability of obtaining x=123 or more individuals with the characterstic? that is , what P(p^ > 0.82)?

P( p^ > 0.82) = _____ (Round four decimals places)

5) What is the probability of obtaining x = 111 of fewer individuals with the characteristic ? that is what is P(p^ < 0.74)?

P(p^ < 0.74) = _______ (Round to four decimal places as needed)

Same as your other post:

http://www.jiskha.com/display.cgi?id=1417003909

(1) The correct phrase that best describes the shape of the sampling distribution is b) Approximately normal because n < 0.05N and np(1-p) > 10. In this case, n=150, N=20,000, and p=0.8. Since n is less than 0.05 times N and np(1-p) is greater than 10, the sampling distribution of p^ is approximately normal.

(2) To determine the mean of the sampling distribution of p^, we use the formula:

Mean of p^ = p = 0.8

So, the mean of the sampling distribution of p^ is 0.8.

(3) To determine the standard deviation of the sampling distribution of p^, we use the formula:

Standard deviation of p^ = √(p*(1-p)/n)

Standard deviation of p^ = √(0.8*(1-0.8)/150)

Standard deviation of p^ ≈ 0.027 (rounded to six decimal places)

So, the standard deviation of the sampling distribution of p^ is approximately 0.027.

(4) To find the probability of obtaining x=123 or more individuals with the characteristic, we need to find P(p^ > 0.82). We can use the normal distribution to approximate this probability.

First, we standardize the value of p^:
Z = (0.82 - p) / √(p*(1-p)/n)

Z = (0.82 - 0.8) / √(0.8*(1-0.8)/150)

Z ≈ 0.54

Next, we look up the probability corresponding to this Z-value in the standard normal distribution table or use a calculator. The probability is approximately 0.2946 (rounded to four decimal places).

So, P(p^ > 0.82) ≈ 0.2946.

(5) Similarly, to find the probability of obtaining x=111 or fewer individuals with the characteristic, we need to find P(p^ < 0.74). We again use the normal distribution to approximate this probability.

First, we standardize the value of p^:
Z = (0.74 - p) / √(p*(1-p)/n)

Z = (0.74 - 0.8) / √(0.8*(1-0.8)/150)

Z ≈ -1.63

Next, we look up the probability corresponding to this Z-value in the standard normal distribution table or use a calculator. The probability is approximately 0.0526 (rounded to four decimal places).

So, P(p^ < 0.74) ≈ 0.0526.