Questions: 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=0.8. Complete parts (a) through (c) below. (b) What is the probability of obtaining x=123 or more individuals with the characteristic? That is, what is P(p̂ ≥ 0.82)? P(p̂ ≥ 0.82)= (Round to four decimal places as needed.) (c) What is the probability of obtaining x=114 or fewer individuals with the characteristic? That is, what is P(p̂ ≤ 0.76)? P(p̂ ≤ 0.76)= (Round to four decimal places as needed.)

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=0.8. Complete parts (a) through (c) below.
(b) What is the probability of obtaining x=123 or more individuals with the characteristic? That is, what is P(p̂ ≥ 0.82)?
P(p̂ ≥ 0.82)= (Round to four decimal places as needed.)
(c) What is the probability of obtaining x=114 or fewer individuals with the characteristic? That is, what is P(p̂ ≤ 0.76)?
P(p̂ ≤ 0.76)= (Round to four decimal places as needed.)

Transcript text: 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 $p=0.8$. Complete parts (a) through (c) below. (b) What is the probability of obtaining $x=123$ or more individuals with the characteristic? That is, what is $P(\hat{p} \geq 0.82)$ ? $P(\hat{p} \geq 0.82)=$ $\square$ (Round to four decimal places as needed.) (c) What is the probability of obtaining $\mathrm{x}=114$ or fewer individuals with the characteristic? That is, what is $\mathrm{P}(\hat{\mathrm{p}} \leq 0.76)$ ? $P(\hat{p} \leq 0.76)=\square$ $\square$ (Round to four decimal places as needed.)

Solution

Solution Steps

Step 1: Calculate the Mean and Standard Deviation

For a sample size $ n = 150 $ from a population with a proportion $ p = 0.8 $, the mean $ \mu $ and standard deviation $ \sigma $ of the sampling distribution of the sample proportion $ \hat{p} $ are calculated as follows:

\[ \mu = p = 0.8 \]

\[ \sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.8 \cdot 0.2}{150}} \approx 0.0327 \]

Step 2: Part (b) Probability $ P(\hat{p} \geq 0.82) $

To find $ P(\hat{p} \geq 0.82) $, we first calculate the z-score:

\[ z = \frac{0.82 - 0.8}{0.0327} \approx 0.6124 \]

Using the standard normal distribution, we find:

\[ P = \Phi(\infty) - \Phi(0.6124) \approx 0 - 0.0 = 0.0 \]

Thus,

\[ P(\hat{p} \geq 0.82) = 0.0 \]

Step 3: Part (c) Probability $ P(\hat{p} \leq 0.76) $

Next, we calculate $ P(\hat{p} \leq 0.76) $ by finding the z-score:

\[ z = \frac{0.76 - 0.8}{0.0327} \approx -1.2247 \]