Suppose a simple random sample of size $$ n=75 $$ is obtained from a population whose size is $$ N=20,000 $$ and whose population proportion with a specified characteristic is $$ p=0.4 $$. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2), C. Approximately normal because $$ n \leq U . U S N $$ and $$ n p(1-p)

Question

Suppose a simple random sample of size $$ n=75 $$ is obtained from a population whose size is $$ N=20,000 $$ and whose population proportion with a specified characteristic is $$ p=0.4 $$. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2), C. Approximately normal because $$ n \leq U . U S N $$ and $$ n p(1-p)<1 U $$. 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 $$ \hat{p} $$. $$ \mu_{p}=0.4 $$ (Round to one decimal place as needed.) Determine the standard deviation of the sampling distritution of $$ \hat{p} $$. $$ \sigma_{\hat{p}}=0.056569 $$ (Round to six decimal places as needed.) (b) What is the probability of obtaining $$ x=36 $$ or more individuals with the characteristic? That is, what is $$ \mathrm{P}(\hat{\mathrm{p}} \geq 0.48) $$ ? $$ \mathrm{P}(\hat{\mathrm{p}} \geq 0.48)=0.0791 $$ (Round to four decimal places as needed.) (c) What is the probability of obtaining $$ \mathrm{x}=21 $$ or fewer individuals with the characteristic? That is, what is $$ \mathrm{P}(\hat{\mathrm{p}} \leq 0.28) $$ ? $$ P(\hat{p} \leq 0.28)= $$ $$ \square $$ (Round to four decimal places as needed.)

UpStudy AI · Accepted Answer

We are given a population with proportion  
$$
p=0.4,
$$  
and a simple random sample of  
$$
n=75.
$$

**Step 1. Mean of the Sampling Distribution**

For the proportion estimator $$\hat{p}$$, the mean is given by:  
$$
\mu_{\hat{p}} = p.
$$  
Thus,  
$$
\mu_{\hat{p}} = 0.4.
$$

**Step 2. Standard Deviation of the Sampling Distribution**

The standard deviation (without finite population correction) is:  
$$
\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}.
$$  
Substitute the values:  
$$
\sigma_{\hat{p}} = \sqrt{\frac{0.4(0.6)}{75}} = \sqrt{\frac{0.24}{75}} = \sqrt{0.0032} \approx 0.056569.
$$

**Step 3. (b) Calculating $$ \mathrm{P}(\hat{p} \geq 0.48) $$**

We first compute the $$z$$-score for $$\hat{p} = 0.48$$:  
$$
z = \frac{0.48-0.4}{0.056569} = \frac{0.08}{0.056569} \approx 1.413.
$$  
Then,  
$$
\mathrm{P}(\hat{p} \geq 0.48) = \mathrm{P}(z \geq 1.413).
$$
From the standard normal distribution table, the probability to the left of $$z=1.413$$ is approximately 0.9209. Therefore,  
$$
\mathrm{P}(z \geq 1.413) = 1 - 0.9209 = 0.0791.
$$

**Step 4. (c) Calculating $$ \mathrm{P}(\hat{p} \leq 0.28) $$**

Now compute the $$z$$-score for $$\hat{p} = 0.28$$:  
$$
z = \frac{0.28-0.4}{0.056569} = \frac{-0.12}{0.056569} \approx -2.121.
$$
Thus,  
$$
\mathrm{P}(\hat{p} \leq 0.28) = \mathrm{P}(z \leq -2.121).
$$
Using the standard normal distribution table, the probability for $$z \leq -2.12$$ is approximately 0.0170. (Rounded to four decimal places.)

**Final Answers:**

- Mean of the sampling distribution:  
  $$
  \mu_{\hat{p}} = 0.4.
  $$
- Standard deviation of the sampling distribution:  
  $$
  \sigma_{\hat{p}} \approx 0.056569.
  $$
- $$ \mathrm{P}(\hat{p} \geq 0.48) \approx 0.0791 $$.
- $$ \mathrm{P}(\hat{p} \leq 0.28) \approx 0.0170 $$.
- Mean of the sampling distribution: $$ \mu_{\hat{p}} = 0.4 $$ - Standard deviation of the sampling distribution: $$ \sigma_{\hat{p}} \approx 0.056569 $$ - Probability of $$ \hat{p} \geq 0.48 $$: $$ 0.0791 $$ - Probability of $$ \hat{p} \leq 0.28 $$: $$ 0.0170 $$

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