Suppose a simple random sample of size $n=1000$ is obtained from a population whose size is $\mathrm{N}=1,500,000$ and whose population proportion with a specified characteristic is $p=0.49$. Complete parts (a) through (c) below.
Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2).
(a) Describe the sampling distribution of $\hat{p}$.
A. Approximately nomal, $\mu_{p}=0.49$ and $\sigma_{\hat{p}} \approx 0.0004$
8. Approximately normal, $\mu_{p}=0.49$ and $\sigma_{\hat{p}} \approx 0.0002$
Vc. Approximately normal, $\psi_{p}=0.49$ and $\sigma_{\hat{p}} \approx 0.0158$
(b) What is the probability of obtaining $x=520$ or more individuals with the characteristic?
$P(x \geq 520)=0.0289$ (Round to four decimal places as needed.)
(c) What is the probability of obtaining $x=450$ or fewer individuals with the characteristio?
$P(x \leq 450)=\square($ Round to four decimal places as needed. $)$
HW Score: $37.5 \%, 3$ of 8 points
Save
Points: 0 of 1
Answer
Therefore, the probability of obtaining 450 or fewer individuals with the characteristic is \(\boxed{0.0}\).
Step 1 ：Given that the sample size (n) is 1000, the population proportion (p) is 0.49, and we are looking for the probability of 450 or fewer successes.
Step 2 ：We first calculate the standard deviation of a proportion using the formula \(\sqrt{\frac{p(1-p)}{n}}\), where p is the population proportion and n is the sample size. This gives us a standard deviation of approximately 0.0158.
Step 3 ：We then standardize the score of 450 using the formula \(z = \frac{x - np}{\sqrt{np(1-p)}}\), where x is the number of successes, n is the sample size, and p is the population proportion. This gives us a z-score of approximately -80.02.
Step 4 ：Finally, we use the standard normal distribution table to find the probability corresponding to the z-score. The probability is 0.0.
Step 5 ：Therefore, the probability of obtaining 450 or fewer individuals with the characteristic is \(\boxed{0.0}\).
