Question 6, 8.2.13
HW Score: $81.9\mathrm{\%},5.73$ of 7 points
Part 3 of 3
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Suppose a simple random sample of size $n=1000$ is obtained from a population whose size is $\mathrm{N}=2,000,000$ and whose population proportion with a specified characteristic is $\mathrm{p}=0.42$. 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 normal, ${\mu }_{\hat{p}}=042$ and ${\sigma }_{\hat{p}}\approx 00156$
B. Approximately normal, ${\mu }_{\hat{p}}=0.42$ and ${\sigma }_{\hat{p}}\approx 0.0003$
C. Approximately normal, ${\mu }_{\hat{p}}=0.42$ and ${\sigma }_{\hat{p}}\approx 0.0002$
(b) What is the probability of obtaining $x=440$ or more individuals with the characteristic?
$P(x\ge 440)={0.1000}^{\mathrm{\top }}$ (Round to four decimal places as needed)
(c) What is the probability of obtaining $x=390$ or fewer individuals with the characteristic?
$P(x\le 390)=◻($ Round to four decimal places as needed)
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Step 1 ：Given that the sample size $n=1000$, the population size $N=2000000$, and the population proportion $p=0.42$.

Step 2 ：The sampling distribution of the proportion is approximately normal if $np$ and $n(1-p)$ are both greater than 5. The mean of the distribution is equal to the population proportion $p$, and the standard deviation is $\sqrt{p(1-p)/n}$.

Step 3 ：Calculate the mean ${\mu }_{\hat{p}}=0.42$ and the standard deviation ${\sigma }_{\hat{p}}=\sqrt{0.42(1-0.42)/1000}\approx 0.0156$.

Step 4 ：For part (b), we can use the normal approximation to the binomial distribution to calculate the probability. The z-score for a value $x$ in a distribution with mean $\mu$ and standard deviation $\sigma$ is given by $z=(x-\mu )/\sigma$. The probability of obtaining a value greater than or equal to $x$ is given by the area under the standard normal curve to the right of the z-score. So, the probability of obtaining $x=440$ or more individuals with the characteristic is $P(x\ge 440)=\boxed{{\displaystyle 0.1000}}$.

Step 5 ：For part (c), similarly, the probability of obtaining a value less than or equal to $x$ is given by the area under the standard normal curve to the left of the z-score. So, the probability of obtaining $x=390$ or fewer individuals with the characteristic is $P(x\le 390)=\boxed{{\displaystyle 0.0273}}$.