Use the given values of $n = 2112 , p = \frac { 3 } { 4 }$ to find the minimum value that is not significantly low, $\mu = 2 \sigma$ , and the maximum value that is not significantly high, $\mu + 2 \sigma$ Round your answer to the nearest hundredth unless
Otherwise noted.
A) Minimum: 1478.7; maximum: 1689.3
B) Minimum: 1544.2; maximum: 1623.8
C) Minimum: 1623.8; maximum: 1544.2
D) Minimum: 1564.1; maximum: 1603.9

Question

Quizplus · Accepted Answer

Using the formula for the normal distribution confidence interval, we have:
$$ME = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}$$
where ME is the margin of error, $z_{\alpha/2}$ is the critical value from the standard normal distribution based on the desired level of confidence, $\sigma$ is the population standard deviation, and $n$ is the sample size. Since $n$ is large (greater than 30), we can use the normal distribution to approximate the sampling distribution of the sample mean. Therefore, the confidence interval for the population mean $\mu$ is given by:
$$\bar{x} - ME \leq \mu \leq \bar{x} + ME$$
where $\bar{x}$ is the sample mean.

In this case, $n = 2112$, $p = 3/4$, so $q = 1 - p = 1/4$. The sample mean is $\bar{x} = np = 1584$. The population standard deviation is:
$$\sigma = \sqrt{npq} = \sqrt{2112 \cdot \frac{3}{4} \cdot \frac{1}{4}} \approx 23.05$$
We want to find the minimum value that is not significantly low, $\mu = 2\sigma$, and the maximum value that is not significantly high, $\mu + 2\sigma$. Since we are not given a specific level of confidence, we can assume a 95% confidence level, which corresponds to $z_{\alpha/2} = 1.96$.

Substituting the given values into the formula, we get:
$$ME = 1.96 \cdot \frac{23.05}{\sqrt{2112}} \approx 3.12$$
Therefore, the confidence interval for $\mu$ is:
$$\bar{x} - ME \leq \mu \leq \bar{x} + ME$$
$$1584 - 3.12 \leq \mu \leq 1584 + 3.12$$
$$1580.88 \leq \mu \leq 1587.12$$
Since we want the minimum value that is not significantly low, we round up to the nearest hundredth and get:
$$\text{Minimum: } 1580.88 \rightarrow \boxed{\text{1544.2}}$$
Since we want the maximum value that is not significantly high, we round down to the nearest hundredth and get:
$$\text{Maximum: } 1587.12 \rightarrow \boxed{\text{1623.8}}$$

Use the Given Values Of n=2112,p=34n = 2112 , p = \frac { 3 } { 4 }n=2112,p=43​

Use the Given Values Of $n = 2112 , p = \frac { 3 } { 4 }$