The problem consists of several questions related to statistics and sampling distributions. Specifically, it includes problems on drawing samples, computing probabilities related to sample means, and constructing confidence intervals.

1. Problem Description

The problem consists of several questions related to statistics and sampling distributions. Specifically, it includes problems on drawing samples, computing probabilities related to sample means, and constructing confidence intervals.

2. Solution Steps

Question 2:

a. We are given a normal population with mean $\mu = 120$ and standard deviation $\sigma = 15$. We have a random sample of size $n = 16$. We need to compute the probability that the sample mean is greater than

1

2

5. The sampling distribution of the sample mean $\bar{x}$ is normal with mean $\mu_{\bar{x}} = \mu = 120$ and standard deviation $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{16}} = \frac{15}{4} = 3.75$.

We want to find $P(\bar{x} > 125)$. We can standardize $\bar{x}$ by calculating the z-score:

$z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{125 - 120}{3.75} = \frac{5}{3.75} = \frac{5}{15/4} = \frac{20}{15} = \frac{4}{3} \approx 1.33$.

So we want to find $P(z > 1.33)$. Using a standard normal table or calculator, we have $P(z > 1.33) = 1 - P(z \le 1.33) = 1 - 0.9082 = 0.0918$.

b. We need to compute the probability that the sample mean is between 116 and

1

2

8. We have $\mu = 120$ and $\sigma_{\bar{x}} = 3.75$.

We want to find $P(116 < \bar{x} < 128)$. We standardize the values:

$z_1 = \frac{116 - 120}{3.75} = \frac{-4}{3.75} = -\frac{16}{15} \approx -1.07$.

$z_2 = \frac{128 - 120}{3.75} = \frac{8}{3.75} = \frac{32}{15} \approx 2.13$.

So we want to find $P(-1.07 < z < 2.13) = P(z < 2.13) - P(z < -1.07)$.

Using a standard normal table, $P(z < 2.13) = 0.9834$ and $P(z < -1.07) = 0.1423$.

$P(-1.07 < z < 2.13) = 0.9834 - 0.1423 = 0.8411$.

Question 3:

a. The appropriate formula to calculate a confidence interval for $\mu$ when the population standard deviation is unknown and the sample size is large ($n = 352 > 30$) is the z-interval:

$\bar{x} \pm z_{\alpha/2} \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $z_{\alpha/2}$ is the z-score corresponding to the desired confidence level.

b. We are given $\bar{x} = 13.4$, $s = 6.8$, and $n = 352$. We want a 95% confidence interval, so $\alpha = 1 - 0.95 = 0.05$ and $\alpha/2 = 0.025$. Thus, $z_{\alpha/2} = z_{0.025} = 1.96$.

The 95% confidence interval is:

$13.4 \pm 1.96 \frac{6.8}{\sqrt{352}} = 13.4 \pm 1.96 \frac{6.8}{18.7617} = 13.4 \pm 1.96(0.3624) = 13.4 \pm 0.7103$.

The interval is $(13.4 - 0.7103, 13.4 + 0.7103) = (12.6897, 14.1103)$.

Question 4:

a. We are given $n = 15$, $\bar{x} = 24$, and $s = 5$. We want to develop a 99% confidence interval for the population mean. Since $n$ is small, we should use the t-distribution.

We have $\alpha = 1 - 0.99 = 0.01$, so $\alpha/2 = 0.005$. The degrees of freedom are $df = n - 1 = 15 - 1 = 14$.

Looking up the t-value for $t_{0.005, 14}$ in a t-table, we get $t_{0.005, 14} = 2.977$.

The confidence interval is:

$\bar{x} \pm t_{\alpha/2, n-1} \frac{s}{\sqrt{n}} = 24 \pm 2.977 \frac{5}{\sqrt{15}} = 24 \pm 2.977 \frac{5}{3.873} = 24 \pm 2.977(1.291) = 24 \pm 3.843$.

The interval is $(24 - 3.843, 24 + 3.843) = (20.157, 27.843)$.

b. We want to know if it's reasonable to conclude that the population mean is greater than

2

8. Since the 99% confidence interval is (20.157, 27.843), and 28 is not within this interval, and is greater than the upper bound of the interval, it is not reasonable to conclude that the population mean is greater than

2

8.

3. Final Answer

Question 2:

a. 0.0918

b. 0.8411

Question 3:

a. $\bar{x} \pm z_{\alpha/2} \frac{s}{\sqrt{n}}$

b. (12.6897, 14.1103)

Question 4:

a. (20.157, 27.843)

b. No, it is not reasonable.