Let's start by solving the questions based on the provided images.

Interval Calculation Formula:

We are asked to find the 95% confidence interval for the mean using the formula: $\left[\mu - z\frac{\sigma}{\sqrt{n}}, \mu + z\frac{\sigma}{\sqrt{n}}\right]$ Where:

• $\mu$ is the population mean,
• $\sigma$ is the population standard deviation,
• $n$ is the sample size,
• $z$ is the z-value corresponding to the 95% confidence level, which is approximately 1.96.

a) Given:

$\mu = 163, \sigma = 17, n = 49$ Substitute the values into the formula to calculate the interval.

b) Given:

$\mu = 1014, \sigma = 18, n = 5$ Substitute these values to calculate the interval.

c) Given:

$\mu = 56, \sigma = 2, n = 20$ We'll compute the confidence interval for this as well. Let me proceed with the calculations.The 95% confidence intervals for each case are as follows:

• a) For $\mu = 163$, $\sigma = 17$, $n = 49$: \text{The 95% confidence interval is from } 158.24 \text{ to } 167.76.

• b) For $\mu = 1014$, $\sigma = 18$, $n = 5$: \text{The 95% confidence interval is from } 998.22 \text{ to } 1029.78.

• c) For $\mu = 56$, $\sigma = 2$, $n = 20$: \text{The 95% confidence interval is from } 55.12 \text{ to } 56.88.

Would you like a more detailed explanation of the calculations or have any further questions?

Here are 5 related questions to explore:

1. What is the significance of the 95% confidence interval in hypothesis testing?
2. How would changing the sample size $n$ affect the width of the confidence interval?
3. Can confidence intervals be used for non-normal distributions?
4. What is the role of the standard deviation $\sigma$ in determining the confidence interval?
5. Why is the z-value approximately 1.96 for a 95% confidence interval?

Tip: The larger the sample size, the narrower the confidence interval, as the margin of error decreases with increasing $n$.