To construct a 95% confidence interval for the mean age of the first arrest, we'll use the following formula:

$\text{Confidence Interval} = \bar{x} \pm z \times \frac{s}{\sqrt{n}}$

Where:

• $\bar{x}$ is the sample mean.
• $z$ is the z-value corresponding to the desired confidence level.
• $s$ is the sample standard deviation.
• $n$ is the sample size.

Step 1: Identify the values

• Sample mean, $\bar{x}$ = 12
• Standard deviation, $s$ = 1.9
• Sample size, $n$ = 25
• For a 95% confidence interval, the z-value ($z$) is approximately 1.96.

Step 2: Calculate the standard error

$\text{Standard Error} = \frac{s}{\sqrt{n}} = \frac{1.9}{\sqrt{25}} = \frac{1.9}{5} = 0.38$

Step 3: Calculate the margin of error

$\text{Margin of Error} = z \times \text{Standard Error} = 1.96 \times 0.38 = 0.7448$

Step 4: Calculate the confidence interval

$\text{Confidence Interval} = \bar{x} \pm \text{Margin of Error} = 12 \pm 0.7448$ $\text{Lower bound} = 12 - 0.7448 = 11.2552$ $\text{Upper bound} = 12 + 0.7448 = 12.7448$

So, the 95% confidence interval is approximately $11.26$ to $12.74$, which is not exactly matching any of the given choices. However, the closest choice would be b) the intervals are 11.20 and 12.80.

Final Answer:

b) the intervals are 11.20 and 12.80

Would you like more details or have any questions?

Here are 5 related questions you might find interesting:

1. How do you interpret a confidence interval in the context of this problem?
2. What happens to the confidence interval if we increase the sample size?
3. How would the confidence interval change if the standard deviation were larger?
4. What is the significance of the z-value in constructing a confidence interval?
5. How would you calculate a confidence interval if the sample size were small (e.g., $n < 30$)?

Tip: The width of the confidence interval depends on both the sample size and the standard deviation; increasing the sample size generally leads to a narrower interval, making the estimate more precise.