To construct a 95% confidence interval for the mean number of personal computers, we can use the following formula for the confidence interval when the population standard deviation ($\sigma$) is known:

$\bar{x} \pm Z \cdot \frac{\sigma}{\sqrt{n}}$

where:

• $\bar{x}$ = sample mean = 1.45
• $Z$ = Z-score corresponding to a 95% confidence level, which is approximately 1.96
• $\sigma$ = population standard deviation = 0.85
• $n$ = sample size = 200

Step 1: Calculate the standard error

$\frac{\sigma}{\sqrt{n}} = \frac{0.85}{\sqrt{200}} \approx \frac{0.85}{14.14} \approx 0.0601$

Step 2: Calculate the margin of error

$Z \cdot \frac{\sigma}{\sqrt{n}} = 1.96 \times 0.0601 \approx 0.1178$

Step 3: Construct the confidence interval

$\text{Lower limit} = 1.45 - 0.1178 \approx 1.33$ $\text{Upper limit} = 1.45 + 0.1178 \approx 1.57$

Thus, the 95% confidence interval for the mean number of personal computers is approximately: $1.33 < \mu < 1.57$

Would you like further explanation on any part of this calculation?

Here are some related questions that may help you explore this topic further:

1. How do we determine the Z-score for different confidence levels?
2. What is the significance of a 95% confidence level?
3. How would the confidence interval change if the sample size increased?
4. What if the population standard deviation were unknown—how would the calculation change?
5. Why do we use the population standard deviation instead of the sample standard deviation in this case?

Tip: Remember that increasing the sample size generally leads to a narrower confidence interval, meaning greater precision in estimating the population mean.