Let's shake on it: A random sample of 12 -ounce milkshakes from 13 fast-food restaurants had the following number of calories.
\begin{tabular}{lllllll}
\hline 591 & 472 & 510 & 580 & 613 & 375 & 700 \\
642 & 473 & 608 & 450 & 476 & 504 & \\
\hline
\end{tabular}
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Assume the population standard deviation is $\sigma=92$.
Part 1 of 3
(a) Explain why it is necessary to check whether the population is approximately nomal before constructing a confidence interval.
It is necessary to check whether the population is approximately normal because the sample size is less than or equal to 30
Part 2 of 3
(b) Following is a dotplot of these data. Is it reasonable to assume that the population is approximately normal?
It is $\quad \mathbf{V}$ reasonable to assume that the population is approximately normai.

Part: $2 / 3$

Part 3 of 3
(c) If appropriate, construct a $99.5 \%$ confidence interval for the mean calorie count for all 12 -ounce milkshakes sold at fast-food restaurants. Round the answers to at least two decimal places.

A $99.5 \%$ confidence interval for the mean calorie count for all 12 -ounce milkshakes sold at fast-food restaurants is $< \mu< \square$.
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Step 1 ：Given the data of calorie counts from 13 different fast-food restaurants, we are asked to construct a 99.5% confidence interval for the mean calorie count for all 12-ounce milkshakes sold at fast-food restaurants. The population standard deviation is given as \(\sigma = 92\).

Step 2 ：To construct the confidence interval, we need to calculate the sample mean and then use the formula for the confidence interval which is: \[\bar{x} \pm Z_{\alpha/2} * \frac{\sigma}{\sqrt{n}}\] where: \(\bar{x}\) is the sample mean, \(Z_{\alpha/2}\) is the Z-score for the desired confidence level (for 99.5% confidence level, \(Z_{\alpha/2}\) is approximately 2.807), \(\sigma\) is the population standard deviation, and \(n\) is the sample size.

Step 3 ：First, calculate the sample mean (\(\bar{x}\)) of the given data. The sample mean is 538.0.

Step 4 ：Next, calculate the margin of error using the formula: \[Z_{\alpha/2} * \frac{\sigma}{\sqrt{n}}\] The margin of error is approximately 71.62.

Step 5 ：Finally, construct the confidence interval by subtracting and adding the margin of error from the sample mean. The lower bound of the confidence interval is \(\bar{x} - \text{margin of error} = 466.38\) and the upper bound is \(\bar{x} + \text{margin of error} = 609.62\).

Step 6 ：\(\boxed{\text{Final Answer: A 99.5% confidence interval for the mean calorie count for all 12 -ounce milkshakes sold at fast-food restaurants is } 466.38 < \mu < 609.62}\)