The following data represent the age (in weeks) at which babies first crawl based on a survey of 12 mothers. The data are normally distributed and $s=10.103$ weeks. Construct and interpret a 95\% confidence interval for the population standard deviation of the age (in weeks) at which babies first crawl. $\begin{array}{llllll}51 & 32 & 45 & 35 & 38 & 25 \\ 47 & 38 & 57 & 25 & 38 & 29\end{array}$ Click the icon to view the table of critical values of the chi-square distribution. Select the correct choice below and fill in the answer boxes to complete your choice. (Use ascending order. Round to three decimal places as needed.) A. There is a $95 \%$ probability that the true population standard deviation is between and B. There is $95 \%$ confidence that the population standard deviation is between and C. If repeated samples are taken, $95 \%$ of them will have the sample standard deviation between and

Step 1 ：Given that the sample size (n) is 12 and the sample standard deviation (s) is 10.103 weeks. We are asked to construct a 95% confidence interval for the population standard deviation.

Step 2 ：First, we calculate the degrees of freedom (df) which is given by n - 1. So, df = 12 - 1 = 11.

Step 3 ：Next, we calculate the critical values of the chi-square distribution. The lower critical value (\(\chi^2_{\alpha/2}\)) is found at the \(\alpha/2\) quantile and the upper critical value (\(\chi^2_{1-\alpha/2}\)) is found at the \(1 - \alpha/2\) quantile of the chi-square distribution with df degrees of freedom. Here, \(\alpha\) is 0.05, so \(\alpha/2\) is 0.025.

Step 4 ：Using a chi-square table or a statistical software, we find that \(\chi^2_{\alpha/2}\) = 3.816 and \(\chi^2_{1-\alpha/2}\) = 21.920.

Step 5 ：Then, we calculate the confidence interval for the population standard deviation. The lower bound is given by \(\sqrt{(df \times s^2) / \chi^2_{1-\alpha/2}}\) and the upper bound is given by \(\sqrt{(df \times s^2) / \chi^2_{\alpha/2}}\).

Step 6 ：Substituting the given values, we get the lower bound as \(\sqrt{(11 \times (10.103)^2) / 21.920}\) = 7.157 and the upper bound as \(\sqrt{(11 \times (10.103)^2) / 3.816}\) = 17.154.

Step 7 ：\(\boxed{\text{Final Answer: There is 95% confidence that the population standard deviation is between 7.157 and 17.154.}}\)