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 $95\mathrm{\%}$ probability that the true population standard deviation is between and
B. There is $95\mathrm{\%}$ confidence that the population standard deviation is between and
C. If repeated samples are taken, $95\mathrm{\%}$ 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 (${\chi }_{\alpha /2}^2$) is found at the $\alpha /2$ quantile and the upper critical value (${\chi }_{1-\alpha /2}^2$) 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 ${\chi }_{\alpha /2}^2$ = 3.816 and ${\chi }_{1-\alpha /2}^2$ = 21.920.

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

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 ：$\boxed{{\displaystyle \text{Final Answer: There is 95\% confidence that the population standard deviation is between 7.157 and 17.154.}}}$