An apple a day: Following are the numbers of grams of sugar per $100 \mathrm{grams}$ of apple in a random sample of 6 Red Delicious apples. Assume the population is normally distributed. Construct a $90 \%$ confidence interval for the standard deviation of the number of grams of sugar. Round the answers to two decimal places. $\begin{array}{llllll}12.3 & 12.5 & 13.3 & 13.5 & 11.8 & 10.6\end{array}$ Send data to Excel A $90 \%$ confidence interval for the standard deviation of the number of grams of sugar is $\square<\sigma<\square$.

Step 1 ：Given the data of the number of grams of sugar per 100 grams of apple in a random sample of 6 Red Delicious apples: \(12.3, 12.5, 13.3, 13.5, 11.8, 10.6\). We are asked to construct a 90% confidence interval for the standard deviation of the number of grams of sugar.

Step 2 ：First, we calculate the sample size \(n\), which is the number of data points. In this case, \(n = 6\).

Step 3 ：Next, we calculate the sample standard deviation \(s\), which is approximately \(1.06\).

Step 4 ：We then determine the significance level for a 90% confidence level, which is \(0.1\).

Step 5 ：We calculate the chi-square values for the lower and upper bounds of the confidence interval. The lower chi-square value is approximately \(1.15\) and the upper chi-square value is approximately \(11.07\).

Step 6 ：Finally, we calculate the confidence interval for the standard deviation. The lower bound of the confidence interval is \(\sqrt{(n - 1) \cdot s^2 / \text{chi2_upper}}\) which is approximately \(0.71\), and the upper bound of the confidence interval is \(\sqrt{(n - 1) \cdot s^2 / \text{chi2_lower}}\) which is approximately \(2.21\).

Step 7 ：So, a 90% confidence interval for the standard deviation of the number of grams of sugar is \(\boxed{0.71}<\sigma<\boxed{2.21}\).