Nutrition: animals. The mouse weights (in grams) of a random sample of 100 mice involved in a nutrition experiment are:
\begin{tabular}{|c|c|c|c|c|c|}
\hline Interval & $41.5-43.5$ & $43.5-45.5$ & $45.5-47.5$ & $47.5-49.5$ & $49.5-51.5$ \\
\hline Frequency & 2 & 5 & 11 & 30 & 17 \\
\hline Interval & $51.5-53.5$ & $53.5-55.5$ & $55.5-57.5$ & $57.5-59.5$ & \\
\hline Frequency & 16 & 13 & 5 & 1 & \\
\hline
\end{tabular}
a. Find the mean of the weight of the mice. $50.22 \mathrm{gms}$ (Type an integer or a decimal. Round to two decimal places.)
b. Find the standard deviation of the weight of the mice. gms (Type an integer or a decimal. Round to two decimal places.)

Step 1 ：Find the midpoint of each interval and multiply it by the frequency of that interval. The midpoints of the intervals are: \(42.5, 44.5, 46.5, 48.5, 50.5, 52.5, 54.5, 56.5, 58.5\). The products of the midpoints and the frequencies are: \(85, 222.5, 511.5, 1455, 857.5, 840, 708.5, 282.5, 58.5\).

Step 2 ：Sum up these products to get \(5020.5\).

Step 3 ：Divide the sum by the total number of mice, which is 100, to find the mean weight of the mice. So, the mean weight of the mice is \(\frac{5020.5}{100} = 50.205\) grams. Rounded to two decimal places, this is \(\boxed{50.21}\) grams.

Step 4 ：To find the standard deviation, first find the variance. The variance is the average of the squared differences from the mean. First, find the squared differences from the mean for each interval. These are: \((42.5-50.21)^2, (44.5-50.21)^2, (46.5-50.21)^2, (48.5-50.21)^2, (50.5-50.21)^2, (52.5-50.21)^2, (54.5-50.21)^2, (56.5-50.21)^2, (58.5-50.21)^2\).

Step 5 ：Multiply each squared difference by the frequency of the corresponding interval. These products are: \(118.848, 656.848, 1615.848, 782.848, 0.848, 338.848, 738.848, 198.848, 69.848\).

Step 6 ：Sum up these products to get \(4520.792\).

Step 7 ：The variance is then \(\frac{4520.792}{100} = 45.20792\).

Step 8 ：Finally, the standard deviation is the square root of the variance, which is \(\sqrt{45.20792} = 6.72\) grams (rounded to two decimal places). So, the standard deviation is \(\boxed{6.72}\) grams.