The lives (in hours of continuous use) of 100 randomly selected flashlight batteries are: \[ \begin{array}{l|c} \text { Interval } & \text { Frequency } \\ \hline 6.95-7.45 & 2 \\ 7.45-7.95 & 8 \\ 7.95-8.45 & 21 \\ 8.45-8.95 & 35 \\ 8.95-9.45 & 20 \\ 9.45-9.95 & 11 \\ 9.95-10.45 & 3 \end{array} \] a. Find the mean of the battery lives. hrs (Type an integer or a decimal. Round to two decimal places.) b. Find the standard deviation of the battery lives. hrs (Type an integer or a decimal. Round to two decimal places.)

Step 1 ：Given the intervals and their corresponding frequencies, we first calculate the midpoints of each interval. The midpoints are calculated as the average of the lower and upper bounds of each interval.

Step 2 ：The midpoints are: \(7.2, 7.7, 8.2, 8.7, 9.2, 9.7, 10.2\)

Step 3 ：We then calculate the mean of the battery lives. The mean is calculated by multiplying each midpoint by its corresponding frequency, summing up these products, and then dividing by the total frequency (which is the total number of batteries, 100 in this case).

Step 4 ：The mean of the battery lives is \(8.74\) hours.

Step 5 ：To find the standard deviation, we first calculate the variance. The variance is the average of the squared differences from the Mean.

Step 6 ：The variance is \(0.4034\)

Step 7 ：The standard deviation is the square root of the variance.

Step 8 ：The standard deviation of the battery lives is \(0.64\) hours.

Step 9 ：Final Answer: The mean of the battery lives is \(\boxed{8.74}\) hours and the standard deviation is \(\boxed{0.64}\) hours.