Answer
$\bar{x}=5.24\%$
$\sigma = 0.053\%$
7
Work Step by Step
The average is given by: $\bar{x}=\frac{\sum x_i}{n}$, from the 10 values reported the average is:
$\bar{x}=5.24\%$
The standard deviation is given by: $\sigma = \sqrt{\dfrac{\sum(x_i-\bar{x})^2}{n-1}}$, which leads to the value of:
$\sigma = 0.053\%$
7 out of 10 student's results are within the range of $5.19\%-5.29\%$ (one standard deviation away from the average).
