The table below shows the amount of time spent playing online games by a group of children in one particular week.

1. Find the class midpoints and use them to calculate the estimated mean.

Midpoints of the intervals are $1, 3, 5, 7, 9, 11$.

$\sum f = 4+9+15+13+7+2=50$

$\sum fx = 4(1)+9(3)+15(5)+13(7)+7(9)+2(11)$

$=4+27+75+91+63+22=282$

Estimated mean:

$\frac{\sum fx}{\sum f}=\frac{282}{50}=5.64$

1. State why the calculated mean from grouped data is not exact.

The data are grouped into class intervals, so the actual time spent by each child is unknown. In calculating the mean, each value in an interval is assumed to be represented by the class midpoint.

1. Find $\sum fx^2$ using the class midpoints.

The class midpoints are $1, 3, 5, 7, 9, 11$.

$\sum fx^2=4(1^2)+9(3^2)+15(5^2)+13(7^2)+7(9^2)+2(11^2)$

$=4+81+375+637+567+242=1906$

Also, $\sum f=4+9+15+13+7+2=50.$

2. Substitute into the standard deviation formula for grouped data.

Using $\bar{x}=5.64$ from part (a)(i),

$\text{standard deviation}=\sqrt{\frac{\sum fx^2}{\sum f}-\bar{x}^2}$

$=\sqrt{\frac{1906}{50}-5.64^2}$

$=\sqrt{38.12-31.8096}$

$=\sqrt{6.3104}=2.512\ldots$