The table below shows the number of books read by 50 students in Class A in semester 1.

1. Find the estimated mean using class midpoints and frequencies.

Class midpoints: $1, 3, 5, 7, 9$

$\text{Estimated mean}=\frac{1(8)+3(21)+5(11)+7(3)+9(7)}{50}$

$=\frac{8+63+55+21+63}{50}$

$=\frac{210}{50}=4.2$

1. Calculate the estimated standard deviation using class midpoints and frequencies.

Class midpoints are $1,3,5,7,9$ and from part (a)(i), $\bar{x}=4.52$.

$\sum fx^2=8(1^2)+21(3^2)+11(5^2)+3(7^2)+7(9^2)$
$=8+189+275+147+567=1186$

$\text{Standard deviation}=\sqrt{\frac{\sum fx^2}{\sum f}-\bar{x}^2}$
$=\sqrt{\frac{1186}{50}-4.52^2}$
$=\sqrt{23.72-20.4304}$
$=2.70\text{ books} \;\text{(to 3 s.f.)}$

1. State the effect on the mean when every value is increased by 2.

Since each student reads $2$ more books, every data value is increased by $2$.

So the mean also increases by $2$:
$4.52+2=6.52$

2. State the effect on the standard deviation when every value is increased by 2.

Adding the same number to every data value shifts the whole distribution but does not change its spread.

So the standard deviation remains unchanged:
$2.70$

1. Compare the standard deviations and interpret the difference in spread.

Class A standard deviation $= 2.70$ books.

Class B standard deviation $= 3.15$ books.

Since $3.15 > 2.70$, Class B has a greater spread in the number of books read. Hence, Class B's data is less consistent than Class A's.