8.4.5 Measures of Dispersion for Ungrouped Data, SPM Paper (Long Questions)

Question 9:

Solution:
(a)
$\begin{array}{l}\text{Mean }\overline{x}=b\\ \frac{1+a+2a+8+9+15}{6}=b\\ 33+3a=6b\\ 3a=6b-33\\ a=2b-11\text{ ——(1)}\\ \\ \text{New median }=\frac{4b}{7}\\ \frac{\left(2a-3\right)+\left(8-3\right)}{2}=\frac{4b}{7}\\ \frac{2a+2}{2}=\frac{4b}{7}\\ 14a+14=8b\\ 7a=4b-7\text{ ——(2)}\\ \\ \text{Substitute (1) into (2),}\\ 7\left(2b-11\right)=4b-7\\ 14b-77=4b-7\\ 10b=70\\ b=7\\ \\ \text{From (1), }a=2(7)-11=3\end{array}$


(b)
$\begin{array}{l}\text{Variance, }{\sigma }^2=\frac{\sum x^2}{N}-{\overline{x}}^2\\ {\sigma }^2=\frac{{\left(-2\right)}^2+{\left(0\right)}^2+{\left(3\right)}^2+{\left(5\right)}^2+{\left(6\right)}^2+{\left(12\right)}^2}{6}-{\left(\frac{-2+0+3+5+6+12}{6}\right)}^2\\ {\sigma }^2=\frac{218}{6}-16=20.333\end{array}$

Question 10:
(b) A sum of certain numbers is 72 with mean of 9 and the sum of the squares of these numbers of 800, is taken out from the set of 20 numbers. Calculate the mean and variance of the remaining numbers.

Solution:
(a)
$\begin{array}{l}\text{Mean }\overline{x}=\frac{\sum x}{N}\\ 8=\frac{\sum x}{20}\\ \sum x=160\\ \\ \text{Standard deviation, }\sigma =\sqrt{\frac{\sum x^2}{N}-{\overline{x}}^2}\\ 3=\sqrt{\frac{\sum x^2}{N}-{\overline{x}}^2}\\ 9=\frac{\sum x^2}{20}-8^2\\ \frac{\sum x^2}{20}=73\\ \sum x^2=1460\end{array}$


(b)
$\begin{array}{l}\text{Sum of certain numbers, }M\text{ is 72 with mean of }9,\\ \frac{72}{M}=9\\ M=8\\ \\ \text{Mean of the remaining numbers}=\frac{160-72}{20-8}=7\frac{1}{3}\\ \text{Variance of the remaining numbers}=\frac{1460-800}{12}-{\left(7\frac{1}{3}\right)}^2\\ =55-53\frac{7}{9}=1\frac{2}{9}\end{array}$