Short Questions (Question 10 & 11)

Question 10:
A set of data consists of twelve positive numbers.
$\text{It is given that }\Sigma {\left(x-\overline{x}\right)}^2=600\text{ and }\Sigma x^2=1032.$
Find
(a) the variance
(b) the mean

Solution:
(a)
$\begin{array}{l}\text{Variance}=\frac{\Sigma {\left(x-\overline{x}\right)}^2}{N}\\ =\frac{600}{12}\\ =50\end{array}$

(b)
$\begin{array}{l}\text{Variance}=\frac{\Sigma x^2}{N}-{\left(\overline{x}\right)}^2\\ 50=\frac{1032}{12}-{\left(\overline{x}\right)}^2\\ {\left(\overline{x}\right)}^2=86-50\\ =36\\ \overline{x}=36\end{array}$

Question 11 (4 marks):
A set of data consists of 2, 3, 4, 5 and 6. Each number in the set is multiplied by m and added by n, where m and n are integers. It is given that the new mean is 17 and the new standard deviation is 4.242.
Find the value of m and of n.

Solution:

$\begin{array}{l}{\displaystyle \sum x}=2+3+4+5+6=20\\ {\displaystyle \sum x^2}=2^2+3^2+4^2+5^2+6^2=90\\ \text{Mean}=\frac{20}{5}=4\\ \text{Variance}=\frac{{\displaystyle \sum x^2}}{n}-{\left(\overline{x}\right)}^2\\ =\frac{90}{5}-4^2=2\\ \\ \text{New mean}=17\\ 4m+n=17\mathrm{..........}\left(1\right)\\ \\ \text{New standard deviation}=4.242\\ m\times \sqrt{2}=4.242\\ m=\frac{4.242}{\sqrt{2}}=2.9995\approx 3\\ \\ \text{Substitute }m=3\text{ into }\left(1\right):\\ 4\left(3\right)+n=17\\ n=5\end{array}$