Short Questions (Question 5 & 6)

Question 5:
A set of data consists of 9, 2, 7, x² – 1 and 4. Given the mean is 6, find
(a) the positive value of x,
(b) the median using the value of x in part (a).

Solution:
(a)
$\begin{array}{l}\text{Mean}=6\\ \frac{9+2+7+x^2-1+4}{5}=6\\ x^2+21=30\\ x^2=9\\ x=\pm 3\\ \text{Positive value of }x=3.\end{array}$

(b)
Arrange the numbers in ascending order
2, 4, 7, 8, 9
Median = 7

Question 6:
A set of seven numbers has a standard deviation of 3 and another set of three numbers has a standard deviation of 4. Both sets of numbers have an equal mean.
If the two sets of numbers are combined, find the variance.

Solution:
$\begin{array}{l}{\overline{X}}_1=\frac{\Sigma X_1}{n_1}\\ m=\frac{\Sigma X_1}{7}\\ \Sigma X_1=7m\\ \\ m=\frac{\Sigma X_2}{3}\\ \Sigma X_2=3m\\ \\ \sigma =\sqrt{\frac{\Sigma X^2}{N}-{\left(\overline{X}\right)}^2}\\ {\sigma }^2=\frac{\Sigma X^2}{N}-{\left(\overline{X}\right)}^2\\ 9=\frac{\Sigma X_1{}^2}{7}-m^2\\ 63=\Sigma X_1{}^2-7m^2\\ \Sigma X_1{}^2=7m^2+63\end{array}$

$\begin{array}{l}\text{Similarly:}\\ 16=\frac{\Sigma X_2{}^2}{3}-m^2\\ 48=\Sigma X_2{}^2-3m^2\\ \Sigma X_2{}^2=48+3m^2\\ \\ \Sigma Y^2=\Sigma X_1{}^2+\Sigma X_2{}^2\\ \Sigma Y^2=7m^2+63+3m^2+48\\ =10m^2+111\\ \\ \Sigma Y=\Sigma X_1+\Sigma X_2\\ \Sigma Y=7m+3m=10m\\ \\ \text{Combine Variance:}\\ {\sigma }^2=\frac{\Sigma Y^2}{N}-{\left(\frac{\Sigma Y}{N}\right)}^2\\ {\sigma }^2=\frac{10m^2+111}{10}-{\left(\frac{10m}{10}\right)}^2\\ =\frac{10m^2+111}{10}-m^2\\ =\frac{10m^2+111-10m^2}{10}\\ =\frac{111}{10}=11.1\end{array}$