Question 5:


Solution:
(a)


(b)
$\begin{array}{l}\text{Interquartile Range}\\ =L_{Q_3}+\left(\frac{\frac{3N}{4}-F}{f_{{}_{Q_3}}}\right)c-L_{Q_1}+\left(\frac{\frac{N}{4}-F}{f_{{}_{Q_1}}}\right)c\\ =59.5+\left(\frac{\frac{3}{4}\left(80\right)-59}{10}\right)10-29.5+\left(\frac{\frac{1}{4}\left(80\right)-7}{18}\right)10\\ =59.5+1-\left(29.5+7.22\right)\\ =23.78\end{array}$

Question 6:
Solution:

Question 3:

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 4:


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

Question 1:
Solution:

$\begin{array}{l}35.5=29.5+\left(\frac{20-\left(4+m\right)}{n}\right)\times 10\\ 6=\left(\frac{16-m}{n}\right)\times 10\end{array}$

Question 2:
Solution:
(a)(i)
$\begin{array}{l}\text{Given mean}=5\\ \frac{\Sigma x}{6}=6\\ \Sigma x=36\end{array}$

(a)(ii)
$\begin{array}{l}\text{Given }\sigma =2.4\\ {\sigma }^2={2.4}^2\\ \frac{\Sigma x^2}{n}-{\overline{X}}^2=5.76\\ \frac{\Sigma x^2}{6}-6^2=5.76\\ \frac{\Sigma x^2}{6}=41.76\\ \Sigma x^2=250.56\end{array}$

(b)(i)

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:

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

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

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

Question 3:
Solution:
$\begin{array}{l}\text{Mean }\overline{x}=\frac{\sum x}{N}\\ \sqrt{p}=\frac{\sum x}{5}\\ \sum x=5\sqrt{p}\\ \\ \text{Standard deviation, }\sigma =\sqrt{\frac{\sum x^2}{N}-{\overline{x}}^2}\\ 2q=\sqrt{\frac{120}{5}-{\left(\sqrt{p}\right)}^2}\\ 4q^2=24-p\\ p=24-4q^2\end{array}$

Question 4:

Solution:
$\begin{array}{l}\text{Variance, }{\sigma }^2=\frac{\sum x^2}{N}-{\overline{x}}^2\\ 6=\frac{1^2+4^2+p^2}{3}-{\left(\frac{1+4+p}{3}\right)}^2\\ 6=\frac{17+p^2}{3}-{\left(\frac{5+p}{3}\right)}^2\\ 6=\frac{17+p^2}{3}-\left[\frac{25+10p+p^2}{9}\right]\\ 6=\frac{51+3p^2-25-10p-p^2}{9}\\ 2p^2-10p+26=54\\ 2p^2-10p+28=0\\ p^2-5p+14=0\\ \left(p-7\right)\left(p+2\right)=0\\ p=-2\text{ (not accepted)}\\ \therefore p=7\end{array}$