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

Question 1:

Solution:

$\begin{array}{l}\text{Given that }\sigma =6\text{, }\Sigma x^2=260.\\ \sigma =6\\ \sqrt{\frac{\Sigma x^2}{n}-{\overline{X}}^2}=6\\ \frac{\Sigma x^2}{n}-{\overline{X}}^2=36\\ \frac{260}{5}-{\overline{X}}^2=36\\ {\overline{X}}^2=16\\ \overline{X}=\pm 4\\ \therefore \text{mean = }\pm 4\end{array}$

(a)
$\begin{array}{l}\text{Given mean = 6}\\ \frac{\Sigma x}{n}=6\\ \frac{\Sigma x}{6}=6\end{array}$  
1 + 3 + 7 + 15 + m + n= 36
$\begin{array}{l}\sigma =6\\ {\sigma }^2=36\\ \frac{\Sigma x^2}{n}-{\overline{X}}^2=36\\ \frac{1+9+49+225+m^2+n^2}{6}-6^2=36\\ \frac{284+m^2+n^2}{6}-36=36\\ \frac{284+m^2+n^2}{6}=72\\ 284+m^2+n^2=432\\ m^2+n^2=148\\ \text{From (a), }m=10-n\\ {\left(10-n\right)}^2+n^2=148\\ 100-20n+n^2+n^2=148\\ 2n^2-20n-48=0\\ n^2-10n-24=0\\ \left(n-6\right)\left(n+4\right)=0\\ n=6\text{ or }-4\end{array}$

(B) Grouped Data (without Class Interval)



Example 2:

Number of child	2	3	4	5	6	7	8
Frequency	6	8	5	3	3	3	2

Find the variance and standard deviation of the data.

(C) Grouped Data (with Class Interval)



Example 3:

Daily Salary(RM)	Number of workers
10 – 14	40
15 – 19	25
20 – 24	15
25 – 29	12
30 – 34	8

Solution:

$\begin{array}{l}\text{Mean }\overline{x}=\frac{\sum fx}{\sum f}\\ \text{Mean of daily salary}=\frac{1815}{100}=18.15\\ \\ \text{Variance, }{\sigma }^2=\frac{\sum f{x}^2}{\sum f}-{\overline{x}}^2\\ \text{Standard deviation, }\sigma \text{ = }\sqrt{\text{variance}}\\ {\sigma }^2=\frac{37185}{100}-{18.15}^2\\ {\sigma }^2=42.43\\ \sigma \text{ = }\sqrt{42.43}\\ \sigma \text{ = 6}\text{.514}\end{array}$

(B) Interquartile Range of Grouped Data (without Class Interval)

$\begin{array}{l}\text{Range}\\ =\text{midpoint of the highest class }-\text{midpoint of the lowest class}\\ \\ \text{Range}=\frac{35+39}{2}-\frac{5+9}{2}\\ =37-7\\ =30\end{array}$

7.1c Median

(B) Grouped Data (without Class Interval)


Example 2:

Solution:
$\text{Median}=T_{\frac{n+1}{2}}=T_{\frac{40+1}{2}}=T_{20\frac{1}{2}}=60\leftarrow (20\frac{1}{2}\text{th term is 60)}$

(C) Grouped Data (with Class Interval)



Example 3:

Find the median.
Solution:
Method 1: using formula



$\begin{array}{l}Step1\\ \text{Median class is given by }{T}_{\frac{n}{2}}=T_{\frac{100}{2}}=T_{50}\\ \therefore \text{ Median class is the class }20-24\\ \\ Step2\\ m=L+\left(\frac{\frac{N}{2}-F}{f_m}\right)c\\ m=19.5+\left(\frac{\frac{100}{2}-33}{26}\right)5\\ m=19.5+3.269=22.77\end{array}$