Question

Find the mean and standard deviation of each of the following probability distributions:
(i)

xi :	3	11
pi :	1/2	1/2

(ii)

xi :	1	3	4	5
pi :	0.4	0.1	0.2	0.3

(iii)

xi :	−5	−4	1	2
pi :	1/4	1/8	1/2	1/8

(iv)

xi :	−1	0	1	2	3
pi :	0.3	0.1	0.1	0.3	0.2

(v)

xi :	1	2	3	4
pi :	0.4	0.3	0.2	0.1

(vi)

xi :	0	1	3	5
pi :	0.2	0.5	0.2	0.1

(vii)

xi :	−2	−1	0	1	2
pi :	0.1	0.2	0.4	0.2	0.1

(viii)

xi :	−3	−1	0	1	3
pi :	0.05	0.45	0.20	0.25	0.05

Answer 1

(i)

xi	pi	pixi	pixi2
3	$\frac{1}{2}$	$\frac{3}{2}$	$\frac{9}{2}$
11	$\frac{1}{2}$	$\frac{11}{2}$	$\frac{121}{2}$
		${\sum }_{}^{}$pixi = $\frac{14}{2}$ = 7	${\sum }_{}^{}$pixi2 = $\frac{130}{2}$ = 65

$$\begin{gathered} \mathrm{Mean}=\sum p_i{x}_i=7 \\ \mathrm{Variance}=\underset{}{\overset{}{\sum p_i{x_i}^2}}-{\left(\mathrm{Mean}\right)}^2 \\ =65-7^2 \\ =65-49 \\ =16 \\ \mathrm{Step}\mathrm{Deviation}=\sqrt{\mathrm{Variance}} \\ =\sqrt{16} \\ =4 \end{gathered}$$


(ii)

xi	pi	pixi	pixi2
1	0.4	0.4	0.4
3	0.1	0.3	0.9
4	0.2	0.8	3.2
5	0.3	1.5	7.5
		${\sum }_{}^{}$pixi = 3	${\sum }_{}^{}$pixi2 = 12

$$\begin{gathered} \mathrm{Mean}=\sum p_i{x}_i=3 \\ \mathrm{Variance}=\underset{}{\overset{}{\sum p_i{x_i}^2}}-{\left(\mathrm{Mean}\right)}^2 \\ =12-3^2 \\ =3 \\ \mathrm{Step}\mathrm{Deviation}=\sqrt{\mathrm{Variance}} \\ =\sqrt{3} \\ =1.732 \end{gathered}$$


(iii)

xi	pi	pixi	pixi2
$-$5	$\frac{1}{4}$	$-\frac{5}{4}$	$\frac{25}{4}$
$-$4	$\frac{1}{8}$	$-\frac{4}{8}$	$\frac{16}{8}$
1	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{1}{2}$
2	$\frac{1}{8}$	$\frac{2}{8}$	$\frac{4}{8}$
		${\sum }_{}^{}$pixi = $-1$	${\sum }_{}^{}$pixi2 = $\frac{74}{8}$

$$\begin{gathered} \mathrm{Mean}=\sum p_i{x}_i=-1 \\ \mathrm{Variance}=\underset{}{\overset{}{\sum p_i{x_i}^2}}-{\left(\mathrm{Mean}\right)}^2 \\ =\frac{74}{8}-{\left(-1\right)}^2 \\ =9.25-1 \\ =8.25 \\ \mathrm{Step}\mathrm{Deviation}=\sqrt{\mathrm{Variance}} \\ =\sqrt{8.25} \\ =2.872 \end{gathered}$$


(iv)

xi	pi	pixi	pixi2
$-$1	0.3	$-$0.3	0.3
0	0.1	0	0
1	0.1	0.1	0.1
2	0.3	0.6	1.2
3	0.2	0.6	1.8
		${\sum }_{}^{}$pixi = 1	${\sum }_{}^{}$pixi2 = 3.4

$$\begin{gathered} \mathrm{Mean}=\sum p_i{x}_i=1 \\ \mathrm{Variance}=\underset{}{\overset{}{\sum p_i{x_i}^2}}-{\left(\mathrm{Mean}\right)}^2 \\ =3.4-1 \\ =2.4 \\ \mathrm{Step}\mathrm{Deviation}=\sqrt{\mathrm{Variance}} \\ =\sqrt{2.4} \\ =1.549 \end{gathered}$$


(v)

xi	pi	pixi	pixi2
1	0.4	0.4	0.4
2	0.3	0.6	1.2
3	0.2	0.6	1.8
4	0.1	0.4	1.6
		${\sum }_{}^{}$pixi = 2	${\sum }_{}^{}$pixi2 = 5

$$\begin{gathered} \mathrm{Mean}=\sum p_i{x}_i=2 \\ \mathrm{Variance}=\underset{}{\overset{}{\sum p_i{x_i}^2}}-{\left(\mathrm{Mean}\right)}^2 \\ =5-2^2 \\ =1 \\ \mathrm{Step}\mathrm{Deviation}=\sqrt{\mathrm{Variance}} \\ =\sqrt{1} \\ =1 \end{gathered}$$


(vi)

xi	pi	pixi	pixi2
0	0.2	0	0
1	0.5	0.5	0.5
3	0.2	0.6	1.8
5	0.1	0.5	2.5
		${\sum }_{}^{}$pixi = 1.6	${\sum }_{}^{}$pixi2= 4.8

$$\begin{gathered} \mathrm{Mean}=\sum p_i{x}_i=1.6 \\ \mathrm{Variance}=\underset{}{\overset{}{\sum p_i{x_i}^2}}-{\left(\mathrm{Mean}\right)}^2 \\ =4.8-1.6^2 \\ =2.24 \\ \mathrm{Step}\mathrm{Deviation}=\sqrt{\mathrm{Variance}} \\ =\sqrt{2.24} \\ =1.497 \end{gathered}$$


(vii)

xi	pi	pixi	${\sum }_{}^{}$pixi2
$-$2	0.1	$-$0.2	0.4
$-$1	0.2	$-$0.2	0.2
0	0.4	0	0
1	0.2	0.2	0.2
2	0.1	0.2	0.4
		${\sum }_{}^{}$pixi = 0	${\sum }_{}^{}$pixi2 =1.2

$$\begin{gathered} \mathrm{Mean}=\sum p_i{x}_i=0 \\ \mathrm{Variance}=\underset{}{\overset{}{\sum p_i{x_i}^2}}-{\left(\mathrm{Mean}\right)}^2 \\ =1.2-0^2 \\ =1.2 \\ \mathrm{Step}\mathrm{Deviation}=\sqrt{\mathrm{Variance}} \\ =\sqrt{1.2} \\ =1.095 \end{gathered}$$


(viii)

xi	pi	pixi	pixi2
$-$3	0.05	$-$0.15	0.45
$-$1	0.45	$-$0.45	0.45
0	0.20	0	0
1	0.25	0.25	0.25
3	0.05	0.15	0.45
		${\sum }_{}^{}$pixi = $-$0.2	${\sum }_{}^{}$pixi2 =1.6

$$\begin{gathered} \mathrm{Mean}=\sum p_i{x}_i=-0.2 \\ \mathrm{Variance}=\underset{}{\overset{}{\sum p_i{x_i}^2}}-{\left(\mathrm{Mean}\right)}^2 \\ =1.6-{\left(-0.2\right)}^2 \\ =1.56 \\ \mathrm{Step}\mathrm{Deviation}=\sqrt{\mathrm{Variance}} \\ =\sqrt{1.56} \\ =1.249 \end{gathered}$$