Answer to Question #129653 in Statistics and Probability for Gazal

Question

Define r.U and Mathematical expectation of a r.U. A discrete r.U. has the following probability distribution.
X : -2 -1 0 1 2 3
P(X) : 0.1 a 0.2 2a 0.3 a
Find i) The value of ‘a’,
ii) its distribution function F(X),
iii) Mean and variance of X.

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} x & -2 & -1 &\ \ 0\ & \ 1\ & 2\ & \ 3 \\ \hline p(x) & 0.1 & a & 0.2 & 2a & 0.3 \ & a \end{array}$$

(i)

$$\displaystyle\sum_{x}p(x)=1$$ $$0.1+a+0.2+2a+0.3+a=1$$ $$4a=0.4$$ $$a=0.1$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} x & -2 & -1 &\ \ 0\ & \ 1\ & 2\ & \ 3 \\ \hline p(x) & 0.1 & 0.1 & 0.2 & 0.2 & 0.3 \ & 0.1 \end{array}$$

(ii)

$F(-2)=F(X\leq-2)=p(-2)=0.1$

$F(-1)=F(X\leq-1)=p(-2)+p(-1)=$

$=0.1+0.1=0.2$

$F(0)=F(X\leq0)=p(-2)+p(-1)+p(0)=$

$=0.1+0.1+0.2=0.4$

$F(1)=F(X\leq1)=p(-2)+p(-1)+p(0)+p(1)=$

$=0.1+0.1+0.2+0.2=0.6$

$F(2)=F(X\leq2)=p(-2)+p(-1)+p(0)+p(1)+p(2)=$

$=0.1+0.1+0.2+0.2+0.3=0.9$

$F(3)=F(X\leq3)=1$

$$F(x) = \begin{cases} 0 & \ \ \ x<-2 \\ 0.1 & -2\leq x<-1 \\ 0.2 & -1\leq x<0 \\ 0.4 &\ \ \ 0\leq x<1\\ 0.6 & \ \ \ 1\leq x<2 \\ 0.9 & \ \ \ 2\leq x<3\\ 1 &\ \ \ 3\leq x \end{cases}$$

(iii)

$$\mu=E(X)=\displaystyle\sum_{x}x\cdot p(x)$$

$$\mu=E(X)=-2\cdot0.1+(-1)\cdot0.1+0\cdot0.2+$$ $$+1\cdot0.2+2\cdot0.3+3\cdot0.1=0.8$$

$$\sigma^2=E(X^2)-\mu^2$$

$$E(X^2)=(-2)^2\cdot0.1+(-1)^2\cdot0.1+(0)^2\cdot0.2+$$ $$+(1)^2\cdot0.2+(2)^2\cdot0.3+(3)^2\cdot0.1=2.8$$

$$\sigma^2=2.8-(0.8)^2=2.16$$