Answer to Question #149040 in Statistics and Probability for Jp

Question

Random variable X has the following probability distribution
X=X -2 -1 0 1
2 3
PAX) 0.1 0.2 2 03 3k
Find
the value of k
(ii) Evaluate P(X2)&P(-2<X2)
Find The Cumulative Distribution Of X
(iv)
Evaluate the Mean of X

Answer 1

$$\begin{matrix} x & -2 & -1& 0 & 1 & 2 & 3 \\ p(x) & 0.1 & k & 0.2 & 2k & 0.3 & 3k \end{matrix}$$

(i)

$$\sum_ip(x_i)=1$$

$$0.1+k+0.2+2k+0.3+3k=1$$

$$6k=0.4$$

$$k=\dfrac{1}{15}$$

(ii)

$$\begin{matrix} x & -2 & -1& 0 & 1 & 2 & 3 \\ \\ p(x) & 0.1 & \dfrac{1}{15} & 0.2 & \dfrac{2}{15} & 0.3 & 0.2 \end{matrix}$$

$$P(X<2)=1-P(X=2)-P(X=3)$$

$$=1-0.3-0.2=0.5$$

$$P(-2<X<2)=P(X=-1)+P(X=0)+P(X=1)$$

$$=\dfrac{1}{15}+0.2+\dfrac{2}{15}=0.4$$

(iii)

$$F(x) = \begin{cases} 0, &\text{for } x<-2 \\ 0.1, &\text{for } -2\leq x<-1 \\ 1/6, &\text{for } -1\leq x<0\\ 11/30, &\text{for } 0\leq x<1\\ 0.5, &\text{for } 1\leq x<2 \\ 0.8, &\text{for } 2\leq x<3 \\ 1, &\text{for } x\geq3 \end{cases}$$

(iv)

$$E(X)=-2(0.1)+(-1)(\dfrac{1}{15})+0(0.2)+1(\dfrac{2}{15})$$

$$+2(0.3)+3(0.2)=\dfrac{16}{15}$$

$E(X)=\dfrac{16}{15}$