Question #149040
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
1
Expert's answer
2020-12-08T06:57:43-0500
x210123p(x)0.1k0.22k0.33k\begin{matrix} x & -2 & -1& 0 & 1 & 2 & 3 \\ p(x) & 0.1 & k & 0.2 & 2k & 0.3 & 3k \end{matrix}

(i)

ip(xi)=1\sum_ip(x_i)=1

0.1+k+0.2+2k+0.3+3k=10.1+k+0.2+2k+0.3+3k=1

6k=0.46k=0.4

k=115k=\dfrac{1}{15}

(ii)


x210123p(x)0.11150.22150.30.2\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)=1P(X=2)P(X=3)P(X<2)=1-P(X=2)-P(X=3)

=10.30.2=0.5=1-0.3-0.2=0.5

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

=115+0.2+215=0.4=\dfrac{1}{15}+0.2+\dfrac{2}{15}=0.4

(iii)


F(x)={0,for x<20.1,for 2x<11/6,for 1x<011/30,for 0x<10.5,for 1x<20.8,for 2x<31,for x3F(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)(115)+0(0.2)+1(215)E(X)=-2(0.1)+(-1)(\dfrac{1}{15})+0(0.2)+1(\dfrac{2}{15})

+2(0.3)+3(0.2)=1615+2(0.3)+3(0.2)=\dfrac{16}{15}

E(X)=1615E(X)=\dfrac{16}{15}



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