Answer to Question #219511 in Statistics and Probability for mimi

Question

Random variable X has the following probability distribution.

x 0 1 3 4 5 6

P .X D x/ 0.1 0.1 0.3 0.2 0.1 0.2

(a) Find P .3  X  5/. (3)

(b) Find E .X/: (5)

(c) Find V ar .X/ (7)

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 3 & 4 & 5 & 6 \\ \hline P(X=x) & 0.1 & 0.1 & 0.3 & 0.2 & 0.1 & 0.2 \\ \end{array}$$

Check

$$0.1+0.1+0.3+0.2+0.1+0.2=1$$

(a)

$$P(3\leq X\leq 5)=P(X=3)+P(X=4)+P(X=5)$$

$$=0.3+0.2+0.1=0.6$$

$$P(3< X<5)=P(X=4)=0.2$$

(b)

$$E[X]=0(0.1)+1(0.1)+3(0.3)+4(0.2)$$

$$+5(0.1)+6(0.2)=3.5$$

(c)

$$E[X^2]=0^2(0.1)+1^2(0.1)+3^2(0.3)+4^2(0.2)$$

$$+5^2(0.1)+6^2(0.2)=15.7$$

$$Var(X)=E[X^2]-(E[X])^2$$

$$=15.7-(3.5)^2=3.45$$