Answer to Question #158179 in Statistics and Probability for xsamdanielx

Question

For the random variable X defined by the probability function below, answer the following questions.

x 2 3 4 5 6 7

f(x) 0.25 0.15 0.10 0.15 0.10 0.25

a. Find the probability that X is even or at least 7.

b. Find and sketch the cdf.

c. Find the expected value and variance of X.

d. Find the expected value and variance of Y=2X+3.

Answer 1

a.

$$P(even\ \cup\ \geq7)=0.25+0.10+0.1+0.25=0.7$$

b.

$$F(2)=f(2)=0.25$$

$$F(3)=f(2)+f(3)=0.25+0.15=0.4$$

$$F(4)=f(2)+f(3)+f(4)=0.25+0.15+0.10$$

$$=0.5$$

$$F(5)=f(2)+f(3)+f(4)+f(5)$$

$$=0.25+0.15+0.10+0.15=0.65$$

$$F(6)=f(2)+f(3)+f(4)+f(5)+f(6)$$

$$=0.25+0.15+0.10+0.15+0.10=0.75$$

$$F(7)=f(2)+f(3)+f(4)+f(5)+f(6)+f(7)$$

$$=0.25+0.15+0.10+0.15+0.10+0.25=1$$

Hence

$$F(x) = \begin{cases} 0, &\text{for } x<2 \\ 0.25, &\text{for } 2\leq x<3 \\ 0.4, &\text{for } 3\leq x<4 \\ 0.5, &\text{for } 4\leq x<5 \\ 0.65, &\text{for } 5\leq x<6 \\ 0.75, &\text{for } 6\leq x<7 \\ 1, &\text{for } x\geq7 \\ \end{cases}$$

c.

$$E(X)=2(0.25)+3(0.15)+4(0.10)+5(0.15)$$

$$+6(0.10)+7(0.25)=4.45$$

$$E(X^2)=2^2(0.25)+3^2(0.15)+4^2(0.10)$$

$$+5^2(0.15)+6^2(0.10)+7^2(0.25)=23.55$$

$$Var(X)=E(X^2)-(E(X))^2=23.55-(4.45)^2$$

$$=3.7475$$

d. $Y=2X+3$

$$E(Y)=E(2X+3)=2E(X)+3$$

$$=2(4.45)+3=11.9$$

$$Var(Y)=Var(2X+3)=2^2Var(X)$$

$$=4(3.7475)=14.99$$