Answer to Question #342976 in Statistics and Probability for Carol

Question

QUESTION 21

Let X represent the number of children in a randomly selected South African household. The

probability distribution of X is given below.

x 1 2 3 4 5

P .x/ 0:25 0:33 0:17 0:15 0:10

(a) What is the probability that a randomly selected South African household will have more than

2 children? (3)

(b) What is the probability that a randomly selected South African household will have between

2 and 4 children (inclusive)? (3)

(c) What is the probability that a randomly selected South African household will have fewer than

4 children? (3)

(d) Calculate the expected number of children in a randomly selected South African household.

(4)

(e) Calculate the variance of the number of children in a randomly selected South African house-

hold. (5)

[18]

Answer 1

(a)

$$P(X>2)=1-0.25-0.33=0.42$$

(b)

$$P(2\le X\le 4)=0.33+0.17+0.15=0.65$$

(c)

$$P(X<4)=1-0.15-0.10=0.75$$

(d)

$$E(X)=0.25(1)+0.33(2)+0.17(3)$$

$$+0.15(4)+0.10(5)=2.52$$

(e)

$$E(X^2)=0.25(1)^2+0.33(2)^2+0.17(3)^2$$

$$+0.15(4)^2+0.10(5)^2=8$$

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

$$=8-(2.52)^2=1.6496$$