Answer to Question #342976 in Statistics and Probability for Carol

Question #342976

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]


1
Expert's answer
2022-05-25T12:16:19-0400

(a)


P(X>2)=10.250.33=0.42P(X>2)=1-0.25-0.33=0.42

(b)


P(2X4)=0.33+0.17+0.15=0.65P(2\le X\le 4)=0.33+0.17+0.15=0.65

(c)


P(X<4)=10.150.10=0.75P(X<4)=1-0.15-0.10=0.75

(d)


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

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



(e)


E(X2)=0.25(1)2+0.33(2)2+0.17(3)2E(X^2)=0.25(1)^2+0.33(2)^2+0.17(3)^2

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

Var(X)=σ2=E(X2)(E(X))2Var(X)=\sigma^2=E(X^2)-(E(X))^2

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




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