Problem 20 There are 1000 households in a town. Specifically, there are 100 hous

Question

Problem 20 There are 1000 households in a town. Specifically, there are 100 households with one member, 200 households with 2 members, 300 households with 3 members, 200 households with 4 members, 100 households with 5 members, and 100 households with 6 members. Thus, the total number of people living in the town is N=100 . 1+200 . 2 + 300 . 3+200-4+100 . 5 + 100 . 6 = 3300. a. We pick a household at random, and define the random variable X as the number of people in the chosen household. Find the PMF and the expected value of X. b. We pick a person in the town at random, and define the random variable Y as the number of people in the household where the chosen person lives. Find the PMF and the expected value of Y.

Explanation / Answer

a. Total number of households = 1000

As X denotes the number of people in a household:

PMF of X:

P(X=1) = 100/1000 = 0.10

P(X=2) = 200/1000 = 0.20

P(X=3) = 300/1000 = 0.30

P(X=4) = 200/1000 = 0.20

P(X=5) = 100/1000 = 0.10

P(X=6) = 100/1000 = 0.10

E(X) = 1*0.10 + 2*0.20 + 3*0.30 + 4*0.20 + 5*0.10 + 6*0.10

= 3.3

b. As Y denotes the number of people in the household in which chosen person lives,

PMF of Y:

P(Y=1) = 100/3300 = 1/33 (As 100 of 3300 people in the population live in house of 1 person)

P(Y=2) = 400/3300 = 4/33

P(Y=3) = 900/3300 = 3/11

P(Y=4) = 800/3300 = 8/33

P(Y=5) = 100/3300 = 1/33

P(Y=6) = 100/3300 = 1/33

E(Y) = 1*1/33 + 2*4/33 + 3*3/11 + 4*8/33 + 5*1/33 + 6*1/33

= 2.3939

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