There are 1000 households in a town. Specifically, there are 100 households with one member, 200 housholds with 2 memeber, 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

[tex]PMF = \left[\begin{array}{ccc}{\frac{100}{1000} = 0.1 \ for X = 1 }\\{\frac{200}{1000} = 0.2 \ for X = 2 }\\{\frac{300}{1000} = 0.3 \ for X = 3 }\\{\frac{200}{1000} = 0.2 \ for X = 4 }\\{\frac{100}{1000} = 0.1 \ for X = 5 }\\{\frac{100}{1000} = 0.1 \ for X = 6 }\end{array}\right[/tex]

Expected value

[tex]E[X] = 3.3 [/tex]

b

[tex]PMF = \left[\begin{array}{ccc}{\frac{100 * 1}{3300} = 0.030 \ for Y = 1 }\\\\{\frac{200 * 2}{3300} = 0.121 \ for Y = 2 }\\\\{\frac{300* 3}{3300} = 0.273 \ for Y = 3 }\\\\{\frac{200 * 4}{3300} = 0.242 \ for Y = 4 }\\\\{\frac{100* 5}{3300} = 0.152 \ for Y = 5 }\\\\{\frac{100 * 6}{3300} = 0.182 \ for Y = 6 }\end{array}\right[/tex]

Expected value

[tex]E(Y) = 3.911[/tex]

Step-by-step explanation:

From question we are told that

The number of households is n = 1000

The number of households with one member is m = 100

The number of households with two member is o = 200

The number of households with three member is p = 300

The number of households with four member is q = 200

The number of households with five member is r = 100

The number of households with six member is r = 100

The total number of people living in the town is N = 3300

Generally the PMF is mathematically represented as

[tex]PMF = \left[\begin{array}{ccc}{\frac{100}{1000} = 0.1 \ for X = 1 }\\{\frac{200}{1000} = 0.2 \ for X = 2 }\\{\frac{300}{1000} = 0.3 \ for X = 3 }\\{\frac{200}{1000} = 0.2 \ for X = 4 }\\{\frac{100}{1000} = 0.1 \ for X = 5 }\\{\frac{100}{1000} = 0.1 \ for X = 6 }\end{array}\right[/tex]

Here X is the random variable as defined in the question

Generally the expected value of X is mathematically represented as

[tex]E[X] = 1 * 0.1\ + \ 2* 0.2\ + \ 3 * 0.3\ + \ 4 * 0.2 \ + \ 5 * 0.1 \ + \ 6 * 0.1[/tex]

=> [tex]E[X] = 3.3 [/tex]

Considering B

Generally the PMF is mathematically represented as

[tex]PMF = \left[\begin{array}{ccc}{\frac{100 * 1}{3300} = 0.030 \ for Y = 1 }\\\\{\frac{200 * 2}{3300} = 0.121 \ for Y = 2 }\\\\{\frac{300* 3}{3300} = 0.273 \ for Y = 3 }\\\\{\frac{200 * 4}{3300} = 0.242 \ for Y = 4 }\\\\{\frac{100* 5}{3300} = 0.152 \ for Y = 5 }\\\\{\frac{100 * 6}{3300} = 0.182 \ for Y = 6 }\end{array}\right[/tex]

Generally the expected value of Y is mathematically represented as

[tex]E(Y) = 0.03 * 1 + 0.121 * 2 + 0.273 * 3 + 0.242 * 4 + 0.152 * 5 + 0.182 * 6[/tex]

=> [tex]E(Y) = 3.911[/tex]