Question

5 In a certain northern Wisconsin town (cough...cough...Oliver), there are 100 families. 30 families have 1 child, 50 families have 2 children, and 20 families have 3 children. The children are then ranked according to age: the oldest has rank 1, second oldest rank 2, third oldest rank 3. Suppose you select a random child from the town. Find the probability mass function, expected value, and variance of the childs rank.

Answer 1

X : Child's Rank

Possible Value of X = 1,2,3

Number of Families = 100

Number of families with one child = 30

Number of children from the families with one child =30

Number of children with rank 1 from the families with one child = 30

Number of families with 2 children = 50

Number of children from the families with 2 children =2 x 50 =100

Number of children with rank 1 from the families with 2 children = 50

Number of children with rank 2 from the families with 2 children = 50

Number of families with 3 children = 20

Number of children from the families with 3 children =3 x 20 = 60

Number of children with rank 1 from the families with 2 children = 20

Number of children with rank 2 from the families with 2 children = 20

Number of children with rank 3 from the families with 2 children = 20

Total number of children from all the 100 families = 30+100+60=190

Total number of children with Rank 1 = 30+50+20 = 100

Total number of children with Rank 2 = 50+20 =70

Total number of children with Rank 3 = 20

X : Child's Rank

A child is randomly selected from the town,

Probability that the child's rank is 1

= P(X=1) = Total number of children with Rank 1 / Total number of children from all the 100 families = 100/190=10/19

Probability that the child's rank is 2

= P(X=2) = Total number of children with Rank 2 / Total number of children from all the 100 families = 70/190=7/19

Probability that the child's rank is 3

= P(X=3) = Total number of children with Rank 3 / Total number of children from all the 100 families = 20/190=2/19

Probability mass function of X:

X	P(X=x)
1	10/19
2	7/19
3	2/19

Expected Value of the child's rank = E(X)

Expected Value of the child's rank = 30/19 = 1.578947368

Variance of the child's rank

Var(X) = 164/361

Variance of the child's rank = 164/361 = 0.454293629