Question

g A group of 50 physical doctors from four countries consists of 5 Americans, 10 British, 15 Chinese and 20 Danes. First, choose one of the 50 doctors at random and let X be the size of that doctor’s group. Next, choose one of the four groups at random and let Y be its size. (With or without replacement do not make any difference here: recall that all random choices are with equal probability, unless otherwise specified.) (a) Write down the probability mass functions for X and Y . (b) Compute EX and EY . (c) Compute V ar(X) and V ar(Y ).

Answer 1

Answer:

a)    x =     5       10      15      20

 P(X) = 0.1     0.2    0.3     0.4

 P(Y) = 0.25  0.25  0.25  0.25

b) E(X) = 15

  E(Y) = 12.5

c) Var(X) = 25

 Var(Y) = 31.25

Explanation:   E(X) = xP(x)  ; E(Y) = xP(Y) ; Var(X) = sum(x^2P(x)) - (E(X))^2   ; Var(Y) = sum(x^2P(y)) - (E(Y))^2

   x =     5       10      15      20

 P(X) = 0.1     0.2    0.3     0.4

 P(Y) = 0.25  0.25  0.25  0.25

b) E(X) = (5 x 0.1) +(10 x 0.2) + (15 x 0.3) + (20 x 0.4)

   = 0.5 + 2 + 4.5 + 8

E(X) = 15

E(Y) = (5 x 0.25) +(10 x 0.25) + (15 x 0.25) + (20 x 0.25)

  = 1.25 + 2.5 + 3.75 + 5

E(Y) = 12.5

c) Var(X) = ( (5^2 x 0.1) + (10^2 x 0.2) +( 15^2 x 0.3) + (20^2 x 0.4) ) - (15^2)

  = 250 - 225

Var(X) = 25

Var(Y)

= ( (5^2 x 0.25) + (10^2 x 0.25) +( 15^2 x 0.25) + (20^2 x 0.25) ) - (12.5^2)

= 187.5 - 156.25

Var(Y) = 31.25