Question

A woman decides to have children until she has her first girl or until she has five children; whichever happens first. Find the expected value and standard deviation of X. Let X represent the number of children she has. Assume that she is equally likely to give birth to a boy or a girl on each child, and that gender is independent between children.

(a) Write out and sketch the complete probability distribution function for X.
(b) Write out and sketch the complete cumulative distribution function for X.
(c) Find the expected value and standard deviation of X.

Answer 1

E(X) = 1.9375

SD(X) = 1.20

Explanation:

Let :

G = girl ; B = Boy

P(G) = 0.5 ; P(B) = 0.5

Sample space :

G, BG, BBG, BBBG, BBBBG, BBBBB

Using the multiplication rule of independence :

P(G) = 0.5

P(BG) = 0.5 * 0.5 = 0.25

P(BBG) = 0.5 * 0.5 * 0.5 = 0.125

P(BBBG) = 0.5 * 0.5 * 0.5 * 0.5 = 0.0625

P(BBBBG) = 0.5 * 0.5 * 0.5 * 0.5 * 0.5 = 0.03125

P(BBBBB) = 0.5 * 0.5 * 0.5 * 0.5 * 0.5 = 0.03125

Creating a probability distribution table :

X = 1, 2, 3, 4, 5

For X = 5 ;

P(BBBBG) + P(BBBBB) = 0.03125 + 0.03125 = 0.0625

X :___ 1 ____ 2 ____ 3 _____ 4 _____ 5

P(X) : 0.5 __ 0.25 _ 0.125 _0.0625 _0.0625

The expected value E(X) :

√(ΣX * P(X)) = (1*0.5)+(2*0.25)+(3*0.125)+(4*0.0625)+(5*0.0625) = 1.9375

The standard deviation SD(x) :

√(ΣX²*p(x) - E(X)²

√((1^2*0.5)+(2^2*0.25)+(3^2*0.125)+(4^2*0.0625)+(5^2*0.0625) - 1.9375^2)

√(5.1875 - 3.75390625)

√1.43359375

SD(X) = 1.197

SD(X) = 1.20