Question

The following data represents the number of children born in individual pregnancies in 1996 and 2006:

# of children	1996 frequency	2006 frequency
one child	3671455	3971276
twins	100750	137085
triplets	5298	6118
quadruplets	560	355
quintuplets or more	81	67
A. Define the random variable x= number of children born in a single pregnancy in 1996 and develop a probablity distribution for the random variable. let x=5 represent quintuplets or more B. Compute expected value and variance for the number of children born in a single pregnancy in 1996 C. define a random variable = number of children born in a single pregnancy in 2006 and develop a probability distribution for the random variable. let y = 5 represent quintuplets or more

Answer 1

A. Random variable x= number of children born in a single pregnancy in 1996

The relative frequency distribution is given as below:

# of children	1996 frequency	1996 Relative Frequency
one child	3671455	0.97176
twins	100750	0.02667
triplets	5298	0.00140
quadruplets	560	0.00015
quintuplets or more	81	0.00002
Total	3778144	1

The probability distribution of X is given as:

X = 1; P(X=1) = 0.97176

X =2; P(X=2) = 0.02667

X = 3; P(X=3) = 0.00140

X = 4; P(X=4) = 0.00015

X = 5; P(X=5) = 0.00002

B.

Expected value of X, E(X) = 1 x 0.97176 + 2 x 0.02667 + 3 x 0.00140 + 4 x 0.00015 + 5 x 0.00002 = 1.03

Variance of X = (1- 1.03)² x 0.97176 + (2 - 1.03)² x 0.02667 + (3 - 1.03)² x0.00140 + (4 - 1.03)² x 0.00015 + (5 - 1.03)² x 0.00002 = 0.03305

C. Let random variable Y = number of children born in a single pregnancy in 2006

The relative frequency distribution is given as below:

# of children	2006 frequency	2006 Relative Frequency
one child	3971276	0.96509
twins	137085	0.03331
triplets	6118	0.00149
quadruplets	355	0.00009
quintuplets or more	67	0.00002
Total	4114901	1

The probability distribution of Y is given as:

Y = 1; P(Y=1) = 0.96509

Y =2; P(Y=2) = 0.03331

Y = 3; P(Y=3) = 0.00149

Y = 4; P(Y=4) = 0.00009

Y = 5; P(Y=5) = 0.00002

Although not asked in the question,

Expected value of Y, E(Y) = 1 x 0.96509 + 2 x 0.03331 + 3 x 0.00149 + 4 x 0.00009 + 5 x 0.00002 = 1.03664

Variance of X = (1- 1.03664)² x 0.96509 + (2 - 1.03664)² x 0.03331 + (3 - 1.03664)² x0.00149 + (4 - 1.03664)² x 0.00009 + (5 - 1.03664)² x 0.00002 = 0.08509

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