Question

) In a study of families with children with disabilities, group of 5 US households were randomly selected. In the table below, the random variable x represents the number of households among 6 that had a child with a disability living there. x P(x) 0 0.02 1 0.15 2 0.29 3 0.31 4 0.23 a) Is this a probability distribution? b) What is the mean? c) What is the standard deviation?

Answer 1

Solution:

Part a

We are given the following probability distribution.

X	P(X)
0	0.02
1	0.15
2	0.29
3	0.31
4	0.23
Total	1

For above distribution, the sum of probabilities is 1, so given distribution is a probability distribution.

Part b

Mean = ∑XP(X)

X	P(X)	XP(X)
0	0.02	0
1	0.15	0.15
2	0.29	0.58
3	0.31	0.93
4	0.23	0.92
Total	1	2.58

Mean = 2.58

Part c

Standard deviation = Sqrt[∑P(X)(X – mean)^2]

X	P(X)	XP(X)	P(X)(X - mean)^2
0	0.02	0	0.133128
1	0.15	0.15	0.37446
2	0.29	0.58	0.097556
3	0.31	0.93	0.054684
4	0.23	0.92	0.463772
Total	1	2.58	1.1236

Standard deviation = Sqrt(1.1236) = 1.06

Standard deviation = 1.06