Homework SourseIn a study of families with children with disabilities a gro
In a study of families with children with disabilities, a group of 6 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
0.539
1
0.351
2
0.095
3
0.014
4
0.001
5
0+
6
0+
Is this a probability distribution?
What is the mean?
What is the standard deviation?
| x | P(x) |
| 0 | 0.539 |
| 1 | 0.351 |
| 2 | 0.095 |
| 3 | 0.014 |
| 4 | 0.001 |
| 5 | 0+ |
| 6 | 0+ |
Solution
In a study of families with children with disabilities, a group of 6 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
0.539
1
0.351
2
0.095
3
0.014
4
0.001
5
0+
6
0+
Is this a probability distribution?
x
P(x)
0
0.539
1
0.351
2
0.095
3
0.014
4
0.001
5
0
6
0
Total
1
Sum of p(x) =1. This is a probability distribution
What is the mean?
X
P(X)
X*P(X)
(X-MEAN)2*P(X)
0
0.539
0
0.185723
1
0.351
0.351
0.05987
2
0.095
0.19
0.189674
3
0.014
0.042
0.081516
4
0.001
0.004
0.011649
5
0
0
0
6
0
0
0
1.000
0.587
0.528431
Mean = sum of X*P(X ) =0.587
What is the standard deviation?
Variance = sum of (X-MEAN)2*P(X) =0.528431
SD = sqrt( 0.528431) = 0.726933
| x | P(x) |
| 0 | 0.539 |
| 1 | 0.351 |
| 2 | 0.095 |
| 3 | 0.014 |
| 4 | 0.001 |
| 5 | 0+ |
| 6 | 0+ |
