a family contain 8 childrens half of which are boys and half of which are girls
if 5 childrens are selected from the family than find the probability
distribution of no. of boys selected. Also find mean, Variance and S.D of
No. of boys selected.
There are 4 boys and 4 girls. Therefore:
in a random selection of 5 children, there are either x = exactly 1 boy, x =
exactly 2 boys, x = exactly 3 boys, or x = exactly 4 boys.
Number of ways to have:
x = exactly 1 boy: Choose the 1 boy 4C1=4 ways and all 4 girls 4C4=1 way. That's 4∙1 = 4 ways.
x = exactly 2 boys: Choose the 2 boys 4C2=6 ways and the 3 girls 4C3=4 ways. That's 6∙4 = 24 ways.
x = exactly 3 boys: Choose the 3 boys 4C3=4 ways and the 2 girls 4C2=6 ways. That's 4∙6 = 24 ways.
x = exactly 4 boys: Choose all 4 boys 4C4=1 way and the 1 girl 4C1=4 ways. That's 1∙4 = 4 ways.
So there are 4+24+24+4 = 56 ways to choose 5 from the 8.
[As a check on the 56, let's calculate the number of ways we can choose 5
children from the 8. --> 8 children CHOOSE 5 = 8C5 = 56. So it checks!]
To find the probability of each, we divide the number of ways by
x f(x) x∙f(x)
1 4 4
2 24 48
3 24 72
4 4 16
------------------
Σ 56 140
We calculate the mean
μ = Σ[x∙f(x)]/Σ[f(x)] = 140/56 = 2.5
x f(x) x-μ (x-μ)² f(x)∙(x-μ)²
1 4 -1.5 2.25 9
2 24 -0.5 0.25 6
3 24 0.5 0.25 6
4 4 1.5 2.25 9
--------------------------------------
Σ 56 30
This is the population, not a sample. So we divide 30/56 and reduce to
15/28.
Variance = σ² = 15/28 = 0.5357142957
S.D. = σ = √(σ²) = √(0.5357142957) = 0.7319250547
Edwin