Please help me with this probibility problem

Question

Let the random variable x represent the number of girls in a family with three children. Assume the probability of a child being a girl is 0.42

 

The table below describes the probability of having x number of girls.
Determine whether the table describes a probability distribution.
If it does, find the mean and standard deviation.

Is it unusual for a family of three children to consist of three girls?

 

X P(x)
0 0.195
1 0.424
2 0.307
3 0.074

Answer 1

yes it's a probability distribution as the percentages sum to 100

mean= .424 + .307(2) + .074(3) = .424 + .614 + .222 = 1.6 children

sd = sqrVar, Var = sum of (Xi' - x bar)/(N-1)

3 children all girls happens (.42)^3 = about 7.4% of the time, so its fairly improbable, sort of rare, but not an outlier

var = (0 -1.6)^2/5 + (1-1.6)^2/5 + (2-1.6)^2/5 + (3-1.6)/5 = 2.56+.36+.16+1.96)/5 = 3.03/5= .606

sd = sqr.606=about .7785

3 children has z score = (3-1.6)/.7785 = about 1.798

you could calculate area under the normal curve = probability, but it's very misleading, as the tails don't exist, no children less than zero, and none greater than three.

You could use the binomial distribution given there are only 2 mutually exclusive disjoint alternatives, boy or girl (ignoring trans gender issues)

p = .42 or possibly .424 as the initial info stating .42 may have been rounded off

q = .58

mean = np = 3x.42 = 1.26

Var = npq = 1.26 x .58 = .7308

sd = sqrVar = sqr.7308 = about .8549

now 3-1.6)/.8549 = z = 1.4/.8549 = about 1.64

then use z tables or a statistics calculator to find the probability = about .48 which is even more misleading as to the likelihood of 3 girls among 3 children

it's not a normal distribution or a binomial approximation to a normal

it's a very skewed distribution with P(3 girls in 3 children) = 1/6)^3 = 1/6^3 = about .0046 = less than 1/2 of 1% chance, so it is rather rare, unlike the normal & binomial calculations