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Question

A researcher reports that when groups of 4 children are randomly selected from a population of couples meeting certain criteria, the probability distribution for the number of girls is as given in the accompanying table

X P(x)
0 0.502
1 0.365
2 0.098
3 0.011
4 0.001


Determine whether a probability distribution is given, if one isn’t described, identify the requirements that are not satisfied. In those cases where a probably distribution is described, find its mean and stand deviation.

I think the mean is 2 and stand dev. is 1.414.

Answer 1

All the possible cases are discussed. The only issue possible is that the probabilities don't add to 1. They come just short.

If it is still a probability distribution, multiply the x by P(x) to get the mean.

1 * .365 + 2 * .098 + 3 * .011 + 4 * .001 = 0.598

The standard dev can be found once you get the variance. An easy way to get the variance is to take x^2 * P(x) and sum those, and then subtract your mean squared...
var(X) = E(X^2) - [E(x)]^2

E(X^2) = 1^2 * .365 + 2^2 * .098 + 3^2 * .011 + 4^2 * .001 =
= 0.872

var(x) = 0.872 - (0.598)^2
= 0.156792

Standard dev = 0.396 (roughly)

I hope the explanation helped! I'm 100% sure about the methodology!