Question

The probability that a woman between the ages of 25 and 29 never married is 40%. i.e. p = 0.40 and q = 1 - 0.40 = 0.60
Given n = 10. The given data follows a Binomial distribution.

Mean = np = 10*0.40 = 4

SD = ?variance = ?npq = ?10*0.4*0.6 =1.549

So, out of a sample of 10 random women in that age group, how many would be usual out of 10 to say never married? (Hint: use the two standard deviation interval)

Answer 1

Let X be the random variable denoting the number of women

out of 10 random women in that age group of 25-29, who

never married.

Hence, X ~ Bin(10, 0.4)

Thus, E(X), = np = 10 * 0.4 = 4, Var(X) = 10 * 0.4 * 0.6 = 2.4,

s.d.(X), = = = 1.5492.

Hence, the usual interval in which the number of women out of

10, who say never married is = [, ]

= [4 - 3.0984, 4 + 3.0984] = [0.9016, 7.0984].

If we approximate the above interval with integers, the usual

number should lie between 1 and 7. (Ans).