Question

The Statistical Bulletin published by Metropolitan Life Insurance Co. reported that 2% of all American births result in twins. If a random sample of 8000 births is taken, find the mean, variance, and standard deviation of the number of births that would be result in twins.

Answer 1

Answer:

Mean = 160 births

Variance = 156.8 births

Standard deviation = 12.522 births

Step-by-step explanation:

The mean number of births resulting in twins is given by the sample size (n=8000 births) multiplied by the probability of twins in any birth (p=2%).

$M = 8,000*0.02=160\ births$

In a binomial distribution like this one, the variance is given by:

$V=n*p*(1-p)\\V = 8,000*0.02*(1-0.02)=156.8\ births$

The standard deviation is simply the square root of the variance:

$SD = \sqrt V=\sqrt156.8\\SD=12.522\ births$