Question

according to government data, the probability that a woman between the ages of 25 and 29 was never married is 40%. in a random survey of 10 women in this age group, what is the mean and standard deviation of the number that never married?
a. mean: 4; standard deviation: 1.55
b. mean: 6; standard deviation: 155
c. mean: 4; standard deviation: 2.4
d. mean: 6; standard deviation: 1.55

Explanation:

Step1: Calculate the mean

For a binomial distribution \(X\sim B(n,p)\), the mean is \(\mu = np\). Here, \(n = 10\) (number of trials, i.e., number of women surveyed) and \(p=0.4\) (probability of success, i.e., probability that a woman was never married). So, \(\mu=np = 10\times0.4=4\).

Step2: Calculate the standard deviation

The formula for the standard deviation of a binomial distribution is \(\sigma=\sqrt{np(1 - p)}\). Substitute \(n = 10\) and \(p = 0.4\) into the formula. First, calculate \(1-p=1 - 0.4=0.6\). Then \(np(1 - p)=10\times0.4\times0.6 = 2.4\). And \(\sigma=\sqrt{2.4}\approx1.55\).

Answer:

A. mean: 4; standard deviation: 1.55