3.21 Married women.

The 2010 American Community Survey estimates that 47.1% of women ages 15 years and over are married.

  1. We randomly select three women between these ages. What is the probability that the third woman selected is the only one who is married?

Assuming the probability of each woman being married is independent, let \(X\) be the random variable having a geometric distribution where \(X\) is the location of the first success.

\[P(X = k) = (1-p)^{k-1}p\] \[P(X = 3) = (1-0.471)^{3-1}(0.471) \approx 0.1318\]

dgeom(2, 0.471) ## note that the R function takes the number of failures (k-1) as its input 
## [1] 0.1318051
  1. What is the probability that all three randomly selected women are married?

Again assuming independence, the probability that all three women are married may be calculated with the general multiplication rule.

\[P(\text{all three are married}) = (0.471)(0.471)(0.471) \approx 0.1045\]

  1. On average, how many women would you expect to sample before selecting a married woman? What is the standard deviation?

For a geometric distribution, the expected value is \(\frac{1}{p}\) and the standard deviation is \(\sqrt{\frac{1-p}{p^2}}\). Therefore:

\[\mu = \frac{1}{0.471} \approx 2.12\]

\[\sigma = \frac{1-0.471}{(0.471)^2} \approx 2.38\]

  1. If the proportion of married women was actually 30%, how many women would you expect to sample before selecting a married woman? What is the standard deviation?

\[\mu = \frac{1}{0.30} \approx 3.33\]

\[\sigma = \frac{1-0.30}{(0.30)^2} \approx 7.78\]

  1. Based on your answers to parts (c) and (d), how does decreasing the probability of an event affect the mean and standard deviation of the wait time until success?

For random variables with a geometric distribution, the mean wait time for the first success is (intuitively) longer for variables with smaller probability. The standard deviation is greater for the wait time of success as well.