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

Source: U.S. Census Bureau, 2010 American Community Survey, Marital Status

Let’s treat the marital status (married or not married) of a randomly selected woman as a Bernoulli random variable with:

\[P(\text{married}) = p = 0.471\] \[P(\text{not married}) = 1-p = 0.529\]

Assuming that each Bernoulli random variable is independent and identically distributed, then the wait time until randomly selecting a success (marital status = married) can be described by the geometric distribution.

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

\[P(\text{1st=not married, 2nd=not married, 3rd=married})\] \[= P(\text{not married}) \cdot P(\text{not married}) \cdot P(\text{married})\] \[= (1-p)^2 \cdot p = 0.132\]

What is the probability that all three randomly selected women are married?

\[P(\text{1st, 2nd, 3rd = married}) = P(\text{married})^3\] \[= p^3 = 0.104\]

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

\[E[\text{# trials before success}] = \mu = \frac{1}{p} = 2.12\] \[\text{variance} = \sigma^2 = \frac{1 - p}{p^2} = 2.38\] \[\text{standard deviation} = \sigma = \sqrt{\text{variance}} = 1.54\]

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?

Now \(p = 0.30\) and \(1 - p = 0.70\).

Then

\[\mu = \frac{1}{p} = 3.33\] \[\sigma = \sqrt{\frac{1-p}{p^2}} = 2.79\]

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?

\[\text{ Lower } P(\text{event}) \implies \text{ Lower } Freq(\text{event})\] This means both

\[\text{Longer wait time for the event} \implies \text{ Higher } \mu\] and

\[\text{ Greater variability in wait times } \implies \text{ Higher } \sigma\]

So decreasing the probability of an event increases the mean and standard deviation of the wait time until success.