Example 4.20 from Statistics, 4.4 Geometric Distribution (Optional)

Geometric Distribution (Optional): example 4.20

The lifetime risk of developing pancreatic cancer is about one in 78 (1.28 percent). Let X = the number of people you ask until one says he or she has pancreatic cancer. Then X is a discrete random variable with a geometric distribution: X ~ G $(\frac{1}{78})$ or X ~ G(.0128).

What is the probability that you ask 10 people before one says he or she has pancreatic cancer?
What is the probability that you must ask 20 people?
Find the (i) mean and (ii) standard deviation of X.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

1
Solution
1. P(x = 10) = geometpdf(.0128, 10) = .0114
2. P(x = 20) = geometpdf(.0128, 20) = .01
3. 1. Mean = μ = $\frac{1}{p}$ = $\frac{1}{.0128}$ = 78
  2. $σ = \sqrt{σ^{2}} = \sqrt{(\frac{1}{p}) (\frac{1}{p} - 1)} = \sqrt{(\frac{1}{.0128}) (\frac{1}{.0128} - 1)} = \sqrt{(78) (78 - 1)} = \sqrt{6,006} = 77.4984 \approx 77$
    The number of people whom you would expect to ask until one says he or she has pancreatic cancer is 78. And you expect that to vary by about 77 people on average.

How did it go?

Solution

Solution