4.4 Geometric distribution  (Page 2/13)

Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says he or she lives within five miles of you. What is the probability that you need to contact four people?

This is a geometric problem because you may have a number of failures before you have the one success you desire. Also, the probability of a success stays the same each time you ask a student if he or she lives within five miles of you. There is no definite number of trials (number of times you ask a student).

Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective?

Let X = the number of computer components tested until the first defect is found.

X takes on the values 1, 2, 3, ... where p = 0.02. X ~ G(0.02)

Find P ( x = 7). P ( x = 7) = 0.0177.

To find the probability that x ≤ 7, follow the same instructions EXCEPT select E:geometcdf(as the distribution function.

The probability that the seventh component is the first defect is 0.0177.

The graph of X ~ G(0.02) is:

The y -axis contains the probability of x , where X = the number of computer components tested.

The number of components that you would expect to test until you find the first defective one is the mean, $\mu \text{ = 50}$ .

The formula for the mean is μ = $\frac{1}{p}$ = $\frac{1}{0.02}$ = 50

The formula for the variance is σ ² = $\left(\frac{1}{p}\right)\left(\frac{1}{p}-1\right)$ = $\left(\frac{1}{0.02}\right)\left(\frac{1}{0.02}-1\right)$ = 2,450

The standard deviation is σ = $\sqrt{\left(\frac{1}{p}\right)\left(\frac{1}{p}-1\right)}$ = $\sqrt{\left(\frac{1}{0.\text{02}}\right)\left(\frac{1}{0.\text{02}}-1\right)}$ = 49.5