### Notation for the Geometric: G = Geometric Probability Distribution Function

*X* ~ *G*(*p*)

Read this as *X is a random variable with a geometric distribution*. The parameter is *p*; *p* = the probability of a success for each trial.

### Example 4.19

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* = .02. *X* ~ *G*(.02)

Find *P*(*x* = 7). There is a formula to define the probability of a geometric distribution $P(x)$. We can use the formula to find $P(x=7)$. But since the calculation is tedious and time consuming, people usually use a graphing calculator or software to get the answer. Using a graphing calculator, you can get $P(x=7)=.0177$. The instruction of TI83, 83+, 84, 84+ is given below.

### Using the TI-83, 83+, 84, 84+ Calculator

Go into 2nd DISTR. The syntax for the instructions are as follows:

**To calculate the probability of a value P(x = value): use geometpdf(p, number)**. Here geometpdf represents geometric probability density function. It is used to find the probability that a geometric random variable is equal to an exact value. p is the probability of a success and number is the value.

**To calculate the cumulative probability P(x ≤ value): use geometcdf(p, number)**. Here geometcdf represents geometric cumulative distribution function. It is used to determine the probability of “at most” type of problem, the probability that a geometric random variable is less than or equal to a value. p is the probability of a success and number is the value.

To find $P(x=7)$, enter 2nd DISTR, arrow down to geometpdf(. Press ENTER. Enter .02,7). The result is $P(x=7)=.0177$.

If we need to find $P(x\le 7)$ enter 2nd DISTR, arrow down to geometcdf(. Press ENTER. Enter .02,7). The result is $(x\le =7)=.1319$.

The graph of *X* ~ *G*(.02) is

The previous probability distribution histogram gives all the probabilities of *X*. The *x*-axis of each bar is the value of *X* = the number of computer components tested until the first defect is found, and the height of that bar is the probability of that value occurring. For example, the *x* value of the first bar is 1 and the height of the first bar is 0.02. That means the probability that the first computer components tested is defective is .02.

The expected value or mean of X is $E(X)=\mu =\frac{1}{p}=\frac{1}{.02}=50$.

The variance of X is ${\sigma}^{2}=(\frac{1}{p})(\frac{1}{p}-1)=(\frac{1}{.02})(\frac{1}{.02}-1)=(50)(49)=\mathrm{2,450}$.

The standard deviation of X is $\sigma =\sqrt{{\sigma}_{}^{2}}=\sqrt{\mathrm{2,450}}=49.5$.

Here is how we interpret the mean and standard deviation. The number of components that you would expect to test until you find the first defective one is 50 (which is the mean). And you expect that to vary by about 50 computer components (which is the standard deviation) on average.

The probability of a defective steel rod is .01. Steel rods are selected at random. Find the probability that the first defect occurs on the ninth steel rod. Use the TI-83+ or TI-84 calculator to find the answer.

### 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*$\left(\frac{1}{78}\right)$ 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*.

*P*(*x*= 10) = geometpdf(.0128, 10) = .0114*P*(*x*= 20) = geometpdf(.0128, 20) = .01-
- Mean =
*μ*= $\frac{1}{p}$ = $\frac{1}{.0128}$ = 78 - $\sigma =\sqrt{{\sigma}^{2}}=\sqrt{\left(\frac{1}{p}\right)\left(\frac{1}{p}-1\right)}=\sqrt{\left(\frac{1}{.0128}\right)\left(\frac{1}{.0128}-1\right)}=\sqrt{\left(78\right)\left(78-1\right)}=\sqrt{\mathrm{6,006}}=77.4984\approx 77$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.

- Mean =

The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate for women in Afghanistan is 12 percent. Let *X* = the number of Afghani women you ask until one says that she is literate.

- What is the probability distribution of
*X*? - What is the probability that you ask five women before one says she is literate?
- What is the probability that you must ask 10 women?
- Find the (i) mean and (ii) standard deviation of
*X*.