Standard Deviation Of Geometric Distribution

Introduction

There are three main characteristics of a geometric experiment.

Repeating independent Bernoulli trials until a success is obtained. Recall that a Bernoulli trial is a binomial experiment with number of trials n = 1. In other words, you keep repeating what yous are doing until the first success. Then you stop. For example, you throw a sprint at a bull's-centre until you hit the bull's-centre. The outset time y'all hit the balderdash's-centre is a success, and then y'all stop throwing the dart. It might take half dozen tries until you hit the balderdash'south-eye. You lot can think of the trials as failure, failure, failure, failure, failure, success, stop.
In theory, the number of trials could continue forever. There must be at least 1 trial.
The probability, p, of a success and the probability, q, of a failure practise non change from trial to trial. p + q = 1 and q = 1 − p. For example, the probability of rolling a three when you throw one fair die is $\frac{one}{6}$ . This is truthful no matter how many times you lot roll the dice. Suppose y'all want to know the probability of getting the kickoff three on the fifth curlicue. On rolls one through iv, you practise not get a face up with a iii. The probability for each of the rolls is q = $\frac{v}{6}$ , the probability of a failure. The probability of getting a three on the fifth roll is $(\frac{5}{half-dozen}) (\frac{5}{6}) (\frac{5}{6}) (\frac{v}{6}) (\frac{1}{6})$ = .0804.

10 = the number of independent trials until the first success.

p = the probability of a success, q = one – p = the probability of a failure.

In that location are shortcut formulas for calculating mean μ, variance σ ^two, and standard deviation σ of a geometric probability distribution. The formulas are given equally below. The deriving of these formulas volition non be discussed in this book.

$μ = \frac{1}{p}, σ^{2} = (\frac{one}{p}) (\frac{i}{p} - 1), σ = \sqrt{(\frac{1}{p}) (\frac{i}{p} - 1)}$

Instance 4.16

Suppose a game has ii outcomes, win or lose. Yous repeatedly play that game until you lot lose. The probability of losing is p = 0.57.

If we let X = the number of games you play until you lose (includes the losing game), then 10 is a geometric random variable. All iii characteristics are met. Each game you lot play is a Bernoulli trial, either win or lose. You would need to play at least one game before yous stop. X takes on the values 1, 2, three, . . . (could go on indefinitely). Since we are measuring the number of games y'all play until you lose, nosotros ascertain a success as losing a game and a failure every bit winning a game. The probability of a success $p = .57$ and the probability of a failure q= 1-p = ane–0.57 = 0.43. Both p and q remain the same from game to game.

If we want to discover the probability that it takes five games until yous lose, and then the probability could be written every bit P(x = 5). Nosotros will explain how to find a geometric probability later in this section.

Effort It 4.16

You throw darts at a board until y'all hitting the centre area. Your probability of hit the center area is p = 0.17. You desire to find the probability that information technology takes 8 throws until you hit the center. What values does 10 have on?

Instance 4.17

A rubber engineer feels that 35 percent of all industrial accidents in her constitute are caused by failure of employees to follow instructions. She decides to look at the accident reports (selected randomly and replaced in the pile after reading) until she finds one that shows an accident caused by failure of employees to follow instructions.

If we permit X = the number of accidents the safety engineer must examine until she finds a report showing an accident caused by employee failure to follow instructions, then Ten is a geometric random variable. All iii characteristics are met. Each accident report she reads is a Bernoulli trial: the blow was either acquired past failure of employees to follow instructions or not. She would need to read at least one accident study earlier she stops. X takes on the values 1, two, 3, . . . (could go on indefinitely). Since we are measuring the number of reports she needs to read until one that shows an accident acquired by failure of employees to follow instructions, we define a success as an accident acquired past failure of employees to follow instructions. If an accident was acquired by another reason, the report is defined as a failure. The probability of a success p = .35 and the probability of a failure $q = 1 - p = 1 - .35 = .65$ . Both p and q remain the same from written report to report.

If we want to find the probability that the safety engineer volition accept to examine at least iii reports until she finds a study showing an blow caused by employee failure to follow instructions, then the probability could be written every bit $p = .35$ . If nosotros want to find how many reports, on average, the safety engineer would expect to wait at until she finds a report showing an accident acquired by employee failure to follow instructions, we need to detect the expected value E(x). We will explicate how to solve these questions later on in this department.

Try It iv.17

An teacher feels that 15 pct of students become below a C on their final examination. She decides to look at concluding exams (selected randomly and replaced in the pile later reading) until she finds one that shows a grade below a C. We want to know the probability that the teacher volition have to examine at least 10 exams until she finds one with a grade beneath a C. What is the probability question stated mathematically?

Example 4.18

Suppose that y'all are looking for a student at your higher who lives within v miles of you. You know that 55 percent of the 25,000 students do live within five miles of you. You randomly contact students from the college until ane 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 accept the one success you want. 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).

a. Allow X = the number of ________ you must inquire ________ one says yes.

Solution 4.18

a. Let X = the number of students you must enquire until i says yes.

b. What values does X take on?

Solution four.18

b. i, 2, 3, . . ., (total number of students)

c. What are p and q?

Solution 4.18

c. p = .five,q = .45

d. The probability question is P(_______).

Solution 4.eighteen

d. P(x = iv)

Try It four.xviii

You lot need to discover a store that carries a special printer ink. You know that of the stores that carry printer ink, 10 percent of them conduct the special ink. Yous randomly call each store until one has the ink you need. What are p and q?

Note for the Geometric: M = Geometric Probability Distribution Function

Annotation for the Geometric: Chiliad = Geometric Probability Distribution Function

X ~ Grand(p)

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

Example four.19

Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the kickoff defect is acquired by the 7th component tested. How many components do you expect to test until one is found to be defective?

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

X takes on the values ane, 2, 3, . . . where p = .02. X ~ Grand(.02)

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

Using the TI-83, 83+, 84, 84+ Computer

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

To calculate the probability of a value P(x = value): apply geometpdf(p, number). Here geometpdf represents geometric probability density function. Information technology 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 summate the cumulative probability P(x ≤ value): utilize geometcdf(p, number). Hither geometcdf represents geometric cumulative distribution function. Information technology is used to determine the probability of "at near" 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 2d DISTR, arrow downwards to geometpdf(. Press ENTER. Enter .02,vii). The issue is $P (x = 7) = .0177$ .

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

The graph of X ~ K(.02) is

This graph shows a geometric probability distribution. It consists of bars that peak at the left and slope downwards with each successive bar to the right. The values on the x-axis count the number of computer components tested until the defect is found. The y-axis is scaled from 0 to 0.02 in increments of 0.005.

Figure 4.2

The previous probability distribution histogram gives all the probabilities of 10. The x-axis of each bar is the value of X = the number of computer components tested until the beginning 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 outset bar is 0.02. That means the probability that the first reckoner components tested is defective is .02.

The expected value or mean of Ten is $E (10) = μ = \frac{1}{p} = \frac{one}{.02} = 50$ .

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

The standard departure of X is $σ = \sqrt{σ_{}^{2}} = \sqrt{ii,450} = 49.5$ .

Hither is how we interpret the mean and standard deviation. The number of components that y'all would expect to test until you observe the get-go lacking i is 50 (which is the mean). And you look that to vary by about 50 computer components (which is the standard deviation) on average.

Effort It 4.nineteen

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

Example 4.20

The lifetime adventure of developing pancreatic cancer is about ane in 78 (1.28 pct). Let 10 = the number of people you ask until one says he or she has pancreatic cancer. Then X is a detached random variable with a geometric distribution: X ~ G $(\frac{one}{78})$ or X ~ G(.0128).

What is the probability that you lot enquire 10 people earlier 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 departure of X.

Solution 4.twenty

P(x = x) = geometpdf(.0128, 10) = .0114
P(x = xx) = geometpdf(.0128, twenty) = .01
1. Mean = μ = $\frac{i}{p}$ = $\frac{1}{.0128}$ = 78
2. $σ = \sqrt{σ^{2}} = \sqrt{(\frac{i}{p}) (\frac{i}{p} - 1)} = \sqrt{(\frac{one}{.0128}) (\frac{1}{.0128} - 1)} = \sqrt{(78) (78 - one)} = \sqrt{half-dozen,006} = 77.4984 \approx 77$ The number of people whom you lot would look to inquire until one says he or she has pancreatic cancer is 78. And you lot wait that to vary by almost 77 people on average.

Try Information technology 4.20

The literacy rate for a nation measures the proportion of people age 15 and over who tin read and write. The literacy rate for women in Afghanistan is 12 per centum. Let 10 = the number of Afghani women you inquire until ane says that she is literate.

What is the probability distribution of Ten?
What is the probability that you lot inquire five women before one says she is literate?
What is the probability that you must ask 10 women?
Notice the (i) mean and (ii) standard deviation of Ten.