section 7.5: binomial and geometric distributions

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Section 7.5: Binomial and Geometric Distributions

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Page 1: Section 7.5: Binomial and Geometric Distributions

Section 7.5: Binomial and Geometric Distributions

Page 2: Section 7.5: Binomial and Geometric Distributions

Properties of a Binomial Experiment

1. The experiment consists of a fixed number of observations, called trials.

2. Each trial can result in one of only two mutually exclusive outcomes, labeled success (S) and failure (F).

3. Outcomes of different trials are independent.

4. The property that a trial results in a success is the same for each trial.

Page 3: Section 7.5: Binomial and Geometric Distributions

Binomial random variable x is defined as

x = the number of successes observed when the experiment is performed

The probability distribution of x is called the binomial probability distribution

Page 4: Section 7.5: Binomial and Geometric Distributions

Binomial Distributions

• Let n be the number of independent trials in a binomial experiment, and let π be the constant probability that any particular trial results in a success. Then

p(x) = P(x successes among n trials)

nxxnx

n xnx ,...,2,1,0)1()!(!

!

Page 5: Section 7.5: Binomial and Geometric Distributions

Example

• Sixty percent of all computer monitors sold by a large computer retailer have a flat panel display and 40% have a CRT display. The type of monitor purchased by each of the next 12 customers will be noted. Define a random variable x by x = number of monitors among these 12 that have a flat panel display

Page 6: Section 7.5: Binomial and Geometric Distributions

• Because x counts the number of flat panel displays, we use S to denote the sale of a flat panel monitor. Then x is a binomial random variable with n = 12 and π = P(S) = .60. The probability distribution of x is given by

12,...,2,1,0)4(.)60(.)!12(!

!12)(

x

xxxp xnx

Page 7: Section 7.5: Binomial and Geometric Distributions

The probability that exactly four monitors are flat panel displays is

042.

)4(.)6)(.495(

)4(.)6(.!8!4

!12

)4()4(

84

84

xPp

Page 8: Section 7.5: Binomial and Geometric Distributions

The mean value and the standard deviation of a binomial random variable are

)1( nandn xx

Page 9: Section 7.5: Binomial and Geometric Distributions

Example

• Newsweek reported that one-third of all credit card users pay their bills in full each month. This figure is, of course, an average across different cards and issuers. Suppose that 30% of all individuals hold Visa cards issued by a certain bank pay in full each month. A random sample of n = 25 cardholders is to be selected. The bank is interested in the variable x = number in the sample who pay in full each month. We can use a binomial distribution with n = 25 and π =.3. We have defined “paid in full” as a success because this is the outcome counted by the random variable x.

Page 10: Section 7.5: Binomial and Geometric Distributions

• The mean value is nπ = 25(.30) = 7.5

• The standard deviation is the square root of 25(.30)(.70) = 5.25 = 2.29

Page 11: Section 7.5: Binomial and Geometric Distributions

Geometric Distributions

Suppose that an experiment consists of a sequence of trials with the following conditions:

1. The trials are independent2. Each trial can result in one of two possible outcomes, success or failure3. The probability of success is the same for all trials

Page 12: Section 7.5: Binomial and Geometric Distributions

• A geometric random variable x is defined as

x = the number of trials until the first success is observed (including the success trial)

The probability distribution of x is called the geometric probability distribution.

Page 13: Section 7.5: Binomial and Geometric Distributions

• Geometric Probability Distribution

If x is a geometric random variable with probability of success = π for each trial, then

p(x) = (1 – π)x-1π x = 1, 2, 3, …

Page 14: Section 7.5: Binomial and Geometric Distributions

Example

• For this problem, π = .4. The probability distribution of x = number of students who must be stopped before finding a student with jumper cables is

p(x) = (.6)x-1(.4) x = 1,2,3,…The probability that three or fewer students must

be stopped is P(x ≤ 3)=p(1) + p(2) + p(3)=(.6)0(.4) + (.6)1(.4) + (.6)2(.4)=.4 + .24 + .144=.784

Page 15: Section 7.5: Binomial and Geometric Distributions

Try This One

• Exit polling has been a controversial practice in recent elections. Suppose that 90% of all registered voters favor banning the release of information from exit polls in presidential elections until after the polls in CA close. A random sample of 25 CA voters is to be selected.

a. What is the probability that more than 20 voters favor the ban?

b. What is the probability that at least 20 voters favor the ban?

c. What are the mean and standard deviation of the number of voters who favor the ban?

Page 16: Section 7.5: Binomial and Geometric Distributions

a. P(x > 20) = .138 + .227 + .266 + .199 + .072 = .902

b. P(x ≥ 20) = .902 + .065 = .967

c. Mean = (25)(.90) = 22.5

Standard Deviation = square root(25(.90)(1-.90) = 1.5