Chapter Discrete Probability Distributions © 2010 Pearson Prentice Hall. All rights reserved 3.

Chapter

Discrete Probability Distributions

© 2010 Pearson Prentice Hall. All rights reserved

3

6-2© 2010 Pearson Prentice Hall. All rights reserved

Section 6.1 Probability Rules

A random variable is a numerical measure of the outcome from a probability experiment, so its value is determined by chance. Random variables are denoted using letters such as X.


A discrete random variable has either a finite or countable number of values. The values of a discrete random variable can be plotted on a number line with space between each point. See the figure.


A continuous random variable has infinitely many values. The values of a continuous random variable can be plotted on a line in an uninterrupted fashion. See the figure.


Determine whether the following random variables are discrete or continuous. State possible values for the random variable.

(a)The number of light bulbs that burn out in a room of 10 light bulbs in the next year.

(b) The number of leaves on a randomly selected Oak tree.

(c) The length of time between calls to 911.

EXAMPLE Distinguishing Between Discrete and Continuous Random Variables

EXAMPLE Distinguishing Between Discrete and Continuous Random Variables

Discrete; x = 0, 1, 2, …, 10

Discrete; x = 0, 1, 2, …

Continuous; t > 0


A probability distribution provides the possible values of the random variable X and their corresponding probabilities. A probability distribution can be in the form of a table, graph or mathematical formula.


The table to the right shows the probability distribution for the random variable X, where X represents the number of DVDs a person rents from a video store during a single visit.

x P(x)0 0.061 0.582 0.223 0.104 0.035 0.01

EXAMPLE A Discrete Probability DistributionEXAMPLE A Discrete Probability Distribution


EXAMPLE Identifying Probability DistributionsEXAMPLE Identifying Probability Distributions

x P(x)0 0.161 0.182 0.223 0.104 0.305 0.01

Is the following a probability distribution?



x P(x)0 0.161 0.182 0.223 0.104 0.305 -0.01




x P(x)0 0.161 0.182 0.223 0.104 0.305 0.04



A probability histogram is a histogram in which the horizontal axis corresponds to the value of the random variable and the vertical axis represents the probability of that value of the random variable.


DVDs Rented at a Video Store

0

0.1

0.2

0.30.4

0.5

0.6

0.7

0 1 2 3 4 5

Number of DVDs Rented

Pro

ba

bili

ty

Draw a probability histogram of the probability distribution to the right, which represents the number of DVDs a person rents from a video store during a single visit.

EXAMPLE Drawing a Probability HistogramEXAMPLE Drawing a Probability Histogram

x P(x)

0 0.06

1 0.58

2 0.22

3 0.10

4 0.03

5 0.01


Compute the mean of the probability distribution to the right, which represents the number of DVDs a person rents from a video store during a single visit.

EXAMPLE Computing the Mean of a Discrete Random Variable

EXAMPLE Computing the Mean of a Discrete Random Variable

x P(x)

0 0.06

1 0.58

2 0.22

3 0.10

4 0.03

5 0.01

( )X x P x 0(0.06) 1(0.58) 2(0.22) 3(0.10) 4(0.03) 5(0.01) 1.49


The following data represent the number of DVDs rented by 100 randomly selected customers in a single visit. Compute the mean number of DVDs rented.


49.1100

... 10021

xxx

X


As the number of trials of the experiment increases, the mean number of rentals approaches the mean of the probability distribution.


Because the mean of a random variable represents what we would expect to happen in the long run, it is also called the expected value, E(X), of the random variable.


EXAMPLE Computing the Expected Value of a Discrete Random Variable

EXAMPLE Computing the Expected Value of a Discrete Random Variable

A term life insurance policy will pay a beneficiary a certain sum of money upon the death of the policy holder. These policies have premiums that must be paid annually. Suppose a life insurance company sells a $250,000 one year term life insurance policy to a 49-year-old female for $530. According to the National Vital Statistics Report, Vol. 47, No. 28, the probability the female will survive the year is 0.99791. Compute the expected value of this policy to the insurance company.

x P(x)

530 0.99791

530 – 250,000 = -249,470

0.00209

Survives

Does not survive

E(X) = 530(0.99791) + (-249,470)(0.00209)

= $7.50


x P(x)0 0.06 -1.43 2.0449 0.1226941 0.58 -0.91 0.8281 0.4802982 0.22 -1.27 1.6129 0.3548383 0.1 -1.39 1.9321 0.193214 0.03 -1.46 2.1316 0.0639485 0.01 -1.48 2.1904 0.021904

Compute the variance and standard deviation of the following probability distribution which represents the number of DVDs a person rents from a video store during a single visit.

EXAMPLE Computing the Variance and Standard Deviation of a Discrete Random Variable

EXAMPLE Computing the Variance and Standard Deviation of a Discrete Random Variable

x P(x)

0 0.06

1 0.58

2 0.22

3 0.10

4 0.03

5 0.01

Xx 2Xx

22 ( )

1.236892X Xx P x

2 ( )Xx P x

1.236892

1.11X


Chapter Discrete Probability Distributions © 2010 Pearson Prentice Hall. All rights reserved 3.

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Transcript of Chapter Discrete Probability Distributions © 2010 Pearson Prentice Hall. All rights reserved 3.