CHAPTER 5 DISCRETE PROBABILITY DISTRIBUTIONS 1. 2.

CHAPTER 5

DISCRETE PROBABILITY DISTRIBUTIONS

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RANDOM VARIABLESTila College is a small private school with 2,500 students. According to the registrar’s office, the frequency and relative frequency distributions of the number of courses taken by all the students in Fall 2010 are shown in the table below.

Table 1: Frequency and Relative Frequency Distribution of the number of Courses Taken by Students

Number of Courses Frequency Relative Frequency

1 150 0.06

2 500 0.20

3 1000 0.40

4 600 0.24

5 250 0.10

f = 2500 = 1.003

Random VariablesNow, suppose we want to conduct an experiment by randomly selecting students from the population of 2,500 students.

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Definition1. The process of randomly selecting a

student from the population is called a random or a chance experiment.

2. Let x = the number of courses taken by the randomly selected student. Then x depends on the outcome of the experiment and can assume any values (1, 2, 3, 4, or 5). x is also called a random variable or a chance variable.

3. A random variable is a variable whose value is the outcome of a random or a chance experiment.

4. A random variable x can be either discrete random variable or continuous random variable.

Discrete Random VariableDefinition A discrete random variable is a random variable that

assumes countable values.

In our example, the number of courses taken by students in Fall 2010 is a discrete random variable because it can assume any value, 1, 2, 3, 4, or 5.

Other examples of discrete random variables are:

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The number of smokers in STA 12 class The number of cars owned by families in your city The number of fish caught on a fishing trip

Continuous Random VariableDefinition A continuous random variable is a random variable

that assumes any values within one or more intervals. Suppose the volume of water used per month per

household in Victorville ranges from 2,000 to 4,000 gallons.

Let’s randomly select a household in the city of Victorville. Let x =the volume of water used per month by the randomly

selected household. Since there are infinitely many values between 2,000 and

4,000 gallons, the random variable, x, is a continuous random variable and can assume infinite number of values.

Graphically,

62,000 4,000x can assume any value on this number line

Discrete and Continuous Random VariablesExample #1

a. The time left on a parking meterb. The number of bats broken by a major

league baseball team in a seasonc. The number of cars in a parking lotd. The total pounds of fish caught on a

fishing tripe. The number of cars crossing a bridge on

a given dayf. The time spent by a physician

examining a patient

Classify each of the following random variables as discrete or continuous.

a. Continuousb. Discrete

c. Discreted. Continuous

e. Discrete

f. Continuous

Solution

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PROBABLITY DISTRIBUTION OF A DISCRETE RANDOM VARIABLE

DefinitionA probability distribution of a discrete random variable lists all possible values that the random variable can assume with their corresponding probabilities.

As discussed in Section 5.1, let x = number of courses taken by

a randomly selected student.

Then, we can write the probability distribution of x as shown to the right.

Note that the probabilities are actual since the relative frequencies represent population. As stated in Chapter 4, if the relative frequencies were for a sample, then the probabilities would have been approximate probabilities.

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Probability Distribution of a Discrete Random Variable

A probability distribution of a discrete random variable must satisfy the following conditions:

a. The probability of each value of x is from 0 to 1.

b. The sum of the probabilities for all values of x is 1.

We can read the probability for any value of x as P(x = 2) = P(2) = 0.20P(x > 3) = P(4)+ P(5) =0.24 + 0.10 = 0.34P(x =< 3) = P(1) + P(2) + P(3) = .06 + .20

+ .40 = 0.66 A probability distribution of a discrete random

variable can be represented by,a. Graphb. Formulac. Table

1)( xP

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1)(0 xP


Example #2Each of the following tables lists certain values of x and their probabilities. Determine whether or not each one satisfies the two conditions required for a valid probability distribution. (1) (2) (3)

x P(x)

5 -.36

6 .48

7 .62

8 .26

x P(x)

1 .27

2 .24

3 .49

x P(x)

0 .15

1 .08

2 .20

3 .50

1. Although the sum of probabilities for all values of x is equal to 1, the probability of x = 5 is negative. Thus, this table does not represent a valid probability distribution.

2. Each probability ranges from 0 to 1, and the sum of the probabilities is 1.00. Thus, this table represents a valid probability distribution.

3. Although each probability lies between 0 and 1, the sum of the probabilities is less than 1.0. Thus, this table does not represent a valid probability distribution.

Solution

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Example #3The following table gives the probability distribution of a discrete random variable x.

Find the following probabilities:

From the given table,

Solution

x 0 1 2 3 4 5

P(x) .03 .17 .22 .31 .15 .12

a. P(x=1) b. P(x 1)c. P(x 3) d. P(0 x 2)e. Probability that x assumes a value < 3f. Probability that x assumes a value > 3g. Probability that x assumes a value in the interval 2 to 4

a. P(x=1) = 0.17b. P(x 1) = P(x = 0) + P(x = 1) = .03 + .17 = .20c. P(x 3) = P(x = 3) + P(x = 4) + P(x = 5)

= .31 + .15 + .12 = .58d. P(0 x 2) = P(x = 0) + P(x = 1) + P(

x = 2) = .03 + .17 + .22 = .42e. P(x < 3) = P(0) + P(1) + P(2) = 0.42f. P(x > 3) = P(x = 4) + P(x = 5) = 0.27g. P(2 x 4) = P(x = 2) + P(x = 3) + P(x = 4) = .22 + .31 + .15 = .68

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Example #4The following table shows the probability distribution for the sum, denoted by x, of the faces on a pair of Nathan’s dice:

Solution

x 2 3 4 5 6 7 8 9 10 11 12

P(x) .065 .065 .08 .095 .11 .17 .11 .095 .08 .065 .065

i. P(x=even number) = P(x=2)+P(x=4)+P(x=6)+P(x=8)+P(x=10)+P(x=12)= .51

ii. P(x=7 or x=11) = P(x=7) + P(x=11) = .17 + .065 = .235

iii. P(x=4 to x=6) = P(x=4)+P(x=5)+P(x=6) = .08 + .095 + .11

= .285

iv. P(x 9) = P(x=9) + P(x=10) + P(x=11) + P(x=12) = .095 + .08 + .065 + .065 = .305

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

2 3 4 5 6 7 8 9 10 11 12

a. Draw a bar graph for this probabilityb. Determine the probability that the sum of the faces on a single roll of Nathan’s dice

is: (i) an even number (ii) 7 or 11 (iii) 4 to 6 (iv) no less than 9

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Example #5According to a survey, 30% of adults are against using animals for research. Assume that this result holds true for the current population of all adults. Let x be the number of adults who are against using animals for research in a random sample of two adults. Obtain the probability distribution for x. Draw a tree diagram for this problem.

x can assume 0, 1, or 2. Thus, the probability distribution is

Solution

x P(x)

0 P(F) P(F)=.7 (.7) = 0.49

1 P(F)P(A)+P(A)P(F)= .7(.3)+.3(.7) =.42

2 P(A) P(A)=.3(.3) = .09

A

F

A

F

A

F

.30

.70

.30

.70

.30

.70

P(AA) = .09

P(AF) = .21

P(FA) = .21

P(FF) = .4913

Let A = against animal research

F = for animal research

MEAN OF A DISCRETE RANDOM VARIABLE

The mean of a discrete variable is denoted by µ and is define as the sum of the product of each value of the discrete variable and the corresponding probabilities. It is also called an expected value and denoted by E(x)

Formal Definition:The mean of a discrete random variable x is the value that is expected to occur per repetition, on average, if an experiment is repeated a large number of times. It is calculated as

µ = E(x) = Σ x P(x)

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STANDARD DEVIATION OF A DISCRETE RANDOM VARIABLE

DefinitionThe standard deviation of a discrete random variable is denoted by and is define as the measure of the spread of its probability distribution. It is computed as,

Basic Formula Short-Cut Formula

Variance

Standard deviation

22 ( )x P x 2 2 2( )x P x

2 ( )x P x

Note A higher value of signifies that x can assume values over a larger range

about the mean

A smaller value of signifies that x can assume values that are closely clustered about the mean.

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2 2( )x P x

Standard Deviation of a Discrete Random VariableExample #7The following table gives the probability distribution of the number of camcorder sold on a given day at an electronic store

Solution

Camcorder sold 0 1 2 3 4 5 6

Probability .05 .12 .19 .30 .20 .10 .04

Calculate the standard deviation for this probability distribution.

X 0 1 2 3 4 5 6

P(x) .05 .12 .19 .30 .20 .10 .04

xP(x) 0 .12 .38 .90 .80 .50 .24

X2 0 1 4 9 16 25 36

x2P(x) 0 .12 .76 2.7 3.2 2.5 1.44

xP(x)=2.94

2x P(x)=10.72

2 2 2( ) 10.72 (2.94) 1.4410x P x16

Standard Deviation of a Discrete Random VariableExample #8

An instant lottery ticket costs $2. Out of a total of 10,000 tickets printed for this lottery, 1000 tickets contain a price of $5 each, 100 tickets have a price of $10 each, and 5 tickets have a price of $1000 each, and 1 ticket has a price of $5000. Let x be the random variable that denotes the net amount a player wins by playing this lottery. Write the probability distribution of x. Determine the mean and standard deviation of x. How will you interpret the values of the mean and standard deviation.

Solution x f P(x) xP(x) X2 x2P(x)

-2 8894 .8894 -1.7788

4 3.5576

3 1000 .1000 0.3000 9 0.9000

8 100 0.0100 0.0800 64 0.6400

998 5 0.0005 0.4990 996004 498.002

4998 1 0.0001 0.4998 24980004 2498.0004

xP(x)=-0.402x P(x)=3001.1

2 2

2

( )

3001.1 ( .40)

$54.78

x P x

0.40 ( )E x

Players are expected to lose, on an average, 40 cents per ticket with a standard deviation of $54.78. 17

Identifying Unusual ResultsRange Rule of Thumb

According to the range rule of thumb, most values should lie within 2 standard deviations of the mean.

We can therefore identify “unusual” values by determining if they lie outside these limits:

Maximum usual value =

Minimum usual value =

2

2

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Example – continued

We found for families with two children, the mean number of girls is 1.0 and the standard deviation is 0.7 girls.

Use those values to find the maximum and minimum usual values for the number of girls.

maximum usual value 2 1.0 2 0.7 2.4

minimum usual value 2 1.0 2 0.7 0.4

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Identifying Unusual ResultsProbabilities

Rare Event Rule for Inferential Statistics

If, under a given assumption (such as the assumption that a coin is fair), the probability of a particular observed event (such as 992 heads in 1000 tosses of a coin) is extremely small, we conclude that the assumption is probably not correct.

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Using Probabilities to Determine When Results Are Unusual

Unusually high: x successes among n trials is an unusually high number of successes if .

Unusually low: x successes among n trials is an unusually low number of successes if .

( or more) 0.05P x

( or fewer) 0.05P x

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Suppose you were tossing a coin to determine whether it favors heads, an supposed 1000 tosses resulted in 501 heads.

a)P(exactly 501 heads in 1000 tosses)=0.0252

Notice 0.0252<0.05 so, is very unlikely.

b) p(501 or more heads in 1000 tosses)=0.487

Since 0.487 in not greater than 0.05 then is not unusually high

CHAPTER 5 DISCRETE PROBABILITY DISTRIBUTIONS 1. 2.

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