Introduction to Business Statistics QM 120 Chapter 5 Discrete - - PowerPoint PPT Presentation

DEPARTMENT OF QUANTITATIVE METHODS & INFORMATION SYSTEMS

Introduction to Business Statistics QM 120 Chapter 5 Discrete random variables and their probability distribution distribution

• Dr. Mohammad Zainal

Spring 2008

Chapter 5: Random Variables

2

Random variables

2

# of PC’s owned Frequency Relative Frequency 120 .12 1 180 .18 2 470 .47 3 230 23 3 230 .23 N = 1000 Sum = 1.000

Let x denote the number of PCs owned by a family. Then x

can take any of the four possible values (0, 1, 2, and 3).

A random

variable (RV) is a variable whose value is determined by the outcome of a random experiment. y p

3

Chapter 5: Random Variables

3

Discrete random variable

A random variable that assumes countable values is called a

discrete random variable.

Examples of discrete RVs:

Number of cars sold at a dealership during a week Number of houses in a certain block Number of fish caught on a fishing trip

g g p

Number of costumers in a bank at any given day

Continuous random variable Continuous random variable

A random variable that can assume any value contained in

• ne or more intervals is called a continuous random variable.
• ne or more intervals is called a continuous random variable.

4

Chapter 5: Random Variables

4

Examples of continuous RVs:

Height of a person Height of a person Time taken to complete a test Weight of a fish

g

Price of a car

E l Cl if h f h f ll i RV di Example: Classify each of the following RVs as discrete or continuous.

Th b f t d t b k d i k

The number of new accounts opened at a bank during a week The time taken to run a marathon The price of a meal in fast food restaurant The price of a meal in fast food restaurant The score of a football game The weight of a parcel

5

Chapter 5: Probability Distribution of a Discrete RV

5

The probability distribution of a discrete RV lists all the

bl l h h RV d h possible value that the RV can assume and their corresponding probabilities. Example: Write the probability distribution of the number of PCs owned by a family.

# of PC’s

• wned

Frequency Relative Frequency

• wned

Frequency 120 .12 1 180 .18 2 470 .47 3 230 .23 N 1000 S 1 000 N = 1000 Sum = 1.000

6

Chapter 5: Probability Distribution of a Discrete RV

6

The following two characteristics must hold for any discrete

b b l d b probability distribution:

The probability assigned to each value of a RV x lies in the range 0

to 1; that is 0 ≤ P(x) ≤ 1 for each x. to 1; that is 0 ≤ P(x) ≤ 1 for each x.

The sum of the probabilities assigned to all possible values of x is

equal to 1.0; that is ΣP(x) = 1.

Example: Each of the following tables lists certain values of x and their probabilities. Determine whether or not each table represents a valid probability distribution.

x P(x) x P(x) x P(x) .08 1 .11 2 39 .25 1 .34 2 28 4 .2 5 .3 6 6 2 .39 3 .27 2 .28 3 .13 6 .6 8 ‐.1

7

Chapter 5: Probability Distribution of a Discrete RV

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Example: The following table lists the probability distribution of a discrete RV x.

6 5 4 3 2 1 x

a) P( 3) b) P( ≤ 2) ) P( ≥ 4) d) P(1 ≤ ≤ 4)

.06 .09 .12 .15 .28 .19 .11 P(x)

a) P(x = 3) b) P(x ≤ 2) c) P(x ≥ 4) d) P(1 ≤ x ≤ 4) e) Probability that x assumes a value less than 4 f) Probability that x assumes a value greater than 2 f) Probability that x assumes a value greater than 2 g) Probability that x assumes a value in the interval 2 to 5

8

Chapter 5: Probability Distribution of a Discrete RV

8

Example: For the following table

5 4 3 2 1

a) Construct a probability distribution table. Draw a graph of

5 4 3 2 1 x 12 16 24 20 8 P(x)

a) Construct a probability distribution table. Draw a graph of the probability distribution. b) Find the following probabilities

• i. P(x = 3)
• ii. P(x < 4)
• iii. P(x ≥ 3)
• iv. P(2 ≤ x ≤ 4)

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Chapter 5: Mean of a discrete RV

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Mean of a discrete RV

The mean μ ‐or expected value E(x)‐ of a discrete RV is the

value that you would expect to observe on average if the i t i t d i d i experiment is repeated again and again

It is denoted by

( ) ( ) E x xp x = ∑

Illustration: Let us toss two fair coins, and let x denote the

number of heads observed We should have the following

( ) ( ) E x xp x = ∑

number of heads observed. We should have the following probability distribution table

2 1 x

Suppose we repeat the experiment a large number of times, say n =4 000 We should expect to have approximately

1/4 1/2 1/4 P(x)

say n =4,000. We should expect to have approximately

10

Chapter 5: Mean of a discrete RV

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1 thousand zeros, 2 thousand ones, and 1 thousand twos. Then th l f ld l the average value of x would equal

Sum of measurements 1,000(0) 2,000(1) 1000(2) 4 000 + + = 4,000 1 1 1 (0) (1) (2) 4 4 2 n ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = + + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Similarly, if we use the , we would have

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

( ) ( ) E x xp x =∑ ( ) 0 (0) 1 (1) 2 (2) 0(1/ 4) 1(1/ 2) 2(1/ 4) E x P P P = + + = + + 1 =

11

Chapter 5: Mean of a discrete RV

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Example: Recall “ number of PC’s owned by a family” l F d h b f PC d b f l

• example. Find the mean number of PCs owned by a family.

12

Chapter 5: Mean of a discrete RV

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Example: In a lottery conducted to benefit the local fire company 8000 tickets are to be sold at $5 each The prize is a company, 8000 tickets are to be sold at $5 each. The prize is a $12,000. If you purchase two tickets, what is your expected gain?

Chapter 5: Mean of a discrete RV

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Example: Determine the annual premium for a $1000 insurance policy covering an event that over a long period of

13

time, has occurred at the rate of 2 times in 100. Let x : the yearly financial gain to the insurance company resulting from the sale of the policy resulting from the sale of the policy C : unknown premium Calculate the value of C such that the expected gain E(x) will Calculate the value of C such that the expected gain E(x) will equal to zero so that the company can add the administrative costs and profit.

Chapter 5: Mean of a discrete RV

14

Example: You can insure a $50,000 diamond for its total value by paying a premium of D dollars. If the probability of theft in

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y p y g p p y a given year is estimated to be .01, what premium should the insurance company charge if it wants the expected gain to equal $1000.

Chapter 5: Standard Deviation of a Discrete RV

15

The standard deviation of a discrete RV x, denoted by σ,

measures the spread of its probability distribution.

15

A higher value of σ indicates that x can assume values over

a larger range about the mean. While, a smaller value indicates that most of the values that can x assume are clustered closely around the mean. Th d d d i i b f d i h f ll i

The standard deviation σ can be found using the following

formula:

2 2

( ) x p x σ μ = −

∑

Hence, the variance σ2 can be obtained by squaring its

standard deviation σ

( ) x p x σ μ = −

∑

standard deviation σ.

Chapter 5: Standard Deviation of a Discrete RV

16

Example: Recall “ number of PC’s owned by a family”

• example. Find the standard deviation of PCs owned by a

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family.

Chapter 5: Standard Deviation of a Discrete RV

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Example: An electronic store sells a particular model of a computer notebook. There are only four notebooks in stocks,

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and the manager wonders what today’s demand for this particular model will be. She learns from marketing department that the probability distribution for x the daily department that the probability distribution for x, the daily demand for the laptop, is as shown in the table.

5 4 3 2 1 x

Find the mean, variance, and the standard deviation of x. Is it lik l th t fi t ill t t b l t ?

.05 .10 .15 .20 .40 P(x)

likely that five or more costumers will want to buy a laptop?

Chapter 5: Standard Deviation of a Discrete RV

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Example: A farmer will earn a profit of $30,000 in case of heavy rain next year, $60,000 in case of moderate rain, and

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$15,000 in case of little rain. A meteorologist forecasts that the probability is .35 for heavy rain, .40 for moderate rain, and .25 for little rain next year Let x be the RV that represents next for little rain next year. Let x be the RV that represents next year’s profits in thousands of dollars for this farmer. Write the probability distribution of x. Find the mean, variance, and the standard deviation of x.

Chapter 5: Standard Deviation of a Discrete RV

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Example: An instant lottery ticket costs $2. Out of a total of 10,000 tickets printed for this lottery, 1000 tickets contain a

19

prize of $5 each. 100 tickets contain a prize of $10 each, 5 tickets contain a prize of $1000 each, and 1 ticket has the prize

• f $5000 Let x be the RV that denotes the net amount player
• f $5000. Let x be the RV that denotes the net amount player

wins by playing this lottery. Write the probability distribution

• f x. Determine the mean and the standard deviation of x.

Chapter 5: Standard Deviation of a Discrete RV

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Example: Based in its analysis of future demand for its products, the financial department a company has determined

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that there is a .17 probability that the company will lose $1.2 million during the next year, a .21 probability that it will lose $ 7 million a 37 probability that it will make a profit of $0 9 $.7 million, a .37 probability that it will make a profit of $0.9 million, and a .25 probability that it will make a profit of $2.3 million. a) Let x be a RV that denotes the profit earned by this company during the next year. Write the probability distribution of x distribution of x.

Chapter 5: Standard Deviation of a Discrete RV

21

b) Find the mean and standard deviation of the probability of part a.

21

Chapter 5: Factorials & Combinations

22

Factorials

The symbol n!, reads as “n factorial,” represents the

22

y p product of all integers from n to 1. In other words,

n! = n(n ‐ 1)(n ‐ 2)(n ‐ 3)…3.2.1

Example: Evaluate 7!

7! = 7.6.5.4.3.2.1 = 5040

Example: Evaluate (12‐4)!

(12‐4)! = (8)! = 8.7.6.5.4.3.2.1 =40,320

Chapter 5: Factorials & Combinations

23

Example: Evaluate (8‐8)!

(8‐8)! = (0)! = 1

23

(8 8)! (0)! 1

Example: Find the value of 8! p By using factorial Table or your calculator. Locate 8 in the column labeled n. Then read the value for n! next to 8.

Chapter 5: Factorials & Combinations

24

Combinations

Combinations give the number of ways x elements can be

24

g y selected from n elements. The notation used to denote the total number of combinations is ) (n

x n x x n

C C = =

It can be found using the following formula

! n )! ( ! ! x n x n C C

x n n x

− = =

Remember: n is always greater than or at least equal to x.

Chapter 5: Factorials & Combinations

25

Example: Three members of a jury will be randomly selected from five people. How many different combinations are

25

possible?

Chapter 5: Factorials & Combinations

26

Using the table of combinations

Values of Table III in Appendix C lists the number of

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combinations of n elements selected x at a time. Example: Evaluate the followings

a) 5C3 b) 8C4 ) C c) 9C5 d) 5C5 e) 5C0 e) 5C0

Chapter 5: The Binomial Probability Distribution

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The most widely used discrete probability distribution.

27

• It is applied to find the probability that an event will occur

x times in n repetitions of an experiment (under certain conditions).

Suppose the probability that a VCR is defective at a factory

is .05. We can apply the binomial probability distribution to fi d th dd f tti tl d f ti VCR t th find the odds of getting exactly one defective VCR out three VCRs.

Chapter 5: The Binomial Probability Distribution

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Binomial experiment

An experiment that satisfies the following four conditions is

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p g called a binomial experiment.

There are n identical trials Each trail has only two possible outcomes. The probabilities of the two outcomes remain constant. The trials are independent The trials are independent.

A success does not mean that the corresponding outcome is A success does not mean that the corresponding outcome is

considered favorable. Similarly, a failure doesnʹt necessarily refer to unfavorable outcome.

Chapter 5: The Binomial Probability Distribution

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Example: Consider the experiment consisting of 10 tosses of a

• coin. Determine whether or not it is a binomial experiment.

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Solution:

Chapter 5: The Binomial Probability Distribution

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The binomial probability distribution and its formula.

The RV x that represents the number of successes in n trials

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p for a binomial experiment is called a binomial RV.

The probability distribution of x is called the binomial

distribution.

Consider the VCRs example. Let x be the number of

p defective VCRs in a sample of 3, x can assume any of the values 0, 1, 2, 3.

Chapter 5: The Binomial Probability Distribution

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For a binomial experiment, the probability of exactly x

successes in n trials is given by the binomial formula

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( ) !

x n x n x

P x C p q − =

where l b f i l

! !( )!

x n x

n p q x n x

−

= −

n = total number of trials, p = probability of success, q = probability of failure 1 – p q = probability of failure 1 – p, x = number of successes in n trials, n – x = number of failures in n trials

Chapter 5: The Binomial Probability Distribution

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Example: Find P(2) for a binomial RV with n = 5 and p =0.1

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Solution:

Chapter 5: The Binomial Probability Distribution

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Example: Over a long period of time it has been observed that a given sniper can hit a target on a single trial with a

33

probability = .8. Suppose he fires four shots at the target. a) What is the probability that he will hit the target exactly two times? b) What is the probability that he will hit the target at least b) What is the probability that he will hit the target at least

• nce?

Chapter 5: The Binomial Probability Distribution

34

Example: Five percent of all VCRs manufactured by a large factory are defective. A quality control inspector selects three

34

VCRs from the production line. What is the probability that exactly one of these three VCRs is defective

Solution:

Chapter 5: The Binomial Probability Distribution

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Example: At the Express Delivery Services (EDS), providing high‐quality service to its customers is the top priority of the

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• management. The company guarantees a refund of all charges

if a package is not delivered at its destination by the specified time It is known from the past that despite all efforts 2% of

• time. It is known from the past that despite all efforts, 2% of

the packages mailed through this company does not arrive on

• time. A corporation mailed 10 packages through EDS.

a) Find the probability that exactly 1 of these 10 packages will not arrive on time b) Find the probability that at most 1 of these 10 packages will not arrive on time not arrive on time.

Chapter 5: The Binomial Probability Distribution

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Example: According to the U.S. Bureau of Labor Statistics, 56%

• f mothers with children under 6 years of age work outside

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their homes. A random sample of 3 mothers with children under 6 years of age is selected. Let x denote the number of mothers who work outside their homes Write the probability mothers who work outside their homes. Write the probability distribution of x and draw a graph of the probability distribution.

Solution:

Chapter 5: The Binomial Probability Distribution

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Using the table of binomial probabilities

We can use the tables of binomial probabilities (Table IV)

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p given in Appendix C to calculate the required probabilities.

P

.95 … .2 .1 .05 x n .0500 … .8000 .9000 .9500 1 .9500 … .2000 .1000 .0500 1 .9500 … .2000 .1000 .0500 1 .0025 … .6400 .8100 .9025 2 .0950 … .3200 .1800 .0950 1 .9025 … .0400 .0100 .0025 2 .0001 … .5120 .7290 .8574 3 .0071 … .3840 .2430 .1354 1 .1354 … .0960 .0270 .0071 2 .8574 … .0080 .0010 .0001 3

Chapter 5: The Binomial Probability Distribution

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Example: Based on data from A peter D. Hart Research Associates’ poll on consumer buying habits and attitudes, It was

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estimated that 5% of American shoppers are status shoppers, that is, shoppers who love to buy designer labels. A random sample of eight American shoppers is selected Using Binomial sample of eight American shoppers is selected. Using Binomial Table, answer the following: a) Find the probability that exactly 3 shoppers are status ) p y y pp shoppers. b) Find the probability that at most 2 shoppers are status shoppers. pp

Chapter 5: The Binomial Probability Distribution

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c) Find the probability that at least 3 shoppers are status shoppers.

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d) Find the probability that 1 to 3 shoppers are status shoppers. e) Let x be the number of status shoppers, Write the probability distribution of x and draw a graph of this probability.

Chapter 5: The Binomial Probability Distribution

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Probability

• f

success and the shape

• f

the binomial distribution

40

distribution

For any number of trials n

The binomial probability distribution is symmetric if p = .5

p .95 .9 … .5 … .1 .05 x n .0000 .0001 … .0625 … .6561 .8145 4

0.25 0.3 0.35 0.4

.0005 .0036 … .2500 … .2916 .1715 1 .0135 .0486 … .3750 … .0486 .0135 2 .1715 .2916 … .2500 … .0036 .0005 3

0.05 0.1 0.15 0.2 P(x)

.8145 .6561 … .0625 … .0001 .0000 4

1 2 3 4 x

Chapter 5: The Binomial Probability Distribution

41 41

The binomial probability distribution is skewed to the right if p is less than .5

p

0.6 0.7

.95 .9 … .5 … .1 .05 x n .0000 .0001 … .0625 … .6561 .8145 4 .0005 .0036 … .2500 … .2916 .1715 1

0.2 0.3 0.4 0.5 P(x)

.0135 .0486 … .3750 … .0486 .0135 2 .1715 .2916 … .2500 … .0036 .0005 3 .8145 .6561 … .0625 … .0001 .0000 4

0.1 1 2 3 4 x

Chapter 5: The Binomial Probability Distribution

42 42

The binomial probability distribution is skewed to the left if p is greater than .5

p

0.6 0.7

.95 .9 … .5 … .1 .05 x n .0000 .0001 … .0625 … .6561 .8145 4 .0005 .0036 … .2500 … .2916 .1715 1

0.2 0.3 0.4 0.5 P(x)

.0135 .0486 … .3750 … .0486 .0135 2 .1715 .2916 … .2500 … .0036 .0005 3 .8145 .6561 … .0625 … .0001 .0000 4

0.1 0.2 1 2 3 4 x

Chapter 5: The Binomial Probability Distribution

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Mean and standard deviation of the binomial distribution

Although we can still use the mean and standard deviations

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g formulas learned in 5.3 and 5.4, it is more convenient and simpler to use the following formulas once the RV x is known to be a binomial RV known to be a binomial RV

and np npq μ σ = =

Chapter 5: The Binomial Probability Distribution

44

Example: Refer to the mothers with children under 6 years of age example, find the mean and the standard deviation of the

44

probability distribution.

Solution: Solution:

Chapter 5: The Binomial Probability Distribution

45

Example: Let x be a discrete RV that posses a binomial

• distribution. Using the binomial formula, find the following

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probabilities. f d a) P(x = 5) for n = 8 and p = .6 b) P(x = 3) for n = 4 and p = .3 c) P(x = 2) for n = 6 and p = .2 c) P(x 2) for n 6 and p .2 Verify your answer by using Table of Binomial Probabilities.

Chapter 5: The Binomial Probability Distribution

46

Example: Let x be a discrete RV that posses a binomial distribution.

46

a) Table of Binomial Probabilities, write the probability distribution for x for n = 7 and p = .3 and graph it. b) What are the mean and the standard deviation of the probability distribution developed in part a?

Chapter 5: The Binomial Probability Distribution

47 47

An experiment that satisfies the following four conditions is

called a binomial experiment.

There are n identical trials Each trail has only two possible outcomes. The probabilities of the two outcomes remain constant. The trials are independent.

Chapter 5: The Hypergeometric Probability Distribution

48

We learned that one of the conditions required to apply the

binomial distribution is that the trials are independent so

48

the probability of the two outcomes remain constant.

What if the probability of the outcomes is not constant? In such cases we replace the binomial distribution by the

hypergeometric probability distribution. yp g p y

Such a case occurs when a sample is drawn without Such a case occurs when a sample is drawn without

replacement from a finite population.

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Chapter 5: The Hypergeometric Probability Distribution

Hypergeometric probability distribution

Let N = total number of elements in the population

49

p p r = number of successes in the population N – r = number of failures in the population n = number of trials (sample size) b f i t i l x = number of successes in n trials n – x = number of failures in n trials Th d th i Th b bili f The mean and the variance are given by The probability of x successes in n trials is given by

r ⎛ ⎞ ⎜ ⎟

( )

r x N r n x N n

C C P x C

− −

=

2

r N r N n − − ⎛ ⎞⎛ ⎞⎛ ⎞ n N μ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠

2

1 r N r N n n N N N σ ⎛ ⎞⎛ ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟⎜ ⎟ − ⎝ ⎠⎝ ⎠⎝ ⎠

50

Chapter 5: The Hypergeometric Probability Distribution

Example: Dawn corporation has 12 employees who hold managerial positions. Of them, 7 are female and 5 are male.

50

The company is planning to send 3 of these 12 managers to a

• conference. If 3 mangers are randomly selected out of 12,

a) find the probability that all 3 of them are female

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Chapter 5: The Hypergeometric Probability Distribution

b) find the probability that at most 1 of them is a female

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Chapter 5: The Hypergeometric Probability Distribution

Example: A case of soda has 12 bottles, 3 of which contain diet

• soda. A sample of 4 bottles is randomly selected from the case

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a) find the probability distribution of x, the number of diet sodas in the sample

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Chapter 5: The Hypergeometric Probability Distribution

b) what are the mean and variance of x?

53

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Chapter 5: The Hypergeometric Probability Distribution

Example: GESCO Insurance company has prepared a final list

• f 8 candidates for 2 positions. Of the 8 candidates, 5 are

54

business majors and 3 are engineers. If the company manager decides to select randomly two candidates from this list, find the probability that the probability that a) both candidates are business majors b) neither of the two candidates is a business major ) j c) at most one of the candidates is a business major

Chapter 5: The Poisson Probability Distribution

55

The Poisson distribution is another discrete probability

distribution that has numerous practical applications.

55

It provides a good model for data that represent the number

• f occurrence of a specified event in a given unit of time or

space.

Here are some examples of experiments for which the RV x

d l d b h P i RV can modeled by the Poisson RV:

The number of phone calls received by an operator during a day

Th b f t i l t h k t t d i h

The number of customer arrivals at checkout counter during an hour The number of bacteria per a cm3 of a fluid The number of machine breakdowns during a given day The number of machine breakdowns during a given day The number of traffic accidents at a given time period

Chapter 5: The Poisson Probability Distribution

56

The

Poisson probability distribution is applied to experiments with random and independent occurrences

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The occurrences are random in the sense they do not follow any

pattern and, hence, the are unpredictable. I d d f th t th

Independence of occurrences means that the occurrence or

nonoccurrence of an event does not influence the successive

• ccurrence of that next event.

The occurrences are always considered with respect to an

interval. h l b l l

The interval may be a time interval, a space interval, or a

volume interval. If th b f (λ) f i i t l

If the average number of occurrences (λ) for a given interval

is known, then by using the Poisson probability we can compute the probability of a certain number of occurrences p p y x in that interval.

Chapter 5: The Poisson Probability Distribution

57

The following three conditions must be satisfied to apply

the Poisson probability distribution

57

x is a discrete RV. The occurrences are random. The occurrences are independent.

The Poisson probability formula is given by

( ) !

xe

P x x

λ

λ

−

=

where λ (pronounced lambda) is the mean number of

• ccurrences

in that interval and the value

• f

e is i l 2 71828 approximately 2.71828.

Chapter 5: The Poisson Probability Distribution

58

The mean number of occurrences, denoted by λ, is called

the parameter of the Poisson probability distribution or the

58

Poisson parameter.

Remember: The interval of λ and x must be of the same

• length. If they are not, the mean λ should be redefined to

k h l F i if λ i i h d make them equal. For instance, if λ was given in hours and we were asked to find the event in minutes, λ needs to be divided by 60 to have both x and λ in the same interval divided by 60 to have both x and λ in the same interval length

Chapter 5: The Poisson Probability Distribution

59

Example: The automatic teller machine (ATM) installed

• utside Mansfield Savings and Loan is used on average by

59

five costumers per hour. The bank closed this ATM for one hour for repairs. What is the probability that during that hour eight customers came to use this ATM? eight customers came to use this ATM?

Solution:

Chapter 5: The Poisson Probability Distribution

60

Example: The average number of traffic accidents on a certain section of highway is two per week. assume that the number of

60

accidents follows a Poisson distribution with λ = 2.

• 1. Find the probability of no accidents on this section of highway

during a 1‐week period.

• 2. Find the probability of at most three accidents on this section
• f highway during a 2‐week period.

Chapter 5: The Poisson Probability Distribution

61

Using the table of Poisson probabilities

The probabilities for Poisson distribution can also be found

61

p using Poisson Probability Table

Chapter 5: The Poisson Probability Distribution

62

Example: On average, two new accounts are opened per day at an Imperial Savings Bank branches. Using Table VI of

62

Appendix C, find the probability that on a given day the number of new accounts opened at this bank will be a) Exactly 6 b) At most 3 c) At least 7

Chapter 5: The Poisson Probability Distribution

63

Mean and standard deviation of the Poisson probability distribution F th P i di t ib ti

63

For the Poisson distribution

μ = λ σ2 = λ σ2 = λ

Example: An auto salesperson sells an average of .9 cars per

• day. Let x be the number of cars sold by this salesperson on
• day. Let x be the number of cars sold by this salesperson on

any given day. Using the Poisson probability table, write the probability distribution of x, draw the probability distribution, d fi d th d t d d d i ti f and find the mean and standard deviation of x