QBM117 Business Statistics Probability Distributions Binomial Distribution 1.

QBM117Business Statistics

Probability Distributions

Binomial Distribution

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Objectives

• To introduce the binomial distribution

• Learn how to recognize an experiment whose outcomes follow a binomial probability distribution

• Learn how to calculate binomial probabilities

• Find the mean and variance of a binomial distribution

• Learn how to use Excel to find binomial probabilities

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Binomial Probability Distribution

• The binomial probability distribution is a discrete probability distribution.

• It is the most widely used distribution for discrete random variables.

• The binomial distribution is used in situations where there are only two possible outcomes.

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Binomial Experiment

A binomial experiment possesses the following properties:

1. The experiment consists of a fixed number n of trials.

2. The result of each trial can be classified into one of two categories: success or failure.

3. The probability p of a success remains constant for each trial.

4. Each trial of the experiment is independent of the other trials.

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Binomial Random Variable

• The random variable in a binomial experiment is the number of successes in the n trials.

• It is called the binomial random variable.

• It is a discrete random variable as the possible values of the random variable are 0,1, 2,…, n.

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Binomial Distribution

• The distribution of a binomial random variable X is the Binomial distribution with parameters n and p, X~B(n,p)

• The parameter n is the number of observations.

• The parameter p is the probability of success on any one observation.

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Example 1

A coin is tossed 10 times and the number of heads is recorded.

- there are n = 10 trials (tosses)- there are two possible outcomes on each

toss: head or tail- the probability of a head is p = 0.5 on each

toss- each toss is independent of the others

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The properties of a binomial experiment are present.

Let X be the number of heads observed in 10 tosses.

X is a binomial random variable and has a binomial distribution, X~B(10,0.05)

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Example 2

500 randomly selected computer chips were produced at a manufacturing facility and each chip was tested to determine whether it was defective. Historically, 1% of the computer chips produced are defective.

- there are n = 500 trials (computer chips tested)- there are two outcomes per trial: the computer chip is either defective or non-defective- the probability of success (computer chip is defective) is p = 0.01 for each computer chip- the trials are independent since the computer chips are randomly selected

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Let X be the number of defective computer chips in the 500 randomly selected computer chips.


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Example 3

1500 people were asked whether they preferred the taste of Coca-Cola over that of other colas. Previous studies have shown that 20% of the population prefers Coca-Cola.

- there are n = 1500 trials (people asked)- there are two outcomes per trial: the person prefers Coca-Cola or does not- the probability of success (person prefers Coca-Cola) is p = 0.20- the trials are independent if the 1500 people

are randomly selected

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Let X be the number of people who prefer Coca-Cola.


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• In a binomial experiment we are interested in computing the probability of obtaining x successes in n trials.

• To do this we need to know the probability distribution.

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Example 4

Genetics says that children receive genes from their parents independently. Each child of a particular pair of parents has a probability 0.25 of having type O blood. This particular pair of parents have 5 children and the number who have type O blood is recorded.

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- There are n=5 trials, as there are 5 children.- The trials are independent as children receive

genes from their parents independently.- There are two possible outcomes for each child;

they either have type O blood or they don’t.- The probability that each child has type O blood is

p=0.25.

Let X be the number of children who have type O blood.

X is a binomial random variable and has a binomial distribution, X~B(5,0.25).

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Each child born to a particular set of parents has a probability 0.25 of having type O blood.

If these parents have 5 children, what is the probability that exactly 2 of them have type O blood?

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If X is the number of children with type O blood

and is a binomial random variable with n = 5 trials

and probability p = 0.25 of a success on each trial.

We want to find

P(exactly 2 children have type O blood) = P(X=2)

Define the events and their probabilities:

S = event of a success P(S) = 0.25

F = event of a failure P(F) = 0.75

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Step 1

Find the probability that a specific 2 of the 5 trials give successes, say the first and the second.

This is the outcome SSFFF.

The multiplication rule of independent events tells us that P(SSFFF) = P(S).P(S).P(F). P(F). P(F)

= 0.250.250.750.750.75 = (0.25)2(0.75)3

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Step 2

Find the number of different ways 2 successes and 3 failures can occur in 5 outcomes.

The possible combinations are

SSFFF SFSFF SFFSF SFFFS FSSFFFSFSF FSFFS FFSSF FFSFS FFFSS

Therefore there are 10 possible ways 2 successes and 3 failures can occur in 5 outcomes.

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The overall probability of 2 successes is therefore

P(X=2) = 10.(0.25)2(0.75)3 = 0.2637

Hence the binomial probability is the number of combinations of x successes in n trials, multiplied by the probability of any specific combination of the x successes.

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The possible values of X are 0, 1, 2, 3, 4 and 5.

We have calculated the probability that X = 2.

To obtain the probability distribution of X we would have to calculate the probabilities that X = 0, X = 1, X = 3, X = 4 and X = 5.

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Calculating Binomial Probabilities

• It is impractical to do this calculation each time we wish to find a probability for a binomial random variable, or to do all the calculations to obtain the probability distribution for a binomial random variable.

• Instead we need a general formula for the probability of obtaining x successes in the n trials, when the probability of success is p.

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• We will observe x successes in the n trials whenever we observe x successes and (n-x) failures.

• The probability of success is p and the probability of failure is q = 1-p.

• Applying the multiplication rule for independent events, the probability that x successes and (n-x) failures occur is pxqn-x.

• We now need to multiply this probability by the number of ways we can get x successes in n trials.

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Counting Rule

• The number of different ways we can choose x objects from n objects is

where n! is read as ‘n factorial’ and is given by

and 0! Is defined to be 1.

!

!( )!nx

nC

x n x

! ( 1)( 2)...2.1n n n n

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Example 4 revisited

The number of ways we can choose 2 children from 5 is

as we calculated before.

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5! 5!10

2!(5 2)! 2!3!C

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Using your calculator

• Combinations can be calculated using your calculator, using the nCr button (Casios).

• Factorials can also be calculated using the

button. It is a 2nd function on most calculators.

nxC

!x !x

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Calculating Binomial Probabilities

• The probability that x successes and (n-x) failures occur is pxqn-x.

• We now need to multiply this probability by the number of ways we can get x successes in n trials.

• Hence the probability of obtaining x successes in n trials is n x n x

xP X x C p q

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• If the random variable X is the total number of successes in the n trials of a binomial experiment that has a probability p of a success on any given trial, the probability distribution of X is given by

(note that p = 1 – q is the probability of failure on any given trial)

• This is known as the binomial probability distribution.

xnxnx qpCxXP )(

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Example 5

National Oil Company conducts exploratory oil drilling operations. In order to fund the operation, investors form partnerships, which provide the financial support necessary to drill a fixed number of oil wells. Each well drilled is classified as a producer well or a dry well. Past experience shows that this type of exploratory operation provides producer wells for 15% of wells drilled. A newly formed partnership has provided the financial support for drilling at 12 exploratory locations.

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- there are n = 12 trials (wells)- there are two outcomes per trial: the well is either

a producer well or it is dry- the probability of success (well is a producer well)

is p = 0.15- We assume that the trials are independent


Define X as the number of producer wells.

X is a binomial random variable with n = 12 and p = 0.15, X~B(12, 0.15).

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What is the probability that all 12 wells will be producer wells?

12 12 012

10

( 12) (0.15) (0.85)

1.2974 10

0.000000000012974

0

P X C

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What is the probability that all 12 wells will be dry?

The probability that all 12 wells are dry is the probability that none of the wells are producer wells.

12 0 120( 0) (0.15) (0.85)

0.1422

P X C

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What is the probability that exactly 1 well will be a producer well?

12 1 111( 1) (0.15) (0.85)

0.3012

P X C

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In order to make the partnership venture profitable, at least 3 of the 12 exploratory wells must be producer wells. What is the probability that the venture will be profitable?

12 2 102

( 3) ( 3) ( 4) ... ( 12)

1- ( 0) ( 1) ( 2)

1 0.1422 0.3012 (0.15) (0.85)

1 0.1422 0.3012 0.2924

1- 0.7358

0.

P X P X P X P X

P X P X P X

C

2642

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Binomial Tables

• An alternative of calculating binomial probabilities is to use the Table 1 in Appendix C of the text (pg 585 full text, pg 583 abridged).

• The probabilities given in this table are cumulative probabilities:

0

( ) ( 0) ( 1) ... ( )

( )k

x

P X k P X P X P X k

P X x

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Example 6

A large bank that issues many loans for General Motors estimates that 70% of the loans are approved within 24 hours of application. If a consumer group takes a random sample of 25 recent General Motors loan applications, what is the probability thata. at most 20 were approved within 24 hours?b. at least 15 were approved within 24 hours?c. at least 12, but no more than 18 are approved

within 24 hours?d. exactly 22 were approved within 24 hours?

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a. The probability that at most 20 were approved within 24 hours is

b. The probability that at least 15 were approved within 24 hours is

910.0)20( XP

902.0

098.01

)14(1)15(

XPXP

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c. The probability that at least 12, but no more than 18 are approved within 24 hours is

653.0

006.0659.0

)11()18()1812(

XPXPXP

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d. The probability that exactly 22 were approved within 24 hours is

024.0

967.0991.0

)21()22()22(

XPXPXP

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Mean and Variance of Binomial Random Variables

• If X is a binomial random variable, the mean and variance of X are

where n is the number of trials, p is the probability of success in any trial, and q=1-p is the probability of failure in any trial.

npqXV

npXE

2)(

)(

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Example 6 revisited

A large bank that issues many loans for General Motors estimates that 70% of the loans are approved within 24 hours of application. If a consumer group takes a random sample of 25 recent General Motors loan applications, what is the expected number of loans that will be approved?

5.17

7.025

)(

npXE

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What is the variance of the number of loans that will be approved?

25.5

3.07.025

)(

npqXV

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Reading for next lecture

• Chapter 5 Section 5.5

Exercises

• 5.31• 5.32• 5.35

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QBM117 Business Statistics Probability Distributions Binomial Distribution 1.

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