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Random Variables and Discrete probability Distributions

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Random Variables and Discrete probability Distributions

Chapter 7

- Many times simple events in the sample space can be assigned numerical values. A random variable can then be defined that takes on these values.
- The random variable reflects a certain characteristic of interest of the experimental outcome.
- Examples:
- The number of cars entering a gas station in the next five minutes.
- The amount of gas filled in a gas tank by a driver.
- The grade in a physics test.
- The number of new employees to be hired next month.

- There are two types of random variables:
- Discrete random variable
- Continuous random variable.

- A random variable is discrete if it can assume only a countable number of values.
- A random variable is continuous if it can assume an uncountable number of values.

0

1

2

3 ...

Discrete random variable

Continuous random variable

Only certain values can be assigned to the random variable

Any value within the range where the variable is defined can be selected

0

Therefore, the number of

values is countable

Therefore, the number of

values is uncountable

- A table, formula, or graph that lists all possible values a discrete random variable can assume, together with the probabilities associated with each value, is called a discrete probability distribution.
- To calculate the probability that the random variable X assumes the value x, P(X = a),
- add the probabilities of all the simple events for which X is equal to ‘a’, or
- Use probability calculation tools (such as tree diagram),

Simple event x Probability

TT01/4

HT11/4

TH11/4

HH21/4

x p(x)

0 1/4

1 1/2

2 1/4

H

HH

(½)(½)=1/4

H

(½)(½)=1/4

HT

T

H

TH

(½)(½)=1/4

T

T

(½)(½)=1/4

TT

- Example 1
- Find the probability distribution of the random variable describing the number of heads that turn-up when a coin is flipped twice.
- Solution

- If a random variable can assume values xi, then the following must be true:

- With the probability distribution we can more conveniently calculate probabilities (see example 2) next.

- Example 2
- A survey reveals the following frequencies for the number of colored TV per household.
Number of TVsNumber of Householdsxp(x)0 1,2180 1218/Total = .012 132,379132379/Total = .3192 37,961 2.374319,3873.1914 7,7144 .076 5 2,8425 .028Total101,5011.000

- A survey reveals the following frequencies for the number of colored TV per household.

Example 2 – continued

Calculate the probability of the following events:

- P(The number of colored TVs is 3) = P(X=3) =.191
- P(The number of colored TVs is two or more) = P(X³2)=P(X=2)+P(X=3)+P(X=4)+P(X=5)=.319+.374+…= .669

- Example 3
- The number of cars a dealer is selling daily were recorded in the last 100 days. This data was summarized as follows:

- Estimate the probability distribution.
- State the probability of selling more than 2 cars a day.

Daily sales Frequency

05

115

235

325

420

100

.35

.25

.20

.15

.05

- The probability of selling more than 2 cars a day is

P(X>2) = P(X=3) + P(X=4) = .25 + .20 = .45

- Example 3 - Solution
From the table of frequencies we can calculate the relative frequencies, which becomes our estimated probability distribution

Daily sales Relative Frequency

05/100=.05

115/100=.15

235/100=.35

325/100=.25

420/100=.20

1.00

X

0 1 2 3 4

- Example 4
- A mutual fund sales person knows that there is 20% chance of closing a sale on each call she makes.
- What is the probability distribution of the number of sales if she plans to call three customers?

- Solution
- Let us use probability rules and probability tree
- Define event Ai = {a sale is made in the iit phone call}.

- Assuming each phone call is independent of all other phone calls, we can assign the following probabilities to each branch on the tree:

(.2)(.2)(.8)= .032

A1 and A2 and A3

A1 and A2 and A3C

A1 and A2C and A3

A1c and A2 and A3

A1 and A1C andA3C

A1C andA2 and A3C

A1C and A2C andA3

A1C andA2C andAC

- XP(x)
- .23 = .008
- 3(.032)=.096
- 3(.128)=.384
- 0.83 = .512

If X represents the number of sales made, then.. (click)

(.2)(.8)(.8)= .128

Describing the PopulationProbability Distribution

- Given a discrete random variable X with values xi, that occurs with probabilities p(xi), the population mean of X is.

- Let X be a discrete random variable with possible values xi that occur with probabilities p(xi), and let E(xi) = m. The variance of X is defined by

- Example 5 Find the mean the variance and the standard deviation for the population of the number of colored TV per household in example 2
- Solution
- E(X) = m = Sxip(xi) = 0p(0)+1p(1)+2p(2)+…=0(.012)+1(.319)+2(.374)+… = 2.084
- V(X) = s2 = S(xi - m)2p(xi) = (0-2.084)2p(0)+(1-2.084)2p(1) + (2-2.084)2+… =1.107
- s = 1.1071/2 = 1.052

Using a shortcut formula

for the variance

- Solution – continued
- The variance can also be calculated as follows:

- The binomial experiment has the following characteristics:
- There are n trials (n is finite and fixed).
- Each trial can result in one out of two outcomes, success or a failure.
- The probability p of a success is the same for all the trials.
- All the trials of the experiment are independent.

- The binomial random variable counts the number of successes in n trials of the binomial experiment.
- The possible values of this count are 0,1, 2, …,n, and therefore the binomial variable is discrete.

The binomial probability distribution is described bythe following closed form formula:

S2

S1

F2

S2

F1

F2

P(S2|S1)

P(F2|S1)

P(S1)=p

P(F1)=1-p

P(S2|F1)

P(F2|F1)

S3

P(S3|S2,S1)

S2

P(S3)=p

S1

P(F3)=1-p

F3

P(F3|S2,S1)

P(S3|F2,S1)

S3

P(S3)=p

P(SFS)=p(1-p)p

P(F2|S1)

P(F2)=1-p

P(F3)=1-p

F2

P(F3|F2,S1)

F3

P(SFF)=p(1-p)2

S3

P(S3|S2,F1)

P(FSS)=(1-p)p2

P(S3)=p

S2

P(S2)=p

P(S2|F1)

P(F3)=1-p

P(F3|S2,F1)

F3

P(FSF)=(1-p)P(1-p)

S3

P(S3|F2,F1)

P(FFS)=(1-p)2p

F1

P(S3)=p

P(F2|F1)

P(F2)=1-p

F2

P(F3)=1-p

P(F3|F2,F1)

F3

P(FFF)=(1-p)3

P(SSS)=p3

P(S2|S1)

P(S2)=p

P(SSF)=p2(1-p)

P(S1)=p

Since the outcome of each trial is independent of the previous outcomes, we can replace the conditional probabilities with the unconditional probabilities.

P(F1)=1-p

P(X = 3) = P(Three successes) = P(SSS) = p3

P(X = 2) = P(Two successes, one Failure) = P(SSF) + P(SFS) + P(FSS) = p2(1-p) + p(1-p)p + (1-p)pp =

p2(1-p) + p2(1-p) + p2(1-p) =

3 p2(1-p).

P(X = 1) = P(One success, two failures) = P(SFF) + P(FSF) + P(FFS) = p(1-p)2 + (1-p)p(1-p) + (1-p)2p =

p(1-p)2 + p(1-p)2 + p(1-p)2 =

3 p(1-p)2

P(X = 0) = P(Three failures) = P(FFF) = (1-p)3

- Example 10
- Pat Statsdud takes a course in statistics, and intends to rely on luck to pass the next quiz.
- The quiz consists on 10 multiple choice questions with 5 possible choices for each question, only one is the correct answer.
- Pat will guess the answer to each question
- Find the following probabilities
- Pat gets no answer correct
- Pat gets two answer correct?
- Pat fails the quiz

- Solution
- Checking the conditions:

- Only one out of two outcomes can occur(An answer can be either correct or incorrect).
- There is a fixed finite number of trials(There are 10 questions in the test, n=10).
- Each answer is independent of the others.
- The probability p of a correct answer does not change from question to question(20% chance that an answer is correct).

- Solution – Continued
- Determining the binomial probabilities:
Let X = the number of correct answers

- Determining the binomial probabilities:

This is called cumulative probability

- Solution – Continued
- Determining the binomial probabilities:
Pat fails the test if she gets less than 5 correct answers.

- Determining the binomial probabilities:

P(X£4)

= p(0) + p(1) + p(2) + p(3) + p(4) = .1074 + .2684 + .3020 + .2013 + .0881 =.9672

E(X) = m = np

V(X) = s2 = np(1-p)

- Example 11
- If all the students in Pat’s class practice the same learning behavior like she does, what is the mean and the standard deviation of the quiz mark?

- Solution
- m = np = 10(.2) = 2.
- s = [np(1-p)]1/2 = [10(.2)(.8)]1/2 = 1.26.

- Example 12
- Records show that 30% of the customers in a shoe store make their payments using a credit card.
- This morning 20 customers purchased shoes.
- Use the binomial table to answer some questions stated in the next few slides.

- Solution
- First we verify that this is a binomial experiment:
- There are two possible outcomes of which only one will take place (paid with the credit card or not)
- There is a finite number of trials (20 customers are observed)
- Customers pay independently
- Each customer has the same probability to pay with a credit card (. 30).

- First we verify that this is a binomial experiment:

p

k

n = 20

.995

- Solution – continued
- Find the probability that at most 11 customers use a credit card.

In what follows we demonstrate the

usage of the Binomial Table

.01……….. .30

0

.

.

11

P(X £ 11) = .995

- To calculate P(X £ 11) when p = 0.3 and n = 20, type in an empty cell the following:=binomdist(11,20,0.3,1). Then click outside the cell.
- To calculate P(X = 11) type the following: =binomdist(11,20,0.3,0)
- To calculate P(X ³ 11) type the following:= 1 - binomdist(10,20,0.3,1)

In what follows we demonstrate how

to use Excel to determine binomial probabilities

p

k

- What is the probability that at least 3 but not more than 6 customers used a credit card?

At least 3 butnot more than 6

Not more than 6

.01……….. .30

0

2

.

6

P(3£X£6)=

P(X=3 or 4 or 5 or 6) =

.035

Click

P(X£6)

- P(X£2)

.573

.608

=.608 - .035 = .573

3 4 5 6

0 1 2 3 4 5 6

- Find the probability that exactly 14 customers did not use a credit card. Let Y be the number of customers who did not use a credit card, while X (as before) the number of those who did use a credit card.P(Y=14) = P(X=6) = P(X £ 6) - P(X £ 5) = .608 - .416 = .192
- Using Excel to solve the last two questions:
- P(3 £ X £ 6) = binomdist(6,20,0.3,1) – binomdist(2,20,0.3,1)
- P(Y = 14) = binomdist(14,20,0.7,0)

- What is the expected number of customers who used a credit card?
E(X) = np = 20(.30) = 6