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Quantitative Analysis for Management

Thirteenth Edition

Chapter 2

Probability Concepts and Applications

Learning Outcomes

Formulate models of random variables to real-world issues.

RANDOM VARIABLES (1 OF 3)

A random variable assigns a real number to every possible outcome or event in an experiment

X = number of refrigerators sold during the day

Discrete random variables can assume only a finite or limited set of values

Continuous random variables can assume any one of an infinite set of values

Random Variables (2 of 3)

TABLE 2.5 Examples of Random Variables

EXPERIMENT	OUTCOME	RANDOM VARIABLES	RANGE OF RANDOM VARIABLES
Stock 50 Christmas trees	Number of Christmas trees sold	X = number of Christmas trees sold	0, 1, 2, . . . , 50
Inspect 600 items	Number of acceptable items	Y = number of acceptable items	0, 1, 2, . . . , 600
Send out 5,000 sales letters	Number of people responding to the letters	Z = number of people responding to the letters	0, 1, 2, . . . , 5,000
Build an apartment building	Percent of building completed after 4 months	R = percent of building completed after 4 months	0 … R … 100
Test the lifetime of a lightbulb (minutes)	Length of time the bulb lasts up to 80,000 minutes	S = time the bulb burns	0 … S … 80,000

Random Variables (3 of 3)

TABLE 2.6 Random Variables for Outcomes That Are Not Numbers

EXPERIMENT	OUTCOME	RANGE OF RANDOM VARIABLES	RANDOM VARIABLES
Students respond to a questionnaire	Strongly agree (SA) Agree (A) Neutral (N) Disagree (D) Strongly disagree (SD)		1, 2, 3, 4, 5
One machine is inspected	Defective Not defective		0, 1
Consumers respond to how they like a product	Good Average Poor		1, 2, 3

PROBABILITY DISTRIBUTIONS (1 OF 4)

For discrete random variables, probability value assigned to each event

Statistics class of 100 students (ref. Table 2.7)

Quiz with five problems with 1 point for each correct answer

Lowest score = 1, highest score = 5

Example follows the three rules:

Events are mutually exclusive and collectively exhaustive

Individual probability values between 0 and 1

Total probability sums to 1

Probability Distributions (2 of 4)

TABLE 2.7 Probability Distribution for Quiz Scores

Developed using relative frequency approach

RANDOM VARIABLE (X) SCORE	NUMBER	PROBABILITY P(X)
5	10	0.1 = 10÷100
4	20	0.2 = 20÷100
3	30	0.3 = 30÷100
2	30	0.3 = 30÷100
1	10	0.1 = 10÷100
Total	100	1.0 = 100÷100

Probability Distributions (4 of 4)

FIGURE 2.4 Probability Distribution for Dr. Shannon’s Class

Central tendency of the distribution is the mean or expected value

Amount of variability is the variance

Expected Value of a Discrete Probability Distribution (1 of 2)

Expected value is a measure of the central tendency of the distribution

where

Xi = random variable’s possible values

P(Xi) = probability of each possible value of the random variable

= summation sign indicating we are adding all n possible values

E(X) = expected value or mean of the random variable

Expected Value of a Discrete Probability Distribution (2 of 2)

For the quiz scores

Variance of a Discrete Probability Distribution (1 of 4)

where

Xi = random variable’s possible values

E(Xi) = expected value of the random variable

[Xi − E(X)] = difference between each value of the random variable and the expected value

E(X) = probability of each possible value of the random variable

Variance of a Discrete Probability Distribution (2 of 4)

For quiz scores

Variance = (5 − 2.9)2(0.1) + (4 − 2.9)2(0.2) + (3 − 2.9)2(0.3)

+ (2 − 2.9)2(0.3) + (1 − 2.9)2(0.1)

= (2.1)2(0.1) + (1.1)2(0.2) + (0.1)2(0.3) + (−0.9)2(0.3) + (−1.9)2(0.1)

= 0.441 + 0.242 + 0.003 + 0.243 + 0.361

= 1.29

Variance of a Discrete Probability Distribution (3 of 4)

Standard deviation is the square root of the variance

where

Variance of a Discrete Probability Distribution (4 of 4)

Standard deviation is the square root of the variance

where

For this example

Using Excel (1 Of 2)

PROGRAM 2.1A Excel 2016 Output for the Dr. Shannon Example

Using Excel (2 of 2)

PROGRAM 2.1B Formulas in an Excel Spreadsheet for the Dr. Shannon Example

PROBABILITY DISTRIBUTION OF A CONTINUOUS RANDOM VARIABLE (1 OF 3)

The fundamental rules for continuous random variables must be modified

The sum of the probability values must still equal 1

The probability of each individual value of the random variable occurring must equal 0 or the sum would be infinitely large

The probability distribution is defined by a continuous mathematical function called the probability density function or just the probability function represented by f (X)

Probability Distribution of a Continuous Random Variable (2 of 3)

FIGURE 2.5 Graph of Sample Density Function

Probability Distribution of a Continuous Random Variable (3 of 3)

For any continuous distribution, the probability does not change if a single point is added to the range of values that is being considered.

The following probabilities are all exactly the same:

P(5.22 < X < 5.26) = P(5.22 < X ≤ 5.26) = P(5.22 ≤ X < 5.26)

= P(5.22 ≤ X ≤ 5.26)

The Binomial Distribution (1 of 3)

Many business experiments can be characterized by the Bernoulli process

The Bernoulli process is described by the binomial probability distribution

Each trial has only two possible outcomes

The probability of each outcome stays the same from one trial to the next

The trials are statistically independent

The number of trials is a positive integer

The Binomial Distribution (2 of 3)

The binomial distribution is used to find the probability of a specific number of successes in n trials

We need to know

n = number of trials

p = the probability of success on any single trial

We let

r = number of successes

q = 1 − p = the probability of a failure

The Binomial Distribution (3 of 3)

The binomial formula is

Probability of r success in n trials

The symbol ! means factorial, and n! = n(n − 1)(n − 2)…(1)

4! = (4)(3)(2)(1) = 24

Also, 1! = 1 and 0! = 0 by definition

Solving Problems with the Binomial Formula (1 of 3)

Find the probability of getting 4 heads in 5 tosses of a coin

n = 5, r = 4, p = 0.5, and q = 1 − 0.5 = 0.5

P(4 successes in 5 trials)

Solving Problems with the Binomial Formula (2 of 3)

TABLE 2.8 Binomial Distribution for n = 5, p = 0.50

NUMBER OF HEADS (r)	PROBABILITY = (0.5)r(0.5)5 − r
0	0.03125 = (0.5)0(0.5)5 − 0
1	0.15625 = (0.5)1(0.5)5 − 1
2	0.31250 = (0.5)2(0.5)5 − 2
3	0.31250 = (0.5)3(0.5)5 − 3
4	0.15625 = (0.5)4(0.5)5 − 4
5	0.03125 = (0.5)5(0.5)5 − 5

Solving Problems with the Binomial Formula (3 of 3)

FIGURE 2.6 Binomial Distribution for n = 5, p = 0.50

The Figure in the text may have a different scale for the values of r, 1 - 6. The scale should be 0 – 5. See pages 36-37 in the text. The PPt slide shows the correct values.

Solving Problems with Binomial Tables (1 of 4)

MSA Electronics is experimenting with the manufacture of a new transistor

Every hour a random sample of 5 transistors is taken

The probability of one transistor being defective is 0.15

What is the probability of finding 3, 4, or 5 defective?

n = 5, p = 0.15, and r = 3, 4, or 5

Solving Problems with Binomial Tables (3 of 4)

TABLE 2.9 (partial) A Sample Table for the Binomial Distribution

We find the three probabilities in the table for

n = 5, p = 0.15, and r = 3, 4, and 5 and add them together

		P
n	r	0.05	0.10	0.15
5	0	0.7738	0.5905	0.4437
	1	0.2036	0.3281	0.3915
	2	0.0214	0.0729	0.1382
	3	0.0011	0.0081	0.0244
	4	0.0000	0.0005	0.0022
	5	0.0000	0.0000	0.0001

P(3 or more defects) = P(3) + P(4) + P(5)

= 0.0244 + 0.0022 + 0.0001

= 0.0267

Solving Problems with Binomial Tables (4 of 4)

Expected value is

Expected value (mean) = np

Variance = np(1 − p)

For the MSA example

Expected value = np 5(0.15) = 0.75

Variance = np(1 − p) = 5(0.15)(0.85) = 0.6375

Using Excel (1 of 2)

PROGRAM 2.2A Excel Output for the Binomial Example

Using Excel (2 of 2)

PROGRAM 2.2B Function in an Excel 2016 Spreadsheet for Binomial Probabilities

IN-CLASS EXERCISE

Summary

Formulated models of random variables to real-world issues.

5 if SA

4 if A

3 if N

2 if D

1 if SD

0 if defective

1 if not defective

3 if good

=2 if average

1 if poor

(

)

(

)

=+++

1122

XXX

X(X)X(X)...X(X)

PPP

i=1

∑

i=1

(

)

(

)

=++++

1122334455

XXX

X(X)X(X)X(X)X(X)X(X)

(5)(0.1) + (4)(0.2) + (3)(0.3) + (2)(0.3

) + (1)(0.1)

2.9

PPPPP

==-

Variance[X(X)](X)

σ = Variance = σ 2

s=Variance=s

square root

standard deviation

Variance

1.291.14

!()!

rnr

45-4

= 0.50.5

4!(5-4)!

5(4)(3)(2)(1)

= (0.0625)(0.5) = 0.15625

4(3)(2)(1)1!

(

!5!

)

–r

(

)

0!5–0!

(

)

1!5–1!

(

)

2!5–2!

(

)

3!5–3!

(

)

4!5–4!

(

)

5!5–5!