© 2011 Pearson Education, Inc - StFXpeople.stfx.ca/asarhan/stat 201/notes/ch 4.pdf · © 2011 Pearson Education, Inc Statistics for Business and Economics Chapter 4 Random Variables

© 2011 Pearson Education, Inc


Statistics for Business and

Economics

Chapter 4

Random Variables &

Probability Distributions


Content

1. Two Types of Random Variables

2. Probability Distributions for Discrete Random Variables

3. The Binomial Distribution

4. Poisson and Hypergeometric Distributions

5. Probability Distributions for Continuous Random Variables

6. The Normal Distribution


Content (continued)

7. Approximating a Binomial Distribution with a Normal Distribution

8. Sampling Distributions

9. The Sampling Distribution of a Sample Mean and the Central Limit Theorem


Learning Objectives

1. Develop the notion of a random variable

2. Learn that numerical data are observed values of either discrete or continuous random variables

3. Study two important types of random variables and their probability models: the binomial and normal model

4. Define a sampling distribution as the probability of a sample statistic

5. Learn that the sampling distribution of follows a normal model

x


Thinking Challenge

• You’re taking a 33 question

multiple choice test. Each

question has 4 choices.

Clueless on 1 question, you

decide to guess. What’s the

chance you’ll get it right?

• If you guessed on all 33

questions, what would be your

grade? Would you pass?


4.1

Two Types of Random Variables


Random Variable

A random variable is a variable that assumes

numerical values associated with the random

outcomes of an experiment, where one (and only

one) numerical value is assigned to each sample

point.


Discrete

Random Variable

Random variables that can assume a countable

number (finite or infinite) of values are called

discrete.


Discrete Random Variable

Examples

Experiment Random

Variable

Possible

Values

Count Cars at Toll

Between 11:00 & 1:00

# Cars

Arriving 0, 1, 2, ..., ∞

Make 100 Sales Calls # Sales 0, 1, 2, ..., 100

Inspect 70 Radios # Defective 0, 1, 2, ..., 70

Answer 33 Questions # Correct 0, 1, 2, ..., 33


Continuous

Random Variable

Random variables that can assume values

corresponding to any of the points contained in

one or more intervals (i.e., values that are

infinite and uncountable) are called continuous.


Continuous Random Variable

Examples

Measure Time

Between Arrivals

Inter-Arrival

Time

0, 1.3, 2.78, ...

Experiment Random

Variable

Possible

Values

Weight 100 People Weight 45.1, 78, ...

Measure Part Life Hours 900, 875.9, ...

Amount spent on food $ amount 54.12, 42, ...


4.2

Probability Distributions for

Discrete Random Variables


Discrete

Probability Distribution

The probability distribution of a discrete

random variable is a graph, table, or formula

that specifies the probability associated with each

possible value the random variable can assume.


Requirements for the

Probability Distribution of a

Discrete Random Variable x

1. p(x) ≥ 0 for all values of x

2. p(x) = 1

where the summation of p(x) is over all possible

values of x.


Discrete Probability

Distribution Example

Probability Distribution

Values, x Probabilities, p(x)

0 1/4 = .25

1 2/4 = .50

2 1/4 = .25

Experiment: Toss 2 coins. Count number of

tails.

© 1984-1994 T/Maker Co.


Visualizing Discrete

Probability Distributions

Listing Table

Formula

# Tails f(x)

Count p(x)

0 1 .25

1 2 .50

2 1 .25

p x n

x!(n – x)! ( )

! = px(1 – p)n – x

Graph

.00

.25

.50

0 1 2

x

p(x)

{ (0, .25), (1, .50), (2, .25) }


Summary Measures

1. Expected Value (Mean of probability distribution)

• Weighted average of all possible values

• = E(x) = x p(x)

2. Variance

• Weighted average of squared deviation about

mean

• 2 = E[(x 2(x 2p(x)

3. Standard Deviation

2 ●


Summary Measures

Calculation Table

x p(x) x p(x) x –

Total (x 2p(x)

(x – 2 (x – 2p(x)

xp(x)


Thinking Challenge

You toss 2 coins. You’re

interested in the number of

tails. What are the expected

value, variance, and

standard deviation of this

random variable, number of

tails? © 1984-1994 T/Maker Co.


Expected Value & Variance

Solution*

0 .25 –1.00 1.00

1 .50 0 0

2 .25 1.00 1.00

0

.50

.50

= 1.0

x p(x) x p(x) x – (x – 2 (x – 2p(x)

.25

0

.25

2.50

.71


Probability Rules for Discrete

Random Variables

Let x be a discrete random variable with probability

distribution p(x), mean µ, and standard deviation .

Then, depending on the shape of p(x), the following

probability statements can be made:

Chebyshev’s Rule Empirical Rule

P x x µ 0 .68

P x 2 x µ 2 34

.95

P x 3 x µ 3 89

1.00


4.3

The Binomial Distribution


Binomial Distribution

Number of ‘successes’ in a sample of n

observations (trials)

• Number of reds in 15 spins of roulette wheel

• Number of defective items in a batch of 5 items

• Number correct on a 33 question exam

• Number of customers who purchase out of 100

customers who enter store (each customer is

equally likely to pyrchase)


Binomial Probability

Characteristics of a Binomial Experiment

1. The experiment consists of n identical trials.

2. There are only two possible outcomes on each trial. We

will denote one outcome by S (for success) and the other

by F (for failure).

3. The probability of S remains the same from trial to trial.

This probability is denoted by p, and the probability of

F is denoted by q. Note that q = 1 – p.

4. The trials are independent.

5. The binomial random variable x is the number of S’s in

n trials.



Distribution !

( ) (1 )! ( )!

x n x x n xn n

p x p q p px x n x

p(x) = Probability of x ‘Successes’

p = Probability of a ‘Success’ on a single trial

q = 1 – p

n = Number of trials

x = Number of ‘Successes’ in n trials

(x = 0, 1, 2, ..., n)

n – x = Number of failures in n trials



Distribution Example

3 5 3

!( ) (1 )

!( )!

5!(3) .5 (1 .5)

3!(5 3)!

.3125

x n xnp x p p

x n x

p

Experiment: Toss 1 coin 5 times in a row. Note

number of tails. What’s the probability of 3 tails?

© 1984-1994 T/Maker Co.


Binomial Probability Table

(Portion) n = 5 p

k .01 … 0.50 … .99

0 .951 … .031 … .000

1 .999 … .188 … .000

2 1.000 … .500 … .000

3 1.000 … .812 … .001

4 1.000 … .969 … .049

Cumulative Probabilities

p(x ≤ 3) – p(x ≤ 2) = .812 – .500 = .312



Characteristics

.0

.5

1.0

0 1 2 3 4 5

X

P(X)

.0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

n = 5 p = 0.1

n = 5 p = 0.5

E(x) np

Mean

Standard Deviation

npq



Thinking Challenge

You’re a telemarketer selling service

contracts for Macy’s. You’ve sold 20

in your last 100 calls (p = .20). If you

call 12 people tonight, what’s the

probability of

A. No sales?

B. Exactly 2 sales?

C. At most 2 sales?

D. At least 2 sales?


Binomial Distribution Solution*

n = 12, p = .20

A. p(0) = .0687

B. p(2) = .2835

C. p(at most 2) = p(0) + p(1) + p(2) = .0687 + .2062 + .2835 = .5584

D. p(at least 2) = p(2) + p(3)...+ p(12) = 1 – [p(0) + p(1)] = 1 – .0687 – .2062 = .7251


4.4

Other Discrete Distributions:

Poisson and Hypergeometric


Poisson Distribution

1. Number of events that occur in an interval

• events per unit

— Time, Length, Area, Space

2. Examples

• Number of customers arriving in 20 minutes

• Number of strikes per year in the U.S.

• Number of defects per lot (group) of DVD’s


Characteristics of a Poisson

Random Variable

1. Consists of counting number of times an event occurs during a given unit of time or in a given area or volume (any unit of measurement).

2. The probability that an event occurs in a given unit of time, area, or volume is the same for all units.

3. The number of events that occur in one unit of time, area, or volume is independent of the number that occur in any other mutually exclusive unit.

4. The mean number of events in each unit is denoted by .


Poisson Probability

Distribution Function

2

p(x) = Probability of x given

= Mean (expected) number of events in unit

e = 2.71828 . . . (base of natural logarithm)

x = Number of events per unit

p x x

( ) !

x

e

–

(x = 0, 1, 2, 3, . . .)


Poisson Probability


.0

.2

.4

.6

.8

0 1 2 3 4 5

X

P(X)

.0

.1

.2

.3

0 2 4 6 8 10

X

P(X)

= 0.5

= 6

Mean

Standard Deviation

E(x)


Poisson Distribution Example

Customers arrive at a

rate of 72 per hour.

What is the probability

of 4 customers arriving

in 3 minutes?

© 1995 Corel Corp.


Poisson Distribution Solution

72 Per Hr. = 1.2 Per Min. = 3.6 Per 3 Min. Interval

-

4 -3.6

( )!

3.6(4) .1912

4!

x ep x

x

ep


Poisson Probability Table

(Portion) x

0 … 3 4 … 9

.02 .980 …

: : : : : : :

3.4 .033 … .558 .744 … .997

3.6 .027 … .515 .706 … .996

3.8 .022 … .473 .668 … .994

: : : : : : :

Cumulative Probabilities

p(x ≤ 4) – p(x ≤ 3) = .706 – .515 = .191


Thinking Challenge

You work in Quality Assurance

for an investment firm. A clerk

enters 75 words per minute

with 6 errors per hour. What is

the probability of 0 errors in a

255-word bond transaction?

© 1984-1994 T/Maker Co.


Poisson Distribution Solution:

Finding * • 75 words/min = (75 words/min)(60 min/hr)

= 4500 words/hr

• 6 errors/hr = 6 errors/4500 words

= .00133 errors/word

• In a 255-word transaction (interval):

= (.00133 errors/word )(255 words)

= .34 errors/255-word transaction


Poisson Distribution Solution:

Finding p(0)*

-

0 -.34

( )!

.34(0) .7118

0!

x ep x

x

ep


Characteristics of a

Hypergeometric

Random Variable

1. The experiment consists of randomly drawing n elements without replacement from a set of N elements, r of which are S’s (for success) and (N – r) of which are F’s (for failure).

2. The hypergeometric random variable x is the number of S’s in the draw of n elements.


Hypergeometric Probability


where . . .

[x = Maximum [0, n – (N – r), …,

Minimum (r, n)] p x

r

x

N r

n x

N

n

µ nr

N 2

r N r n N n N 2 N 1


Hypergeometric Probability


N = Total number of elements

r = Number of S’s in the N elements

n = Number of elements drawn

x = Number of S’s drawn in the n elements


4.5

Probability Distributions for

Continuous Random Variables


Continuous Probability

Density Function

The graphical form of the probability distribution for a

continuous random variable x is a smooth curve


Continuous Probability

Density Function

This curve, a function of x, is denoted by the symbol f(x)

and is variously called a probability density function

(pdf), a frequency function, or a probability

distribution.

The areas under a probability

distribution correspond to

probabilities for x. The area A

beneath the curve between two

points a and b is the probability

that x assumes a value between a and b.


4.6

The Normal Distribution


Importance of

Normal Distribution

1. Describes many random processes or

continuous phenomena

2. Can be used to approximate discrete

probability distributions

• Example: binomial

3. Basis for classical statistical inference


Normal Distribution

1. ‘Bell-shaped’ &

symmetrical

2. Mean, median, mode

are equal x

f ( x )

Mean

Median

Mode


Probability Density Function

where

µ = Mean of the normal random variable x

= Standard deviation

π = 3.1415 . . .

e = 2.71828 . . .

P(x < a) is obtained from a table of normal probabilities

f (x) 1

2e

1

2

x

2


Effect of Varying

Parameters ( & )


Normal Distribution

Probability

c d x

f ( x )

Probability is

area under

curve! P(c x d) f (x)

c

d

dx?


Standard Normal Distribution

The standard normal distribution is a normal

distribution with µ = 0 and = 1. A random variable

with a standard normal distribution, denoted by the

symbol z, is called a standard normal random variable.


z = 0

= 1

1.96

Z .04 .05

1.8 .4671 .4678 .4686

.4738 .4744

2.0 .4793 .4798 .4803

2.1 .4838 .4842 .4846

The Standard Normal Table:

P(0 < z < 1.96)

.06

1.9 .4750

Standardized Normal

Probability Table (Portion)

Probabilities

.4750

Shaded area

exaggerated



P(–1.26 z 1.26)

z = 0

= 1

–1.26

Standardized Normal Distribution

Shaded area exaggerated

.3962

1.26

.3962 P(–1.26 ≤ z ≤ 1.26)

= .3962 + .3962

= .7924



P(z > 1.26)

z = 0

= 1


1.26

P(z > 1.26)

= .5000 – .3962

= .1038

.3962

.5000



P(–2.78 z –2.00)

= 1

= 0

–2.78 z –2.00

.4973

.4772



P(–2.78 ≤ z ≤ –2.00)

= .4973 – .4772

= .0201



P(z > –2.13)

z = 0

= 1

–2.13



P(z > –2.13)

= .4834 + .5000

= .9834

.5000 .4834


x

f(x)

Non-standard Normal

Distribution

Normal distributions differ by

mean & standard deviation.

Each distribution would

require its own table.

That’s an infinite

number of tables!


Property of Normal Distribution

If x is a normal random variable with mean μ and

standard deviation , then the random variable z,

defined by the formula

has a standard normal distribution. The value z describes

the number of standard deviations between x and µ.

z x µ


Standardize the

Normal Distribution

Normal

Distribution

x

One table!

= 0

= 1

z

Standardized Normal

Distribution

z x


Finding a Probability Corresponding

to a Normal Random Variable

1. Sketch normal distribution, indicate mean, and shade

the area corresponding to the probability you want.

2. Convert the boundaries of the shaded area from x

values to standard normal random variable z

z x µ

Show the z values under corresponding x values.

3. Use Table IV in Appendix A to find the areas

corresponding to the z values. Use symmetry when

necessary.


z = 0

= 1

.12

Standardized Normal

Distribution


.0478

Non-standard Normal μ = 5, σ = 10:

P(5 < x < 6.2)

Normal

Distribution

x = 5

= 10

6.2

z

x

6.2 5

10 .12


z = 0

= 1

-.12

Standardized Normal

Distribution


P(3.8 x 5)

Normal

Distribution

x = 5

= 10

3.8

.0478


z

x

3.8 5

10 .12


0

= 1

-.21 z .21

Standardized Normal

Distribution


P(2.9 x 7.1)

5

= 10

2.9 7.1 x

Normal

Distribution

.1664

.0832 .0832


z

x

2.9 5

10 .21

z

x

7.1 5

10 .21



P(x 8)

x = 5

= 10

8

Normal

Distribution

z = 0 .30

Standardized Normal

Distribution

= 1

.3821

.5000

.1179


z

x

8 5

10 .30


= 0

= 1

.30 z .21

Standardized Normal

Distribution


P(7.1 X 8)

= 5

= 10

8 7.1 x

Normal

Distribution

.1179 .0347

.0832


z

x

7.1 5

10 .21

z

x

8 5

10 .30


Normal Distribution Thinking

Challenge You work in Quality Control for

GE. Light bulb life has a normal

distribution with = 2000 hours

and = 200 hours. What’s the

probability that a bulb will last

A. between 2000 and 2400

hours?

B. less than 1470 hours?


Standardized Normal

Distribution

z = 0

= 1

2.0

Solution* P(2000 x 2400)

Normal

Distribution

x = 2000

= 200

2400

.4772

z

x

2400 2000

200 2.0


z = 0

= 1

–2.65

Standardized Normal

Distribution

Solution* P(x 1470)

x = 2000

= 200

1470

Normal

Distribution

.0040 .4960

.5000

z

x

1470 2000

200 2.65


Finding z-Values

for Known Probabilities

What is Z, given

P(z) = .1217?

Shaded area

exaggerated

z = 0

= 1

?

.1217

Standardized Normal

Probability Table (Portion)

Z .00 0.2

0.0 .0000 .0040 .0080

0.1 .0398 .0438 .0478

0.2 .0793 .0832 .0871

.1179 .1255

.01

0.3 .1217

.31


Finding x Values

for Known Probabilities

Normal Distribution

x = 5

= 10

?

.1217


Shaded areas exaggerated

z = 0

= 1

.31

.1217

8.1

x z 5 .31 10


4.8

Approximating a

Binomial Distribution with a

Normal Distribution


Normal Approximation of


1. Useful because not all

binomial tables exist

2. Requires large sample

size

3. Gives approximate

probability only

4. Need correction for

continuity

n = 10 p = 0.50

.0

.1

.2

.3

0 2 4 6 8 10

x

p(x)


.0

.1

.2

.3

0 2 4 6 8 10

x

p(x)

Why Probability

Is Approximate

Binomial Probability:

Bar Height

Normal Probability: Area Under

Curve from 3.5 to 4.5

Probability Added

by Normal Curve

Probability Lost by

Normal Curve


Correction for Continuity

1. A 1/2 unit adjustment to

discrete variable

2. Used when approximating

a discrete distribution

with a continuous

distribution

3. Improves accuracy

4.5

(4 + .5)

3.5

(4 – .5) 4


Using a Normal Distribution to

Approximate Binomial

Probabilities

1. Determine n and p for the binomial distribution,

then calculate the interval:

If interval lies in the range 0 to n, the normal

distribution will provide a reasonable

approximation to the probabilities of most

binomial events.

3 np 3 np 1 p




Probabilities 2. Express the binomial probability to be

approximated by the form

For example,

P x a or P x b P x a

P x 3 P x 2

P x 5 1 P x 4

P 7 x 10 P x 10 P x 6




Probabilities 3. For each value of interest a, the correction for

continuity is (a + .5), and the corresponding

standard normal z-value is

z a .5 µ




Probabilities 4. Sketch the approximating normal distribution and

shade the area corresponding to the event of

interest. Using Table IV and the z-value (step 3),

find the shaded area.

This is the approximate

probability of the

binomial event.


.0

.1

.2

.3

0 2 4 6 8 10

x

P(x)

Normal Approximation Example

3.5 4.5

What is the normal approximation of p(x = 4)

given n = 10, and p = 0.5?


Normal Approximation Solution

1. Calculate the interval:

2. Express binomial probability in form:

P x 4 P x 4 P x 3

np 3 np 1 p 10 0.5 3 10 0.5 1 0.5

5 4.74 0.26, 9.74

Interval lies in range 0 to 10, so normal

approximation can be used



3. Compute standard normal z values:

z a .5 n p

n p 1 p

3.5 10 .5

10 .5 1 .5 0.95

z a .5 n p

n p 1 p

4.5 10 .5

10 .5 1 .5 0.32


= 0 = 1

-.32 z -.95


.1255

.3289

- .1255

.2034

.3289

4. Sketch the approximate normal distribution:



.0

.1

.2

.3

0 2 4 6 8 10

x

p(x)

5. The exact probability from the binomial formula is

0.2051 (versus .2034)


4.10

Sampling Distributions


Parameter & Statistic

A parameter is a numerical descriptive measure

of a population. Because it is based on all the

observations in the population, its value is almost

always unknown.

A sample statistic is a numerical descriptive

measure of a sample. It is calculated from the

observations in the sample.


Common Statistics & Parameters

Sample Statistic Population Parameter

Variance s2 2

Standard

Deviation s

Mean x

Binomial

Proportion p p ̂


The sampling distribution of a sample statistic

calculated from a sample of n measurements is

the probability distribution of the statistic.

Sampling Distribution


Developing

Sampling Distributions

• Population size, N = 4

• Random variable, x

• Values of x: 1, 2, 3, 4

• Uniform distribution

© 1984-1994 T/Maker Co.

Suppose There’s a Population ...


Population Characteristics

xi

i1

N

N 2.5

Population Distribution Summary Measure

.0

.1

.2

.3

1 2 3 4

P(x)

x


All Possible Samples

of Size n = 2

Sample with replacement

1.0 1.5 2.0 2.5

1.5 2.0 2.5 3.0

2.0 2.5 3.0 3.5

2.5 3.0 3.5 4.0

16 Samples

1st

Obs

1,1 1,2 1,3 1,4

2,1 2,2 2,3 2,4

3,1 3,2 3,3 3,4

4,1 4,2 4,3 4,4

2nd Observation

1 2 3 4

1

2

3

4

2nd Observation

1 2 3 4

1

2

3

4

1st

Obs

16 Sample Means



of All Sample Means

1.0 1.5 2.0 2.5

1.5 2.0 2.5 3.0

2.0 2.5 3.0 3.5

2.5 3.0 3.5 4.0

2nd Observation

1 2 3 4

1

2

3

4

1st

Obs

16 Sample Means Sampling Distribution

of the Sample Mean

.0

.1

.2

.3

1.0 1.5 2.0 2.5 3.0 3.5 4.0

P(x)

x


Summary Measure of

All Sample Means

X

xi

i1

N

N

1.0 1.5 ... 4.0

16 2.5


Comparison

Population Sampling Distribution

2.5x

.0

.1

.2

.3

1 2 3 4

2.5

.0

.1

.2

.3

1.0 1.5 2.0 2.5 3.0 3.5 4.0

P(x)

x

P(x)

x


4.11

The Sampling Distribution of

a Sample Mean and the

Central Limit Theorem


Properties of the Sampling

Distribution of x

1. Mean of the sampling distribution equals mean of

sampled population*, that is,

x E x .

2. Standard deviation of the sampling distribution equals

Standard deviation of sampled population

Square root of sample size

x

n.That is,


Standard Error of the Mean

The standard deviation is often referred to

as the standard error of the mean.

x


Theorem 4.1

If a random sample of n observations is selected from

a population with a normal distribution, the sampling

distribution of will be a normal distribution. x


n =16x = 2.5

Sampling from

Normal Populations

• Central Tendency

• Dispersion

– Sampling with

replacement

= 50

= 10

x

n = 4x = 5

x

= 50 - x


Population Distribution

x

xn


Standardizing the Sampling

Distribution of x

Standardized Normal

Distribution

= 0

= 1

z

z x

x

x

x

nSampling

Distribution

x x

x


Thinking Challenge

You’re an operations analyst

for AT&T. Long-distance

telephone calls are normally

distributed with = 8 min.

and = 2 min. If you select

random samples of 25 calls,

what percentage of the

sample means would be

between 7.8 & 8.2 minutes?

© 1984-1994 T/Maker Co.


Sampling Distribution Solution*

Sampling

Distribution

8

x

= .4

7.8 8.2 x 0

= 1

–.50 z .50

.3830

Standardized Normal

Distribution

.1915 .1915

z x

n

7.8 8

225

.50

z x

n

8.2 8

225

.50


Sampling from

Non-Normal Populations

• Central Tendency

• Dispersion

– Sampling with

replacement

Population Distribution


n =30x = 1.8

n = 4x = 5

= 50

= 10

x

x

= 50 - x

x

xn



Consider a random sample of n observations selected

from a population (any probability distribution) with

mean μ and standard deviation . Then, when n is

sufficiently large, the sampling distribution of will be

approximately a normal distribution with mean

and standard deviation The larger the

sample size, the better will be the normal approximation

to the sampling distribution of .

x x

x n .

x



x

As sample

size gets

large

enough

(n 30) ...

sampling

distribution

becomes almost

normal.

x

xn


Central Limit Theorem Example

SODA

The amount of soda in cans of a

particular brand has a mean of 12

oz and a standard deviation of .2

oz. If you select random samples

of 50 cans, what percentage of

the sample means would be less

than 11.95 oz?


Central Limit Theorem Solution*

Sampling

Distribution

12

x = .03

11.95 x 0

= 1

–1.77 z

.0384

Standardized Normal

Distribution

.4616

z x

n

11.9512

.250

1.77



Key Ideas

Properties of Probability Distributions

Discrete Distributions

1. p(x) ≥ 0

2.

Continuous Distributions

1. P(x = a) = 0

2. P(a < x < b) = area under curve between a and b

p x 1all x


Key Ideas

Normal Approximation to Binomial

x is binomial (n, p)

P x a P z a .5 µ


Key Ideas

Methods for Assessing Normality

1. Histogram


Key Ideas


2. Stem-and-leaf display

1 7

2 3389

3 245677

4 19

5 2


Key Ideas


3. (IQR)/S ≈ 1.3

4. Normal probability plot


Key Ideas

Generating the Sampling Distribution of x

Documents

© 2011 Pearson Education, Inc - StFXpeople.stfx.ca/asarhan/stat 201/notes/ch 4.pdf · © 2011 Pearson Education, Inc Statistics for Business and Economics Chapter 4 Random Variables