chapter 5: probability distributions - governors state … · 2009-01-05 · chapter 5: probability...

1Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 5Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.

Chapter 5:Probability Distributions

Hildebrand, Ott and GrayBasic Statistical Ideas for Managers

Second Edition


2

Learning Objectives for Ch. 5

• Understanding the counting techniques needed for sequences and combinations.

• Understanding that a binomial random variable counts the number of successes in a fixed number of trials with each trial being a success or failure.• Assumptions needed to use the binomial.

• Understanding that a Poisson random variable counts the number of occurrences of an event in a unit of time, area or volume.• Assumptions needed to use the Poisson.


Learning Objectives for Ch. 5

• Understanding that a normal random variable measures a characteristic of interest and has a bell-shaped distribution.

• Learning how to calculate probabilities for binomial, Poisson and normal random variables.

• Understanding how to use a normal probability plot to determine if data is from a normal distribution.


Section 5.1Counting Possible Outcomes


5.1 Counting Possible Outcomes

• Under the classical interpretation of probability:

• We need ways to count the number of outcomes.

Number of favorable outcomesTotal number of outcomes

P(Event) =



• Preliminary Concept – Factorials • The factorial symbol is “!”• Definition of n!

n! = n (n - 1) (n - 2)…1Example:

3! = (3)(2)(1) = 6 • By definition, 0! = 1.



• One consideration in counting techniques• Order matters ⇒ sequences• Order doesn’t matter ⇒ subsets

Example:Consider the letters a and b. If order matters, there are 2 sequences:

(a,b) and (b,a) If order does not matter, there is only 1 subset:

{a,b}



• Number of sequences • Rule: The number of sequences of k objects that

can be formed from a set of r distinct objects, denoted rPk, is: rPk = (r) (r - 1)…(r – k + 1)

Example: The number of sequences of 2 lettersformed from the 4 letters a, b, c, d, is: (4) (3) = 12The sequences are: (a,b) (a,c) (a,d) (b,c) (b,d) (c,d)(b,a) (c,a) (d,a) (c,b) (d,b) (d,c)



• Number of subsets or combinations • Rule: The number of subsets of k objects

that can be formed from a set of r distinct objects, denoted rCk, is:

• Notation: Use rCk or ( )

rCk =_______k! (r – k)!

r!

rk



Example: The number of subsets of 2 letters formed from the 4 letters a, b, c, d is:

rCk = ( ) = = 6

The subsets are: {a,b} {a,c} {a,d} {b,c} {b,d} {c,d}

rk

_______2! (4-2)!

4!



Exercise 5.67:Several states now have a Lotto game. A player chooses 6 distinct integers in the range 1 to 40. If exactly those 6 numbers are selected as the winning numbers, the player receives a very large prize. What is the probability that a particular set of 6 numbers will be drawn? You may wish to think of the 6 numbers drawn as the “success” numbers.

First approach: Order matters (even though it doesn’t)

Total number of outcomes = 40P6 = (40)(39)(38)(37)(36)(35)

Number of favorable outcomes = 6P6 = (6)(5)(4)(3)(2)(1) = 6!

P(Winning) = 6!/ [(40)…(35)] = .00000026052657



• Another perspective of the first approach

P(Winning) = 6! / [(40) ··· (35)]

=

=

= P(W1) P(W2/W1) ··· P(W6/W1 and ··· W5)

Where Wi ≡ {The ith number is a winning number}

)35)(36)(37)(38)(39)(40()1)(2)(3)(4)(5)(6(

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

351

395

406



Second approach: Order doesn’t matter (and it really doesn’t)

Total number of outcomes =

Number of favorable outcomes =

P (Winning) = 6!/[(40)…(35)] = .00000026052657

Moral: You can’t lose if you don’t play!

40C6 = _____6! 34!

40!

6C6 · 34C0

____ . _____6! 0! 0! 34!

34!= = 16!


Section 5.2The Binomial Distribution


5.2 The Binomial Distribution

• Examples of a Bernoulli Trial1. A coin toss results in a head (H) or a tail (T).

2. A bit sent through a digital communications channel is entered as either 0 or 1 and received either correctly or incorrectly.

3. An audited account is either current (C) or delinquent (D).

4. A consumer is either aware (A) of a particular product or not aware (N).

5. A flight reservation is either a show (S) or no-show (N).



• Features of a Bernoulli Trial:

• Only 2 possible outcomes for each trial,characterized as:Success (S) or Failure (F)

• π denotes P(S)(1 – π) denotes P(F).



• Bernoulli R.V. and Probability Distribution

Let Y = 1, if trial results in S

= 0, if trial results in F

ππ

11

)(0 −

yPy Y



• Graphical representation of a Bernoulli probability distribution

PY(y)

y0 1

The distribution is skewed when π ≠ .5

1 – π

π


• E(Y) = 0(1 - π) + 1(π ) = π

• V(Y) = Σ (y - µ)2 PY(y) = [0 – π]2 (1 – π) + [1 – π]2 π= π (1 – π) [π + (1 – π) ]= π (1 – π)



• Examples of Binomial Random Variable:1. Toss a coin 10 times. Let Y denote the number of heads in the 10

tosses.2. For the next 3 bits transmitted through a digital communications

channel, let Y be the number of bits received that are in error.3. 20 accounts are randomly selected from a population of several

thousand accounts and are audited. Let Y be the number of delinquent accounts in the sample.[The sampling has to be with replacement for the probability of success to remain constant. In reality, the sampling is done without replacement.]

4. 100 randomly selected consumers are surveyed as part of a marketresearch study. Let Y denote the number of these consumers who are aware of a particular product.

5. Out of 50 flight reservations made, let Y be the number of passengers who show.



• Features of a Binomial Experiment:• There are n Bernoulli trials [each one results in

S or F].• The probability of a success, π = P(S), remains

constant over the n trials; [P(F) = 1 - π ].• The trials are independent.

• The binomial random variable is the total number of successes in n trials, where the ordering is unimportant.




• Binomial Probability Distribution

•

• The expression for PY(y) can be used tocalculate probabilities for a binomial randomvariable.

• What is the basis for the expression for PY(y)?

y n - yY

nP (y) (1 - ) , y 0,1,...,

ynπ π

⎛ ⎞= =⎜ ⎟

⎝ ⎠

!y)-(n!y !n

yn

=⎟⎟⎠

⎞⎜⎜⎝

⎛



Example: Y denotes number of bits in error in next 3 transmitted where P(Error) = π

Found by using principles of Chapter 3.

Outcomes

E, E, E

E, E, O

E, O, E

O, E, E

E, O, O

O, E, O

O, O, E

O, O, O

y y

3

2

1

0

Probability y

π3

3 π2 (1- π)

3 π (1- π) 2

(1- π) 3

From PY(y) y

π3 (1-π)0 = π3

π2 (1- π)1 = 3 π2 (1- π)

π1 (1- π)3 –1 = 3 π (1- π ) 2

(1- π ) 3

⎟⎟⎠

⎞⎜⎜⎝

⎛33

⎟⎟⎠

⎞⎜⎜⎝

⎛23

⎟⎟⎠

⎞⎜⎜⎝

⎛13



• Calculation of Probabilities • Use the binomial probability distribution formula • Instead of actually calculating the probabilities, we can

look them up in a table. Table 1 at the end of Hildebrand, Ott & Gray gives the probabilities for n = 2(1) 10 (2) 20, 50, 100 and π = .05(.05).50.

• We can also use software (MINITAB; EXCEL’sBINOMDIST function)

• Two obvious cases

P[0 successes] =nn0

)-1()-1(0n

πππ =⎟⎟⎠

⎞⎜⎜⎝

⎛

P[n successes] = n0

)-1(nn

πππ =⎟⎟⎠

⎞⎜⎜⎝

⎛ n



• Mean and Variance of a Binomial Random Variable

E[Y] = nπ

V(Y) = σ2 = nπ (1 - π)



• An easy way to find E(Y) and V(Y)

Y = total number of successes in n trials= Number of successes on 1st trial

+ Number of successes on 2nd trial + …+ Number of successes on nth trial

E(Y) = π + π + …. + π = nπ

V(Y) = π (1 – π) + π (1 – π)+ … + π (1 – π) = nπ (1 – π)



Exercise 5.61 [Revised so that number of potential customers is 50.]

Executives at a soft drink company wish to test a new formulation of their chief product. The new drink is tested in comparison to the current one. Each of 50 potential customers is given a cup of the current formulation and a cup of the new one. The cups are labeled H and K to avoid bias. Each customer indicates a preference. Assume that, in fact, the customers can't detect a difference and are, in effect, guessing. Define Y to be the number (out of 50) indicating preference for the new formulation.



a.What probability distribution should apply to Y? Do the assumptions underlying that distribution seem plausible in this context?

• Each of the 50 customers is a Bernoulli trial (either prefers new product or does not).

• If customers are guessing, the probability of preference for new product is 0.5.

• Reasonable to assume trials are independent.• Let Y be the number of customers who indicate a

preference for the new product.

Then Y is binomial with n = 50 and π = 0.5.



A graph of the probability distribution of Y follows.

The graph is symmetric because π = 0.5

y

P(Y

=Y

)

50403020100

0.12

0.10

0.08

0.06

0.04

0.02

0.00

Probability Distribution of Y



b. Find the mean and standard deviation of Y.

µy = nπ = (50)(0.5) = 25

σ2 = nπ (1-π) = 12.5

σ = 3.54



c. (Cont’d)Find the probability that the number of customers preferring the new brand is within 2 standard deviations of the mean.

P[µ – 2σ ≤ Y ≤ µ + 2σ ] = P[25 – 2(3.54) ≤ Y ≤ 25 + 2(3.54)]

= P[ 17.93 ≤ Y ≤ 32.08]

= P[ 18 ≤ Y ≤ 32]

= P[Y=18] + P[Y=19] + … + P[Y=32]

= .0160 + .0270 + … + .0160 (From Table 1)

= .9672Most of the time (97%), we should observe between 18 and 32 customers indicating a preference for the new product if, in fact, they are guessing.



d. (Cont’d)In one such test, 12 people preferred the new formulation. Find the probability that 12 or fewer would prefer the new formulation if the customers can’t detect a difference. What, if anything, can you infer about consumer preferences from the results of the taste test.

P(Y ≤ 12) = .0001 (from Table 1)

If the hypothesis that the people can’t detect a difference is correct, P(Y ≤ 12) is very small [ <.05]. Since this probability is very small,it implies the hypothesis that the people can’t detect a difference isincorrect! Or, π ≠ .5

Why were the cups labeled H and K?

Studies have shown that people have no preference for either ofthese letters, as opposed to the letters A and B.


Section 5.3The Poisson Distribution


5.3 The Poisson Distribution

• Named for Simeon D. Poisson (1781-1840)

• Examples of a Poisson random variable• The number of work-related injuries per month at a

manufacturing plant.• The number of e-mail messages arriving at a personal

computer in one hour.• The number of network errors per day on a local area

network.

• The Poisson random variable is the number of occurrences in a given unit.



• Features of a Poisson ExperimentFor a unit of time, area or volume • Probability that an event occurs in a given unit is the

same for all units.• Probability of two or more events occurring at same

time is 0.• The occurrence of the event in one unit is independent

of the number that occur in other units.

• The expected number of occurrences in each unit is denoted by µ.



• Poisson Probability Distribution

• Calculation of Probabilities• Use formula for pY (y) • Use Table 2 for µ = 0.1(0.1)5 and

5.5(0.5)10 and 11(1)20 • Use software

...2,1,0)!(

)( ==−

yy

eyPy

Yµµ



• Mean and Variance for a Poisson Random Variable

• E(Y) = µ

• Var(Y) = µ



Exercise 5.29:Suppose that the number of defaults on home mortgage loans at National Mortgage Company follows a Poisson distribution with an average of 8.2 defaults per month.

a. Compute the probability of exactly 12 defaults at NMC next month.

P(Y = 12) =

= 0.0529925 { From Minitab

)!12()2.8()12(122.8−

=ePY



A graph of the probability distribution of Y, the number of defaults per month follows.

The probability distribution quickly tapers off to .005 or less for y ≥ 16.

y

P(Y

=y)

9080706050403020100

0 .14

0 .12

0 .10

0 .08

0 .06

0 .04

0 .02

0 .00

P r o ba b i l i ty D is tr ib uti on o f Y



b. What is the chance of at least one default next week?P(Y ≥ 1) = 1 – P(Y = 0) = 1 - .00027

= 0.99973c. Because of poor economic times, NMC believes that the

average number of defaults may have increased from 8.2 per month. Last month, there were 15 defaults. If the average number of defaults has not changed from 8.2, find P(Y ≥ 15).

P(Y ≥ 15) = 1 – P(Y ≤ 14) = 1 - .9791= .0209

⇒ Since P(Y ≥ 15) is small, this implies µ has changed from 8.2.


Section 5.4The Normal Distribution


5.4 The Normal Distribution

Continuous Random Variables in General

• Examples of continuous random variables:

• Stock market returns • Quality characteristics of finished products

(such as net contents)• Heights of males; heights of females• Age at time of death



Continuous Random Variables in General (Cont’d)

• Features of a continuous random variable:

• The possible values are uncountable.• The probability that the random variable takes on

a specific value is 0.• Only an interval of values has a nonzero

probability.




• The probability for an interval of values will be shown as the area under the pdf.

a by

)(yfY

P(a< Y < b)




• Details:• It doesn’t matter whether endpoints are included

in the interval:

P[a < Y < b] = P[a ≤ Y < b] = P[a < Y ≤ b] = P[a ≤ Y ≤ b]

Why? P[Y = a] = P[Y = b] = 0.

• Data are never continuous!



• The Standard Normal Random Variable

• The probability distribution of a standard normal random variable Z is shown below:

fz (z)

z0



• E(Z) = µz = 0 {The curve is symmetric around 0}

V(Z) = σz2 = 1

• Other Properties:

Total area under the curve is 1.

The curve is symmetric around 0.P(Z > 0) = 0.5



• Determination of probabilities for a standardnormal random variable:

• Use Table 3 (area from 0 to a right-hand value z)

• Use software



Exercise 5.30: Suppose that Z represents astandard normal randomvariable. i. Find P(Z ≤ -2.42).

P(Z ≤ -2.42)

= 0.5 - P( 0 ≤ Z ≤ 2.42)

= 0.5 - .4922 (from Table 3)

= 0.0078

( )Zf z

z0-2.42



g. Find P(-1.07 ≤ Z ≤ 2.33) P(-1.07 ≤ Z ≤ 2.33)

= P(-1.07 ≤ Z ≤ 0)+ P(0 ≤ Z ≤ 2.33)

= 0.3577 + 0.4901(from Table 3)

= 0.8478

( )Zf z

z-1.07 0 2.33



Exercise 5.31:For the standard normal random variable Z, solve the following equation for k.a. P(Z ≥ k) = .01

From Table 3, P(0 ≤ Z ≤ 2.33) = 0.4901

P(Z ≥ 2.33) = .01

⇒ k = 2.33 ( )Zf z

z

.01

k



• Normal Random Variables in General

• The probability distribution is mound-shaped.• µy is the expected value of the distribution.• σy is the standard deviation of the distribution.

( )Yf y

yyµ

yσ



• Standardize Y to find areas under the normal curve of Y.

{Procedure for standardizing Y}

Now use Table 3.

• The standardized variable Z measures how many standard deviations Y is above or below its mean.

Y

Y

YZ µσ−

=



Exercise 5.41:A potato chip packaging plant has a process line that fills 12 ounce bags of potato chips. At the current setting of the machine, the quality control engineer knows that the actual distribution of weights in the bags follows a normal distribution with a mean of 12.0 ounces and a standard deviation of 0.18 ounces.

a. What percentage of all bags filled contain exactly 12ounces?

P(Y = 12) = 0, since the probability at a point is 0.



b. What percentage of all bags filled contain more than 12.4 ounces?

P(Y > 12.4)

= P(Z > )

= P( Z > 2.22)

= 0.5 – 0.4868

= 0.0132

• 12.4 is 2.22 standard deviations from 12.0.

18.0124.12 −

y

)(yfY

12 12.4



c. Find the 60th percentile of the actual weights of 12-ounce bags of potato chips.

Find k so that P(Y< k) = .60Standardizing

P(Z< ) = .60

From Table 3, P(Z < 0.253) = .60,

Set = 0.253

k = 12 +(0.18)(0.253)

k = 12.046 ounces

1 20 .1 8

k −

1 20 .1 8

k −

y

)(yfY

12

.60



d. Management is concerned when 12-ounce bags of potato chips contain less than 11.75 ounces. The quality control engineer can set the filling machine so that actual mean filling weight is whatever he chooses, but the standard deviation always remains at 0.18 ounces. What mean filling weight should he set the machine to if he wants only 1% of all bags to contain less than 11.75 ounces?

Find µ so that P (Y < 11.75) = .01

.01 = P(Y < 11.75) = P(Z < )

.01 = P(Z < -2.33) from Table 3

Set = -2.33

µ = 11.75 + (2.33)(0.18) = 12.17 ounces

11.750.18

µ−

11.750.18

µ−


Section 5.5Checking Normality


5.5 Checking Normality

• Many of the statistical techniques in later chapters assume that the data is from a normal distribution.

• Chapter 2 presented several graphical techniques that could be useful in assessing whether or not the data is from a normal distribution.

• For example, is a histogram mound-shaped? The answer to this question is facilitated by superimposing a normal distribution over the histogram.



Example:Consider the returns for ^DJI first presented in Chapter 1. The histogram with a normal distribution superimposed follows.

R^DJI

Freq

uenc

y

1050-5-10

10

8

6

4

2

0

Mean -0.3414StDev 5.287N 35

Histogram for R^DJI with Normal Distribution SuperimposedNormal



Conclusion:At first glance, it appears that the normal distribution is nota good fit. However, the shape of the histogram is determined by the number of class intervals and their width. So, this may not be the best approach.

R^DJI

Freq

uenc

y

1050-5-10

10

8

6

4

2

0

Mean -0.3414StDev 5.287N 35

Histogram for R^DJI with Normal Distribution SuperimposedNormal



• Another approach for assessing normality is the Normal Probability Plot.• The data are arranged in ascending order. • Each data value, y(i), is assigned a cumulative

relative frequency, pi:

• Think of 0.5 as a correction factor.• Other correction factors are sometimes used.

100( 0.5)i

ipn−

=



• For example, if the data set has 25 observations, then

p1 = 2.00, p2 = 6.00,…, p25 = 98.00 • The percentage of the observations less than or

equal to y(1) is 2.00%.• The percentage of the observations less than or

equal to y(2) is 6.00%.• (y(i), pi) are plotted on a graph where the vertical

axis is scaled so that if the data is from a normal distribution, the resulting plot should be approximately linear.



• Appearance of NPP’s for data from a distribution that is not normal.• Right-skewed data plot as a curve, with the slope

getting flatter as one moves to the right. • Left-skewed data plot as a curve, with the slope getting

steeper as one moves to the right.• Data from symmetric distributions with more tail area

than the normal plot as an S-shape, with the slope steepest at both ends.

• The straight line drawn through the points can assist in assessing linearity. It can also be misleading if a few of the points are outliers.

• In the following examples, the sample size is fixed at 25. This value for n was arbitrarily chosen.



Example:What does the NPP look like for data from a standard normal distribution?

z

Perc

ent

3210-1-2-3

99

95

90

80

70

60504030

20

10

5

1

Mean

0.878

0.07199StDev 1.285N 25AD 0.196P-Value

Probability Plot of zNormal



Conclusion:Since the plotted points are nearly linear, conclude that the data came from a normal distribution.

z

Perc

ent

3210-1-2-3

99

95

90

80

70

60504030

20

10

5

1

Mean

0.878

0.07199StDev 1.285N 25AD 0.196P-Value

Probability Plot of zNormal



Example:What does the NPP look like for data from a normal distribution with µ = 100 and σ = 10?

y

Perc

ent

1301201101009080

99

95

90

80

70

60504030

20

10

5

1

Mean

0.781

103.8StDev 10.11N 25AD 0.231P-Value

Probability Plot of yNormal



Conclusion:Since the plotted points are nearly linear, conclude that the data came from a normal distribution.

y

Perc

ent

1301201101009080

99

95

90

80

70

60504030

20

10

5

1

Mean

0.781





Example:A uniform distribution is one that is of uniform or constant height for the range of y values. For the interval from –3 to +3, a uniform distribution has height of (1/6). What does the NPP look like for data from a uniform distributionthat ranges from –3 to +3 ?

y

Perc

ent

5.02.50.0-2.5-5.0

99

95

90

80

70

60504030

20

10

5

1

Mean

0.025

0.1511StDev 1.944N 25A D 0.844P-Value




Conclusion:Because the plot is S-shaped with the slope steepest at both ends, conclude that the data came from a symmetric distribution with more probability in each tail than the normal distribution.

y

Perc

ent

5.02.50.0-2.5-5.0

99

95

90

80

70

60504030

20

10

5

1

Mean

0.025

0.1511StDev 1.944N 25AD 0.844P-Value




Example:What does the NPP look like for data from a distribution that is skewed to the right with E(Y) = 927 and σY = 871?

y

Perc

ent

3000200010000-1000

99

95

90

80

70

60504030

20

10

5

1

Mean

<0.005





Conclusion:Since the plot is curved with the slope getting flatter as one moves to the right, conclude that the data came from a right-skewed distribution.

y

Perc

ent

3000200010000-1000

99

95

90

80

70

60504030

20

10

5

1

Mean

<0.005

904.9StDev 800.1N 25A D 1.229P-Value




Example:Consider the returns for R^DJI. What does the NPP tell us?

R^DJI

Perc

ent

1050-5-10-15

99

95

90

80

70

60504030

20

10

5

1

Mean

0.760

-0.3414StDev 5.287N 35AD 0.240P-Value

NPP for R^DJINormal



Conclusion:Because the NPP is linear, conclude that the R^DJI are normally distributed. However, it’s a different story for the RIBM data. The NPP for RIBM follows.

RIBM

Perc

ent

403020100-10-20-30

99

95

90

80

70

60504030

20

10

5

1

Mean

0.038

0.4368StDev 12.30N 35AD 0.784P-Value

NPP for RIBMNormal



• Procedure to obtain a Normal Probability Plot using Minitab:

Suppose the data to be analyzed are stored in C1Click on Stat Basic Statistics Normality TestEnter “C1” in box for “Variable”Select “Percentile Lines” option. The default option is “None”Select “Tests for Normality” option. The default option is “Anderson-Darling”Enter “Title” for plotClick on “OK”


Keywords: Chapter 5

•Factorial•Sequences•Combinations•Bernoulli trials•Binomial random variable

•Binomial probability distribution

•Poisson random variable•Poisson probability distribution

•Normal random variable•Standard normal probability distribution

•Normal probability distribution

•Normal probability plot


Summary of Chapter 5

• The counting techniques needed for sequences and combinations.

• A binomial random variable counts the number of successes in n trials, with each trial being a success or failure.

• A Poisson random variable counts the number of occurrences of an event over a specified length of time.

• A normal random variable measures the characteristic of interest and the probability distribution is bell-shaped.


Summary of Chapter 5

• Computing probabilities for the binomial, Poisson and normal random variables.

• Assessing normality of data by the normal probability plot.

chapter 5: probability distributions - governors state … · 2009-01-05 · chapter 5: probability...

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