Download Chapter 6: Discrete Distributions

Chapter 6: Discrete Distributions
Chapter Objectives
When you finish this chapter you should be able to






define a discrete random variable and a probability distribution.
solve problems by using the concepts of expected value and variance.
explain common discrete probability models and their parameters.
recognize the appropriate discrete model to use from the problem context.
find event probabilities for discrete models by using Excel, formulas or tables.
find the expected values of random variables transformed by adding or multiplying by a
constant.
Quiz Yourself
True/False Questions
T F 1. A table, formula, or graph that shows all possible values a random variable can assume, together
with their associated probabilities, is referred to as discrete probability distribution.
T F 2.
T F 3.
Faculty rank (professor, associate professor, assistant professor, and lecturer) is an example of
discrete numerical data.
The responses to the question “How many movies have you seen in the last month?” are values
from a discrete variable.
T F 4.
For each trial in a binomial random variable the probability of success is constant.
T F 5.
A binomial distribution with π =0.2 is always right skewed.
T F 6.
A Poisson distribution will be left-skewed for small values of its parameter, μ.
T F 7.
In a Poisson distribution, the mean and variance are equal.
T F 8.
If π remains constant in a Binomial distribution, an increase in n will not change the mean.
Multiple Choice Questions
1.
Which of the following is a characteristic of every binomial distribution?
A.
B.
C.
D.
2.
Each outcome is dependent on the previous outcome.
The probability of success increases from trial to trial.
Each outcome is mutually exclusive.
The outcome of a trial depends on the number of trials.
Ten bolts were selected randomly from a production line and the diameter of each was measured.
Why do these diameters not have a binomial distribution?
A.
B.
C.
D.
They have equal probability of success.
The diameters of the bolts are not independent of one another.
Diameter is a continuous random variable.
They do have a binomial distribution.
3.
Which of the following cannot generate a Poisson distribution?
A.
The number of people at a movie theatre on Friday evening
B.
The number of customers arriving at an ATM in a minute.
C.
The number of defects in a new automobile.
D.
The number of goals in the World Cup soccer game.
4.
The local police department must write, on average, 5 tickets a day to keep department revenues at
budgeted levels. Suppose the number of tickets written per day follows a Poisson distribution
with a mean of 5.5 tickets per day. Interpret the value of the mean.
A.
B.
C.
D.
5.
A multiple-choice test has 20 questions, and there are 4 choices for each question. A student
comes to class with four different colored balls of identical shape and size in her pocket. She has
decided to draw one ball from her pocket per question and allow the color of the ball to select the
answer for each question. What type of probability distribution can be used to figure out her
chance of getting 10 questions right?
A.
B.
C.
D.
6.
The number of tickets that is written most often is 5.5 tickets per day
Half of the days have less than 5.5 tickets written and half of the days have more than 5.5
tickets written
If we sampled all days, the arithmetic average or expected number of tickets written
would 5.5 tickets per day.
The mean cannot be interpreted.
Hypergeometric distribution
Poisson distribution
Normal distribution
Binomial distribution
The Information Commons in the main library has 150 personal computers. The probability that
any one of them will require repair on a given day is 0.02. To find the probability that exactly 25
of the computers will require repair, one will use what type of probability distribution?
A.
B.
C.
D.
Binomial distribution
Poisson distribution
Normal distribution
Hypergeometric distribution
Buyers for large department stores typically purchase costume jewelry in assortments rather than by the
piece. A buyer is interested in ordering a ring that is made in 5 sizes, and comes in lots of 50 rings. The
sizes and the proportion of rings manufactured in that size are in the table below. Use this information to
answer the next THREE questions.
7.
Size
5
6
7
8
9
Probability
0.1
0.2
0.4
0.2
0.1
What is the probability that, at most, a shipment received will contain 10 size six rings?
A.
B.
C.
D.
8.
=BINOMDIST(10,50,0.2,1)
=1-BINOMDIST(10,50,0.2,1)
=BINOMDIST(11,50,0.2,0)
=1-BINOMDIST(10,50,0.2,0)
What is the probability that a shipment of 50 rings will have more than 20 but no more than 30
size 7 or size 8 rings?
A.
B.
C.
D.
9.
What is the expected number of size 9 rings in a 50 ring shipment?
A.
10.
B.
9
C.
5
D.
3.5
be symmetric only if n is large.
always be symmetric.
be right-skewed.
be left-skewed.
Two international students were talking in the wake of the Singapore Air disaster a few years ago.
The first student said she had been intending to fly Singapore Air for the semester break, but
wasn’t going to now that there had been an accident! The second said, “But it should be extra safe
now. The chances of it crashing again so soon are very small.” If the likelihood of an airplane
crashing follows a Poisson distribution, what are these students forgetting about a Poisson random
variable?
A.
B.
C.
D.
12.
7
Whenever π = 0.5, the binomial distribution will:
A.
B.
C.
D.
11.
=BINOMDIST(10,50,0.6,1)
=BINOMDIST(20,50,0.6,1)+BINOMDIST(30,50,0.6,1)
=BINOMDIST(30,50,0.6,1)-BINOMDIST(19,50,0.6,1)
=BINOMDIST(30,50,0.6,1)-BINOMDIST(20,50,0.6,1)
the Poisson random variable is discrete.
the number of trials must be at least 30.
the outcome of each trial is one of two mutually exclusive categories.
the results of one trial is not dependent on the results of the other trials.
Which of these parameters associated with a binomial experiment will produce a probability
distribution with the smallest standard deviation, given n is the same for each situation?
A.
0.1.
π = 0.5
B.
π = 0.4
C.
π = 0.3
(1- π) =
D.
The Big Red Arcade and Pizza Palace caters to young teens and sells tokens to play arcade games for 25¢
each and slices of pizza for 50¢ each. The distribution of sales of pizza slices per customer and sales of
arcade tokens per customer is below. Use this information to answer the next THREE questions.
# Slices of Pizza
Probability
# Arcade Games
Probability
13.
1
2
3
4
5
0.05
0.10
0.20
0.30
0.15
0.20
0
1
2
3
4
5
0.01
0.04
0.15
0.20
0.25
0.35
What is the total expected per customer revenue for pizza slice sales and arcade token sales?
A.
14.
0
$6.69
B.
$2.60
C.
$2.42
D.
$1.34
If the expected number of arcade games played by the typical customer is 2 games, what is the
standard deviation of the number of arcade games played per customer?
A.
1.43
B.
2.10
C.
2.96
D.
4.43
15.
What is the probability that revenues from pizza slice sales will be greater than $1.00 per
customer?
A.
16.
B.
0.65
C.
0.85
D.
0.43
Past history shows that 70% of college students prefer Pepsi. Groups of nine students are selected.
Describe the histogram that would represent this distribution. The
histogram would be:
A.
B.
C.
D.
17.
0.95
left-skewed with ten columns.
right-skewed with nine columns.
left-skewed with nine columns.
right-skewed with ten columns.
Kunlakarn has decided to invest in oil paintings as a hedge against inflation. She is particularly
interested in paintings by Picasso or Constable. Below is a table of information about the value of
paintings by these artists in differing states of the Fine Art Market.
Values of one painting by Picasso or Constable
in different states of the Fine Art Market.
Probability of Fine Art Market State
Picasso
300K
Depressed
0.25
400K
Stable
0.45
500K
Expanding
0.30
Constable
500K
400K
700K
Kunlakarn wishes to purchase 2 paintings by one of the artists and 1 by the other. Assuming that
the values of paintings by Picasso are independent of the values of paintings by Constable, if she
wishes to minimize the risk of her investment, which collection of paintings should she buy?
A.
B.
C.
D.
One Picasso and Two Constables
Two Picassos and One Constable
There is no difference in the risk between the two collections.
Insufficient information to tell.
Solved Problems From Text
6.1
A is a probability distribution, since the sum of P(x) is 1 and all probabilities are nonnegative,
while B and C are not
probability distributions since the sum of P(x) is .95 for B and 1.30 for C.
6.2
a.
b.
E(X) = 70, V(X) = 100,  = 10. The worksheet is:
x
P(x)
xP(x)
xE(X)
(x-E(X))2
P(x)[xE(X)]2
60
70
80
90
Total
0.40
0.30
0.20
0.10
1.00
24.00
21.00
16.00
9.00
Σ=E(X)=70.00
-10.00
0.00
10.00
20.00
100
0
100
400
40.00
0.00
20.00
40.00
Σ=V(X)=100.00
Distribution is skewed to the right. Here is the graph of the distribution, with the long right tail.
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
60
6.8
80
90
Apply the formulas for mean and standard deviation of a Uniform distribution .


6.13
70
ab
2
Using a = 0000 and b = 9999,   (0  9999) / 2  4999.5
(b  a)  12  1
12
Using a = 0000 and b = 9999,   [(9999  0  1)2  1]/12  2886.75
First, define success and assign it a value of 1. Now define failure as everything that is NOT
success and assign it a value of 0. Remember, in order to be a valid Bernoulli trial, there can only
be two mutually exclusive and collectively exhaustive outcomes. Your answers may vary which
would switch 0 and 1.
a.
b.
c.
d.
1 = correct, 0 = incorrect
1= insured, 0 = uninsured
1 = busy, 0 = not busy
1 = lost weight, 0 no weight loss
6.15
Apply the formulas for the mean and standard deviation of a binomial distribution.
E ( X )    n *  and   n *  * (1   )
a.
 = (8)(.1) = 0.8,   (8)(.1)(1  .1)  0.8485
b.
 = (10)(.4) = 4,   (10)(.4)(1  .4)  1.5492
c.
 = (12)(.5) = 6,   (12)(.5)(1  .5)  1.7321
d.
 = (30)(.9) = 27,   (30)(.9)(1  .9)  1.6432
e.
 = (80)(.7) = 56,   (80)(.7)(1  .7)  4.0988
f.
 = (20)(.8) = 16,   (20)(.8)(1  .8)  1.7889
Using the Excel BINOMDIST function =BINOMDIST(k,n,π,0), where k = the value of X you
want the probability for, n =number of trials, π = probability of success and 0 means you want
only the probability for the one X value,
6.16
a.
b.
c.
d.
6.23
P(X = 2) =BINOMDIST(2, 8, .10, 0) = .1488
P(X = 1) =BINOMDIST(1, 10, .40, 0) = .0403
P(X = 3) =BINOMDIST(3, 10, .40,0) = .0015
P(X = 5) =BINOMDIST(5, 9, .90, 0) = 0.0074
Remember, the mean and variance of a Poisson are identical and equal to lambda, λ. Thus, in each
case above, the
a.
b.
c.
d.
e.
mean = λ and the standard deviation =
 = 1,  = 1.0 and  =1
 = 2,  = 2.0 and  =1.414
 = 4,  = 4.0 and  =2.0
 = 9,  = 9.0 and  =3
 = 12,  = 12.0 and  =3.464
.
6.24
The Excel formula for calculating Poisson probabilities is =POISSON(x,λ,0), where, as in the
binomial, x is the value of the random variable in which you are interested, λ is the value of the paramater,
and 0 indicates that you only what the probability associated with a specific value of the random
variable.
 = 0.1, P(X = 2) =POISSON(2, 0.1, 0) = .24377
 = 2.2, P(X = 1) =POISSON(1, 2.2, 0) =.00452
 = 1.6, P(X = 3) = POISSON(3, 1.6, 0) = .13783
 = 4.0, P(X = 6) = POISSON(6, 4.0, 0) = .10420
 = 12.0, P(X = 10) =POISSON(10, 12, 0) = .10484
a.
b.
c.
d.
e.
6.46
a.
b.
c.
E(Height in cm) =  = 2.54*E(Height in inches) = 2.54*79.6 = 202.184 cm
(Height in cm) = SQRT(V(2.54*Height in inches)) = SQRT((2.542 )*(3.242)) =
SQRT(6.4516*10.4976) = SQRT(67.72631616) = 8.2296cm
The mean: that the mean of a variable transformed by multiplying it by a constant is
equal to the mean of the original variable times the constant.
The standard deviation: that the standard deviation of a variable transformed by
multiplying it by a constant is equal to the square root of the variance of the random
variable times the square of the constant.
Quiz Yourself Answers
True/False
1 T 6 F
2 F 7 T
3 T 8 F
4 T
5 F
1
2
3
4
5
C
C
A
C
D
Multiple Choice
6 B
7 A
8 D
9 C
10 B
11
12
13
14
15
D
D
A
B
B
16
17
A
B