A Very Nice Practice Midterm
Exam
A representative sampling of the
first half of prob and stat
- Complete with follow-on questions to enhance the
learning process
Need to know in general

sample spaces and random events


counting methods



Probabilities, means and variances, modes and medians
joint distributions



Probabilities, means and variances
continuous random variables


Combinations, permutations, fundamental counting rule
condition probabilities, total prob rule, & Bayes theorem
discrete random variables


Unions, intersections, and complements
covariance and correlation
marginal and conditional distributions
Descriptive statistics
Question 1
The chief executive officer (CEO) of the Combinatorial
Company, Mr. Hi N. Mitey, will be visiting 3 of his 5 plants
this week. The plant manager of the Homestead Plant is
betting that the CEO will not visit him this week and is
therefore delaying preparations for his visit. If Hi N. Mitely
selects his itinerary randomly, what is the probability that
the Homestead Plant will not be visited this week?
 4 1 
  
favorable  3  0  4
Prob 


total
10
 5
 
 3
order of visits is not important
Alternate approach
favorable  4  3 2  2
Prob 


total
 5 4  3 5
Follow-up to Question 1

The chief executive officer (CEO) of the
Combinatorial Company decides not to visit
the plants. Instead he will send his 3 vicepresidents. Each VP selects a plant at random
to visit. What is the probability that the
Homestead Plant is not selected?
favorable 43 64
Prob 
 3 
 .512
total
5 125
Bonus Question #1

Faculty advisors are assigned randomly to new students entering the
Engineering Management program from among five full-time faculty
members. Three new students are admitted on a particular day. The
office staff member, Dizzy Dunce, assigns (randomly) each student a
faculty advisor. (a) What is the probability that each student is
assigned to a different advisor? (b) What is the probability that all
three students are assigned to one faculty member?
(a) (5)(4)(3) / 53 = .48
(b) 5 / 53 = 1/25 = .04 (5 ways to select the same faculty member)
Question 2

Eighty percent of all students taking probability and
statistics pass the course while seventy-five percent of
those taking operations research pass the course.
However, of those that have passed probability and
statistics, 90 percent will pass operations research.
What fraction of students passes both courses? What
fraction will pass at least one course?
Let A = the event, student passes prob/stat
B = the event, student passes operations research
Given: P(A) = .80; P(B) = .75; P(B|A) = .90
Required: P(A  B) = ? and P(A  B) = ?
P(A  B) = P(B|A) P(A) = (.90) (.80) = .72
P(A  B) = P(A) + P(B) - P(A  B) = .80 + .75 - .72 = .83
Question 3



The Rockweed Aircraft Corporation is redesigning the
instrument panel for its new F-222 steam propelled fighter.
After interviewing a large number of pilots, they decided to
replace certain gauges with idiot (warning) lights.
One such light warns of engine trouble. There is a probability
of .01 that there will be engine trouble for a single mission of
this aircraft.




Given there is engine (boiler) trouble, there is a .99 probability
that the idiot light turns on.
If there is no engine trouble during the mission, there is a .98
probability that the idiot light will not turn on.
What is the probability that the engine light turns on during a
mission.
If the engine light turns on during a mission, what is the
probability of engine trouble?
Question 3 (continued)
Let T = the event, engine trouble
O = the event, engine light turns on
Given: P(T) = .01; P(O|T) = .99; P(Oc|Tc) = .98
Required: P(O) and P(T|O)
The Total Probability Rule (TPR):
P(O) = P(O|T) P(T) + P(O|Tc) P(Tc)
P(O) = .99 (.01) + .02 (.99) = .0297
Then Bayes:
P(T|O) = P(O|T) P(T) / P(O) = .99 (.01) / .0297 = .333
Question 4
The table summarizes the GPA and statistics aptitude test results
of students admitted to the grad engineering programs.
Based upon these historical numbers, determine the relative
frequency probability that an individual applying for a grad
engineering program:
(a) will score low on the aptitude test and have a GPA above 3.5
(b) will score high on the aptitude test if individual has a GPA
between 3.0 and 3.5
(c) Show that test scores and GPA are not independent events
Entering Grade Point Average (GPA)
Aptitude
High
test results Medium
Low
Below 3.0
3.0 – 3.5
Above 3.5
0
17
13
5
40
5
15
3
2
Question 4
(a) will score low on the aptitude test and have a GPA above 3.5
(b) will score high on the aptitude test if individual has a GPA
between 3.0 and 3.5
(c) Show that test scores and GPA are not independent events
(a) P(score low and GPA > 3.5) = 2/100
(b) P(score high | 3<GPA<3.5) = 17/60
(c) P(High|GPA<3) = 0  P(High) = 30/100
Entering Grade Point Average (GPA)
Below 3.0
3.0 – 3.5
Above 3.5
0
17
13
30
5
40
5
50
Low
15
3
2
20
total
20
60
20
Aptitude
High
test results Medium
Question 5
The probability that a missile fired at a terrorist target will
destroy the target is .6. Assuming independence, what is the
minimum number of missiles to be launched against the
target in order to have at least a 90 percent chance of
destroying the target?
Let X = the discrete random variable, f ( x)  .6 .4  x 1 , F ( x)  1  .4  x
the number of missiles fired until
1
Pr{ X  1}  F (1)  1  .4   .60
the target is destroyed.

X ~ Geo(.6)
Pr{ X  2}  F (2)  1  .4   .84
2
Pr{ X  3}  F (3)  1  .4   .936
3
Pr{ X  4}  F (4)  1  .4   .9744
4
Pr{ X  5}  F (5)  1  .4   .98976
5
Follow-on Question 5

The probability that a missile fired at a terrorist target will
destroy the target is .6. Assuming independence, what is the
probability of destroying the target with the 3rd missile?
What is the expected number of missiles needed to destroy
the target?
Let X = the discrete random variable,
the number of missiles fired until
the target is destroyed.
X ~ Geo(.6)
f ( x)  .6 .4 
x 1
, F ( x)  1  .4 
Pr{ X  3}  f (3)  .6 .4 
1
E[ x]   1.667
.6
31
x
 .096
Bonus Question #2

The Catastrophic Construction Company provides emergency repair of
facilities damaged during natural disasters such as hurricanes and
tornados. They respond primary to requests from FEMA (Federal
Emergency Management Agency). The number of such requests per
year is a random variable best described by the following probability
mass function (PMF):
2
x  4

f ( x) 
31
, x  0,1, 2,3, 4,5
E[X] = 0 (16/31) + 1(9/31) + 2(4/31) + 3(1/31) + 4(0/31) + 5(1/31) = 25/31 = .806
E[X2] = 0 (16/31) + 1(9/31) + 4(4/31) + 9(1/31) + 16(0/31) + 25(1/31) = 1.903
V[X] = 1.903 - .8062 = 1.253,  = 1.12
note correction
Question 6
The Loose Screw Company manufactures nuts and bolts for sale
to wholesale distributors. The delivery time (lead-time) for
orders in days is given by the following Probability Density
Function (PDF):
f(x) = x/300 , 5  x  25
Find the probability that delivery will be within one standard
deviation of the mean.
Question 6
x
x
z
z
x 2  25
f ( x) 
, 5  x  25; F ( x)  
dz 

300
300
600 5
600
5
x
25
x
x
15625  125
dx


 17.222
5 300 900
900
5
25
E[ X ] 
2
3
25
x
x
390625  625
dx


 325
5 300 1200
1200
5
25
E[ X 2 ] 
2
3
4
 2  325  17.2222  28.402,   5.3294
Pr{17.222  5.3294  x  17.222  5.3294}  Pr{11.89  x  22.55}
22.552  25 11.892  25
 F (22.55)  F (11.89) 

 .80584  .19395  .612
600
600
f(x) = x/300 , 5  x  25
The Geometry of Question 6
25/300
f(x)
20/300
5/300
x
5
25
Area = (20)(5/300) + (1/2)(20)(20/300) =1
F ( x) 
5
1
 x 5
x

5

x

5

 


300
2
 300 
(5)(2)  x  5    x  5  x  5  x 2  25


600
600
x
Question 7

Given the following joint probability mass function,
find the correlation between X and Y.
X/Y
0
1
totals
E[Y] =
E[XY] =
covar =
y
0
1
2
f(y)
.3
.4
.3
E[Y] = 0 (.3) + 1(.4) + 2(.3) = 1.0
V[Y] = 0 (.3) + 1(.4) + 4(.3) - 12 = .6
X/Y
0
1
2
0
.1
.3
.2
1
.2
.1
.1
0
0.1
0.2
1
0.3
0.1
2 totals
0.2 0.6
0.1 0.4
0.3
0.4
0.3
1 E[Y^2] =
1.6
0.3
-0.1 Cor[XY] = -0.263523
E[X]
E[X^2}
0.4
0.4
V[X] =
0.24
V[Y] =
0.6
Question 8

Ms. Ima Borne Loser makes correct decisions 10 percent of
the time. How many (independent) decisions must she
make to have at least a 90 percent chance of making at
least one correct decision? Hint: let Ei = the event, the ith
decision is correct.
Given: P( Ei )  .10, therefore P( Eic )  .9
Required : P( E1  E2  ...  En )  .9, find n
P( E1  E2  ...  En )  1  P( E1c  E2c  ...  Enc )
 1  P( E1c )  P( Enc )  1  .9n  .9
.9n  .1
ln .1
n ln .9  ln .1 or n 
 21.85; n  22
ln .9
Question 8 – alternate approach
Ms. Ima Borne Loser makes correct decisions 10 percent of
the time. How many (independent) decisions must she
make to have at least a 90 percent chance of making at
least one correct decision?
Let X = a discrete RV, the number of correct decisions in n
trials


X
B(n,.1)
Pr  X  1  1  Pr  X  0  1  (1  .1) n  1  .9n  .9
n
10
15
20
25
Pr{X  1}
.6513
.7941
.8784
.9282
n
21
22
Pr{X  1}
.8906
.9015
Question 9
Thirty percent of the Fly-By-Nite Airline Company’s flights are
delayed. If Mr. I. N. Hurrie is scheduled on 5 flights as he
travels to each of the company’s 4 technical centers, what is
the probability of no more than one flight being delayed?
Assume Independence.
Let X = a discrete RV, the number of delayed flights among the 5
X ~ B(5, .3)
Pr{X  1} = f(0) + f(1) = F(1) = .528 (from the Prob Calculator)
What is the expected number of delayed flights? E[X] = (5)(.3) = 1.5
Question 10

Given the following CDF, find the mean of the
probability distribution.
x2
F ( x) 
, 0  x  10
100
dF ( x) x
f ( x) 

dx
50
10
3 10
x
x
1000
E[ X ]   x dx 

 6.67
50
150 0
150
0
Follow-on Question 10

Given the following CDF, find the variance of the
probability distribution.
x2
F ( x) 
, 0  x  10
100
10
4
10
x
x
10000
E[ X ]   x
dx 

 50
50
200 0
200
0
2
2
V [ X ]  50   6.6667   5.55511;   2.3569
2
Follow-on to the Follow-on of
Question 10

Given the following CDF, find the mean and variance
of the probability distribution.
Isn’t this just a right
triangular distribution
with b = 10?
x2
x
F ( x) 
, 0  x  10; f ( x) 
100
50
2x
f ( x)  2 0  x  b
b
x2
F ( x)  2
2
Var[X]
=
2.357
b
from Prob Calc
2
Beta ( =2 =1)
E[ X ]  b
3
Question 11

Professor I. Do Little’s research (NSF Grant) on the migration
pattern of the coastal plain swamp sparrow has determined
that their migration (flying) time is normally distributed with a
mean of 23 days and a variance of 36 days. Compute the
probability that a particular swamp sparrow will migrate
between 30 to 36 days.
X n(23,36)
Pr{30  X  35}  F (35)  F (30)
35  23 
30  23 


 Pr  z 
  Pr  z 
  Pr  z  2  Pr  z  1.17
6 
6 


 .9772  .8783  .0989
Follow-on Question 11


Professor I. Do Little’s research (NSF Grant) on the migration
pattern of the coastal plain swamp sparrow has determined
that their migration (flying) time is normally distributed with a
mean of 23 days and a variance of 36 days.
90 percent of the sparrow population will complete their
migration in how many days?
X n(23,36)
Pr{ X  x}  .90
x  23 

 Pr  z 
  .90; z  1.2816
6 

x    z  23  1.2816  6   30.6896 days
Question 12

Ted E. Bare, an engineering student, has observed that the
number of times the campus police check the parking lot has
a Poisson distribution with a mean of once every 4 hours. If
Ted E. is parked illegally during a 3-hour evening class, what
is the probability he will get ticketed?
Let X = a discrete RV, the number of times in a 3-hour period, the
Campus police will check the parking lot.
X
Pois (  3 x .25)
Pr  X  1  1  f (0)  1 
e
.75
.75
0!
0
 1  e .75  .5276
Follow-on to Question 12

At break time, one and half hours into the class, Ted checks
and sees that he did not receive a ticket as yet. What is the
probability that he will get ticketed by the time class is out?
Let X = a discrete RV, the number of times in a 1.5-hour period, the
Campus police will check the parking lot.
X
Pois (  1.5 x .25)
Pr  X  0 
e
.375
.375
0!
Prob get ticket
= 1 - .6873 = .3127
0
 e .375  .6873
Also, let Y = a continuous RV, the time to the next arrival
of the campus police
Y ~ Exp( = .25) and F(y) = 1 – e-.25y
Therefore Pr{Y<1.5} = 1 – F(1.5) = 1 - e-.25(1.5) = .3127
Question 13

Dawn E. Brook is an undergraduate student who works parttime in the Engineering Management office. Among her duties
is to make coffee for the faculty and staff. The time it takes
the old coffee pot to make coffee is a random variable having a
Weibull distribution with  = 2 and  = 20 minutes. If the
coffee pot perks for less than 16 minutes, the coffee will be too
weak and if the coffee pot perks for more than 23 minutes, the
coffee will be too strong. What is the probability that Dawn E.
Brook will brew a pot of coffee to the faculty’s liking?
X
Weib(2, 20)
 Pr 16  X  23  F (23)  F (16)  .73353  .47271  .261
Alternate Question 13

Dawn E. Brook is an undergraduate student who works parttime in to the Engineering Management office. Among her
duties is to make coffee for the faculty and staff. The time it
takes the old coffee pot to make coffee is a random variable
having the following cumulative distribution (CDF):
F ( x)  1  e


 x 
 
 20 
2
, x0
If the coffee pot perks for less than 16 minutes, the coffee will
be too weak and if the coffee pot perks for more than 23
minutes, the coffee will be too strong. What is the probability
that Dawn E. Brook will brew a pot of coffee to the faculty’s
liking? F (23)  F (16)  .73353  .47271  .261
What is the mean brewing time? X
Weib(2, 20)
E[ X ]  17.72 min . from Prob Calc
Question 14

Laye Z. Jones has been given three tasks to complete within
the next 8 hours. The time for Laye Z. to complete each task
based upon past performance is normally distributed with the
following parameters:
What is the probability that Laye Z.
will complete all 3 task on time.
Y  X1  X 2  X 3
Y
Task Mean
Variance
1
3.4 hr.
2.4
2
4.8 hr.
3.8
3
2.5 hr.
1.8
Totals 10.7
8.0
n(10.7,8)
8  10.7 

Pr Y  8  Pr  z 
  Pr  z  .9546  .1699
8 

100
E[ X ] 

0
 3x 2
x 6
 10
100
4

3
3
x
3
 dx  6  x dx 
6
10
4
x
10

0
Question 15

100

0
300
 75 days
4
5000
 66.67  75
75
The time it takes to complete an engineering design project in
days is a random variable having the following probability
density function (PDF):
 3x 2
for 0  x  100 days

f ( x)   106
0
otherwise

The profit resulting from completion of the project
is given by $5,000 / X. Find the expected profit.
100
E[Pr ofit ] 

0

15, 000 x
2 x 106
2 100
0
100
2
15000
 5000   3x 
x dx

  6  dx 
6

10 0
 x   10 

15, 000
 $75
2 x 100
Bonus Question #3

Given the following very fine probability density function (PDF) where
the random variable X is Professor Domkoff ’s driving time in hours to
school, find the (a) mean and the (b) probability that the driving time is
no more than 90 minutes.
2
f ( x)  2 , 1  x  2
x
2
2x
2
   2 dx  2 ln x 1  2 ln 2  2 ln1  1.3863 hr
x
1
x
2
2
2 2
2
F ( x)   2 dz 

  2
z
z 1
x 1
x
1
x
F (1.5)  2 
2
 .667
1.5
Problem 16 – Descriptive Stats &
point estimation
The following random sample was obtained by measuring the time
in (working) hours to complete a particular construction job.
Treating the data as continuous, answer the following questions:
(a) Find a sample estimate for the population mean
(b) Find a sample estimate for the population variance
(c) Is the data skewed left, right, or almost symmetrical?
(d) Find the sample interquartile range.
(e) Find the sample median.
(f) Based only on a histogram, which of the following distributions
are not likely models for the population distribution?
Normal
Weibull
Rectangular
Exponential
Triangular
A Truly Great Set of Data
83.8
64.2
89.8
82.3
90.4
63.3
40.4
104
108
96.6
65.4
98.1
46.8
86.9
71.8
56.2
72.1
73.7
77.2
113
135
56.4
99.8
64.6
95.7
85.3
75.8
88.5
71.7
72
99.7
49.1
98.9
85.2
110
68.4
123
58.6
74.1
67.9
66.5
44.5
55.6
136
91.7
Task Time in Hours
30.9
76.7
36.8
48
71.5
Problem 16
sample size
mean
variance
std dev
median
1st quartile
3rd quartile
interquatile range
Mimimum
Maximum
Range
Skewness
Kurtosis
50
78.420
569.696
23.868
74.95
64.3
95.7
31.4
30.9
135.7
104.8
0.3234
-0.0294
sturges rule =
interval width =
6.606601
17.46667
6
17
18
16
14
12
10
8
6
4
2
0
47
64
81
98
115
132
149
From the histogram rule out the
rectangular and exponential
- also compare mean and std. dev.
some skewness to the right
Keys to successful completion
of the midterm

Work every problem



if unable to complete – submit a partial or intermediate answer
state any assumptions, identify type of problem
anything submitted correctly about problem will earn some credit


Apply common sense to your answers



probabilities cannot exceed one, variances are positive
does the magnitude of the number make sense
Use the computer and Prob Calculator
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no credit for simply restating the problem
saves time, less chance of an error
Start by defining events or RV’s
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write down what’s given
follow up with what is required
Key Things to know
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Recognize problem type
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Recognize types of random events
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random events versus random variables
dependent versus independent events
conditional versus unconditional probability
total probability problem (weighted average)
Bayes problem
Recognize probability distributions
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theoretical density and distribution functions
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i.e. Poisson or uniform PMF; exponential or Weibull PDF
distinguish between binomial and geometric
joint versus marginal versus condition distributions
The Important Distributions
The discrete ones
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Uniform
Binomial
Geometric
Poisson
The continuous ones
 Rectangular
 Normal
 Exponential
 Weibull
 Triangular
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Left triangular
Right triangular
Key things to expect
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A combinatorial problem or an a priori probability
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A random event problem
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find mean, median, variance, probabilities, median
A joint distribution – most likely discrete
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conditional or Bayes
independent
A binomial, geometric or both
A normal distribution problem
A general distribution
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# favorable /total
use of counting methods
find marginal, conditional – probability, mean, variance, correlation
A problem requiring integration
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find a mean or probability