Probability and Statistics
 Basic Probability
 Binomial Distribution
 Statistical Measures
 Normal Distribution
Probability and Statistics
1
FE Reference Handbook
 Published by the National Council of Examiners for
Engineering and Surveying (NCEES)
 Available electronically at exam
 Only reference material allowed at exam
 Free preview copy (PDF) available: ncees.org
Probability and Statistics
2
Probability of an Event
Event … a possible outcome of a trial (experiment)
𝐴 = event
0 ≤ Pr(𝐴) ≤ 1
Examples:
𝐴
sun rising in the east tomorrow morning
getting heads when flipping a coin
baby being born as female
Dr. Kinman winning Dancing with the Stars
Probability and Statistics
Pr(𝐴)
1
0.5
≅ 0.5
0
3
Probability as a Percentage
Pr 𝐴 = 0.7
can also be stated as
The probability of 𝐴 is 70%.
Probability and Statistics
4
Equally Likely Events
We can infer the probabilities of events when all events are
equally likely.
Examples:
experiment
flipping a coin
example probability
Pr(heads) =
tossing a die
Selecting a card from a complete deck*
1
2
1
Pr 1 =
6
1
Pr(2♠) =
52
*but no jokers in the deck
Probability and Statistics
5
Complement of an Event
𝐴 is the complement of 𝐴
Pr 𝐴 = 1 − Pr(𝐴)
Pr 𝐴 + Pr 𝐴 = 1
𝐴
𝐴
Venn diagram
Examples:
𝐴
sun rising in the east tomorrow morning
getting heads when flipping a coin
Probability and Statistics
Pr(𝐴) Pr(𝐴)
1
0
0.5
0.5
6
Composite Event
Composite event formed from 2 or more component events
Examples:
Component events
Composite event
𝐴 = Jack solves problem
𝐵 = Jill solves problem
𝐴 = pump works
𝐵 = pipe intact
Probability and Statistics
𝐴 or 𝐵
𝐴 and 𝐵
7
𝐴 and 𝐵
Pr(𝐴 and 𝐵) = Pr 𝐴 ∙ Pr 𝐵
∗
Example:
𝐴 = pump works, Pr 𝐴 = 0.9
𝐵 = pipe intact, Pr 𝐵 = 0.8
Pr(𝐴 and 𝐵) = Pr 𝐴 ∙ Pr 𝐵 = 0.9 0.8 = 0.72
*This
is true as long as the 𝐴 and 𝐵 are statistically independent.
Probability and Statistics
8
𝐴 or 𝐵
Pr(𝐴 or 𝐵) = Pr 𝐴 + Pr 𝐵 − Pr(𝐴 and 𝐵)
Example:
𝐴 = Jack solves problem, Pr 𝐴 = 0.6
𝐵 = Jill solves problem, Pr 𝐵 = 0.6
Pr(𝐴 or 𝐵) = Pr 𝐴 + Pr 𝐵 − Pr(𝐴 and 𝐵)
= 0.6 + 0.6 − (0.6)(0.6)
= 0.84
Probability and Statistics
9
Pr(𝐴 or 𝐵) Alternate Solution
Complement of (𝐴 or 𝐵) → (𝐴 and 𝐵)
Pr(𝐴 or 𝐵) = 1 − Pr 𝐴 and 𝐵
Pr(𝐴 or 𝐵) = 1 − Pr(𝐴) ∙ Pr(𝐵)
Example:
𝐴 = Jack doesn’t solve problem, Pr 𝐴 = 0.4
𝐵 = Jill doesn’t solve problem, Pr 𝐵 = 0.4
Pr(𝐴 or 𝐵) = 1 − Pr 𝐴 ∙ Pr 𝐵
= 1 − (0.4)(0.4)
= 0.84
Probability and Statistics
10
𝐴 or 𝐵 or 𝐶
Pr(𝐴 or 𝐵 or 𝐶) = 1 − Pr(𝐴) ∙ Pr(𝐵) ∙ Pr(𝐶)
Example:
𝐴 = Moe has a watch,
𝐵 = Larry has a watch,
𝐶 = Curly has a watch,
Pr 𝐴 = 0.8
Pr 𝐵 = 0.7
Pr 𝐶 = 0.6
Pr(someone has a watch) = 1 − (0.2)(0.3)(0.4)
= 0.976
Probability and Statistics
11
𝐴 or 𝐵 with Pr(𝐴 and 𝐵) = 0
(𝐴 and 𝐵 are mutually exclusive)
𝐴
𝐵
Pr(𝐴 or 𝐵) = Pr 𝐴 + Pr 𝐵 , when Pr(𝐴 and 𝐵) = 0
Example: Roll one die and get …
𝐴 = face 1 is up,
𝐵 = face 2 is up,
Pr 𝐴 = 1 6
Pr 𝐵 = 1 6
Pr(𝐴 or 𝐵) = Pr 𝐴 + Pr 𝐵 = 1 6 + 1 6 = 1 3
Probability and Statistics
12
Basic Probability
from NCEES, FE Reference Handbook
Probability and Statistics
13
A coin is flipped twice. What is the probability that we get heads
both times?
A. 1 4
B. 1 2
st
𝐴 = 1 toss heads
C. 3 4
𝐵 = 2nd toss heads
D. 1
Pr(𝐴 and 𝐵) = Pr(𝐴) ∙ Pr(𝐵)
Pr(𝐴 and 𝐵) =
1 1
∙
2 2
=
1
4
A
Probability and Statistics
14
A die is tossed. What is the probability that the result is an odd
number?
A. 1 6
B. 1 3
𝐴 = face 1 is up
C. 1 2
𝐵 = face 3 is up
D. 1
𝐶 = face 5 is up
Events 𝐴, 𝐵 and 𝐶 are mutually exclusive:
Pr(𝐴 or 𝐵 or 𝐶) = Pr 𝐴 + Pr 𝐵 + Pr 𝐶
1
1
1
= + +
=
6
1
2
6
6
C
Probability and Statistics
15
A coin is flipped twice. What is the probability that there is at
least one head?
A.
B.
C.
D.
𝐴 = 1st toss is heads
𝐵 = 2nd toss is heads
WRONG:
Pr(𝐴 or 𝐵) = Pr 𝐴 + Pr 𝐵
1
1
Pr(𝐴 or 𝐵) = + = 1
2
1 4
1 2
3 4
1
2
D (WRONG)
This is wrong because events 𝐴 and 𝐵 are not mutually exclusive!
Probability and Statistics
16
A coin is flipped twice. What is the probability that there is at
least one head?
A. 1 4
B. 1 2
C. 3 4
D. 1
The complement of at least one head is “no heads”.
Pr(at least one head) = 1 − Pr(no heads)
Pr no heads = Pr(1st toss tails) ∙ Pr(2nd toss tails)
1 1
1
= ∙ =
2 2
4
Pr(at least one head) = 1
1
−
4
=
3
4
Probability and Statistics
C
17
A coin is flipped twice. What is the probability that either the
1st toss is heads or the 2nd toss is tails?
The event “1st toss heads” and the event
“2nd toss tails” are not mutually exclusive.
A.
B.
C.
D.
1 4
1 2
3 4
1
The complement of the desired composite event is
1 1
1
“1st toss tails and 2nd toss heads”, whose probability is ∙ = .
2 2
4
The desired composite event therefore has the probability 1 −
1
3
= .
C
4
4
Probability and Statistics
18
From a standard deck of cards (with no jokers), 4 cards are
selected at random. What is the probability that all 4 are
aces?
A. 1.4 × 10−7
B. 3.7 × 10−6
C. 8.9 × 10−6
D. 4.3 × 10−5
Pr(1st is ace) ∙ Pr 2nd is ace ∙ Pr 3rd is ace ∙ Pr 4th is ace =
4 52 ∙
∙
∙
3 51
1 49
2 50
≅ 3.7 × 10−6
B
Probability and Statistics
19
Probability and Statistics
 Basic Probability
 Binomial Distribution
 Statistical Measures
 Normal Distribution
Probability and Statistics
20
Factorial 𝑛!
0! = 1
1! = 1
2! = 2 ∙ 1 = 2
3! = 3 ∙ 2 ∙ 1 = 6
4! = 4 ∙ 3 ∙ 2 ∙ 1 = 24
⋮
𝑛! = 𝑛 ∙ 𝑛 − 1 ∙ 𝑛 − 2 ∙ ⋯ 1
Probability and Statistics
21
Combinations
from NCEES, FE Reference Handbook
Probability and Statistics
22
Computing 𝐶(𝑛, 𝑟)
𝑛!
𝑛 ∙ 𝑛 − 1 ∙ 𝑛 − 2 ∙ ⋯ (𝑛 − 𝑟 + 1)
𝐶 𝑛, 𝑟 =
=
𝑟! 𝑛 − 𝑟 !
𝑟!
examples:
5!
5
𝐶 5,1 =
= =5
1! 5 − 1 ! 1!
9!
9∙8∙7∙6
𝐶 9,4 =
=
= 126
4! 9 − 4 !
4!
special cases:
𝐶 𝑛, 0 = 1
𝐶 𝑛, 1 = 𝑛
Probability and Statistics
23
Binomial Distribution
from NCEES, FE Reference Handbook
Probability and Statistics
24
Pascal’s Triangle: 𝐶(𝑛, 𝑟) from a Diagram
1
𝑛 =1
1
𝑛 =2
1
𝑛 =3
1
𝑛 =4
𝑛 =5
𝑟 =0
1
1
2
3
4
5
1
1
3
6
10
1
4
10
1
5
1
𝑟 =1
Probability and Statistics
25
When Binomial Distribution is Used
1. Binary outcomes: 𝐴 or 𝐵 with Pr 𝐴 + Pr 𝐵 = 1
2. Repeated trials of with binary outcomes
3. Underlying probabilities Pr 𝐴 and Pr 𝐵 do not
change.
Probability and Statistics
26
Examples where Binomial Distribution is Used
1.
What is the probability of 𝑟 heads in 𝑛 flips of a coin?
Pr(heads) + Pr(tails) = 1
2.
A device comes off an assembly line and works with
probability 𝑝, and doesn’t work with probability 1 − 𝑝.
What is the probability that 𝑟 of 𝑛 such devices work?
𝑝+ 1−𝑝 =1
3.
What is the probability that a die lands with the 1 face up
𝑟 times in 𝑛 tosses?
Pr(1) + Pr(not 1) = 1
Probability and Statistics
27
A coin is flipped 4 times. What is the probability of (exactly) 3
heads?
1
Pr(heads):
A.
1
𝑝=
2
1
𝑞=
2
B.
Pr(tails):
𝑛 = 4 and 𝑟 = 3
C.
D.
8
1
4
3
8
1
2
Pr(3 heads) = 𝐶(4,3) ∙ 𝑝3 ∙ 𝑞1
1
1
1
1
1
1
2
3
4
3
6
or
1
𝐶 4,3 =
4∙3∙2
3!
=4
1
4
1
Pr(3 heads) = 4 ∙
B
Probability and Statistics
1 3
2
∙
1 1
2
=
28
1
4
A device comes off an assembly line and works with probability 0.8,
and doesn’t work with probability 0.2. What is the probability that
exactly 1 of 4 such devices works?
A. 0.256
Pr(good) = 0.8
B. 0.408
Pr(bad) = 0.2
C. 0.500
𝑛 = 4 and 𝑟 = 1
D. 0.742
Pr(1 good) = 𝐶(4,1) ∙ (0.8)1 ∙ (0.2)3
𝐶 𝑛, 1 = 𝑛 ⇒ 𝐶 4,1 = 4
Pr(1 good) = 4 ∙ 0.8
1
∙ 0.2
3
= 0.256
A
Probability and Statistics
29
A die is tossed 10 times. What is the probability that the die lands
with the 1 face up exactly one time?
Pr(1) =
1
6
A.
B.
C.
D.
5
6
Pr(not 1) =
𝑛 = 10 and 𝑟 = 1
Pr(one time) = 𝐶(10,1) ∙
0.184
0.230
0.323
0.417
1 1 5 9
( ) ∙( )
6
6
𝐶 10,1 = 10
Pr(one time) = 10 ∙
C
1 1
6
∙
5 9
6
= 0.323
Probability and Statistics
30
Ten percent of the parts in a large bin are bad. If 5 parts are
selected at random, what is the probability that at least 4 of the
selected parts will be good?
A. 0.631
Pr(good) = 0.9
B. 0.720
Pr(bad) = 0.1
C. 0.853
𝑛=5
D. 0.919
Consider 𝑟 = 4 and 𝑟 = 5
Pr(4 good) = 𝐶(5,4) ∙ (0.9)4 ∙ (0.1)1
Pr(5 good) = 𝐶(5,5) ∙ (0.9)5 ∙ (0.1)0
𝐶 5,4 = 5 and 𝐶 5,5 = 1
“4 good” and “5 good” are mutually exclusive:
Pr(at least 4 good) = Pr(4 good) + Pr(5 good)
Pr(at least 4 good) = 0.919
Probability and Statistics
D
31
A coin is flipped 7 times. What is the probability that the number
of heads is fewer than 7?
Pr(heads) = 0.5
Pr(tails) = 0.5
𝑛=7
Pr(fewer than 7 heads) =
6
𝑟=0 𝐶(7, 𝑟)
∙
1 7
2
Pr(fewer than 7 heads) = 1 − Pr(7 heads)
Pr(7 heads) = 𝐶(7,7) ∙
A.
B.
C.
D.
0.889
0.956
0.992
0.999
← hard
← easy
1 7
2
𝐶 7,7 = 1
Pr(fewer than 7 heads) = 0.992
Probability and Statistics
C
32
Probability and Statistics
 Basic Probability
 Binomial Distribution
 Statistical Measures
 Normal Distribution
Probability and Statistics
33
The “Middle” of a Set of Measured Values
Mean: the average of the numbers
Mode: the value that occurs most often
Median: the middle value
Example:
Measured values: 17, 9, 12, 14, 13, 18, 12, 15
Reordered values: 9, 12, 12, 13, 14, 15, 17, 18
Mean = (9 + 12 + 12 + 13 +14 + 15 + 17 + 18)/8 = 13.75
Mode = 12
Median = 13.5
Probability and Statistics
34
Mean
from NCEES, FE Reference Handbook
Probability and Statistics
35
Sample Variance
from NCEES, FE Reference Handbook
Probability and Statistics
36
Sample Variance for a Set of Measured Values
Example:
Measured values: 17, 9, 12, 14, 13, 18, 12, 15
𝑛 =8
𝑋 = 13.75
sample variance =
1
𝑛−1
𝑛
𝑖=1
sample standard deviation =
𝑋𝑖 − 𝑋
2
= 8.50
sample variance = 2.92
Probability and Statistics
37
Population Variance
from NCEES, FE Reference Handbook
Probability and Statistics
38
Standard Deviation
from NCEES, FE Reference Handbook
Probability and Statistics
39
Sample Variance vs. Population Variance
For both variances we calculate the difference between each value
and a mean, then we square the differences and sum them, then
we divide by a number.
Sample Variance
The mean is the sample mean, which is the
average of the sample values.
We divide by 𝑛 − 1.
Population Variance The mean is a model mean.
We divide by 𝑛.
Probability and Statistics
40
We have measured the following values:
17, 9, 12, 14, 13, 18, 12, 15
The mean has been modeled as 12.5. What is the population
variance?
A. 8.5
B. 9.0
C. 9.5
D. 9.9
𝑛 =8
μ = 12.5
population variance =
1
𝑛
𝑛
𝑖=1
𝑋𝑖 − 𝜇
2
= 9.0
Probability and Statistics
B
41
Linear Regression (Least-Squares Straight Line)
from NCEES, FE Reference Handbook
Probability and Statistics
42
Find the slope of the linear regression of the following data:
x
1.19
y
2.79
2.35
1.96
3.29
5.03
5.13
6.25
A.
B.
C.
D.
1.097
1.565
1.972
2.281
𝑥 = 2.196
𝑦 = 4.801
𝑆𝑥𝑥 = 2.280
𝑆𝑥𝑦 = 3.568
𝑎 = 1.364 y-intercept
𝑏 = 1.565 slope
𝑦 = 𝑎 + 𝑏𝑥
Probability and Statistics
B
43
𝑦 = 𝑎 + 𝑏𝑥
Probability and Statistics
44
Probability and Statistics
 Basic Probability
 Binomial Distribution
 Statistical Measures
 Normal Distribution
Probability and Statistics
45
Normal (Gaussian) Distribution
from NCEES, FE Reference Handbook
Probability and Statistics
46
Typical Problem with Normal Distribution
A physical quantity (for example, a pressure or temperature) has
been measured many times. The quantity is thought to be
unchanging, but the measured values are different because of noise
in the measurement process. The measured values will often be
modeled as having a normal distribution with mean 𝜇 and variance
𝜎 2 (or, equivalently, a standard deviation of 𝜎). We want to answer
questions about the next (measured) value, 𝑋𝑛+1 .
Probability and Statistics
47
Some Problems with Normal Distribution
Type I Problem:
𝑑
A
𝑑
A
𝜇
Pr 𝐴 = 𝑅 𝑑 𝜎
B
𝑑
Type II Problem:
𝜇
Pr 𝐵 = 𝑅 𝑑 𝜎
B
𝑑
Type III Problem:
Pr(𝐴 or 𝐵) = 2𝑅 𝑑 𝜎
𝜇
Probability and Statistics
48
Type I Problem:
𝑅 𝑥 = area under the curve to the right of 𝑥
Pr(𝑋𝑛+1 > 𝜇 + 𝑑) = 𝑅(𝑑 𝜎)
Type II Problem:
Pr(𝑋𝑛+1 < 𝜇 − 𝑑) = 𝑅(𝑑 𝜎)
Probability and Statistics
49
Type III Problem:
𝑅 𝑥 = area under the curve to the right of 𝑥
Pr(𝑋𝑛+1 < 𝜇 − 𝑑 or 𝑋𝑛+1 > 𝜇 + 𝑑) = 2𝑅(𝑑 𝜎)
Probability and Statistics
50
Unit Normal Distribution Table
from NCEES, FE Reference Handbook
Probability and Statistics
51
A set of measured values are modeled as having a normal
distribution with mean 5.0 and variance 4.0. What is the probability
that a new value will be larger than 5.8?
A. 0.15
B. 0.22
Type I Problem
C. 0.28
D. 0.34
𝜇 = 5.0
𝜎 = 4.0 = 2.0
𝑑 = 5.8 − 𝜇 = 0.8
𝑑
𝑥 = = 0.4
𝜎
𝑅 0.4 = 0.3446
Probability(𝑋𝑛+1 > 5.8) ≅ 0.34
Probability and Statistics
D
52
A set of measured values are modeled as having a normal
distribution with mean 7.0 and standard deviation 2.7. What is the
probability that a new value will be smaller than 6.2?
A.
B.
C.
D.
Type II Problem
𝜇 = 7.0
𝜎 = 2.7
𝑑 = 𝜇 − 6.2 = 0.8
𝑑
𝑥 = = 0.3
0.17
0.25
0.38
0.45
𝜎
𝑅 0.3 = 0.3821
Probability(𝑋𝑛+1 < 6.2) ≅ 0.38
Probability and Statistics
C
53
A set of measured values are modeled as having a normal
distribution with mean 5.6 and variance 4.0. What is the probability
that a new value will be at least 0.4 away from the mean (in either
direction)?
A. 0.84
Type III Problem
B. 0.90
C. 0.95
𝜇 = 5.6
D. 0.99
𝜎 = 4.0 = 2.0
𝑑 = 0.4
𝑑
𝑥 = = 0.2
𝜎
2𝑅 0.2 = 0.8415
Probability(𝑋𝑛+1 < 5.2 or 𝑋𝑛+1 > 6.0) ≅ 0.84
Probability and Statistics
A
54
A set of measured values are modeled as having a normal
distribution with mean 5.6 and variance 4.0. What is the probability
that a new value will be within 0.4 of the mean?
𝜎 = 4.0 = 2.0
𝑑 = 0.4
𝑑
𝑥 = = 0.2
𝜎
2𝑅 0.2 = 0.8415
Pr(within 0.4 of mean) = 1 − 2𝑅 0.2 ≅ 0.16
A.
B.
C.
D.
0.10
0.16
0.20
0.25
B
Probability and Statistics
55
Confidence Interval
from NCEES, FE Reference Handbook
Probability and Statistics
56
Parameter for Calculation of Confidence Interval
from NCEES, FE Reference Handbook
Probability and Statistics
57
We have a set of 100 measured values for a physical quantity. We
believe a normal distribution is the correct model for these data
and that the variance is 9.0. However, the population mean has not
yet been determined. We estimate this mean by computing the
sample mean from the data, and this estimated mean is 8.3. If we
want a confidence level of 95%, what is the confidence interval for
the mean?
A. 7.512 → 9.088
B. 7.712 → 8.888
𝑛 = 100
C. 7.912 → 8.688
𝑋 = 8.3
D. 8.112 → 8.488
𝜎 = 9.0 = 3.0
𝑍𝛼
2
= 1.96 (for 95% confidence level)
𝑋 − 𝑍𝛼
𝜎
2 𝑛
≤ 𝜇 ≤ 𝑋 + 𝑍𝛼
7.712 ≤ 𝜇 ≤ 8.888
Probability and Statistics
𝜎
2 𝑛
B
58