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TRANSCRIPT
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Chapter 3:
Probability
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Section 3.1: Basic Ideas
Definition: An experiment is a process that
results in an outcome that cannot be predicted
in advance with certainty.
Examples:
rolling a die
tossing a coin
weighing the contents of a box of cereal.
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Sample Space
Definition: The set of all possible outcomes of an experiment is called the sample space for the experiment.
Examples: For rolling a fair die, the sample space is {1, 2, 3, 4, 5, 6}.
For a coin toss, the sample space is {heads, tails}.
Imagine a hole punch with a diameter of 10 mm punches holes in sheet metal. Because of variation in the angle of the punch and slight movements in the sheet metal, the diameters of the holes vary between 10.0 and 10.2 mm. For this experiment of punching holes, a reasonable sample space is the interval (10.0, 10.2).
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More Terminology
Definition: A subset of a sample space is called an
event.
A given event is said to have occurred if the
outcome of the experiment is one of the
outcomes in the event. For example, if a die
comes up 2, the events {2, 4, 6} and {1, 2, 3}
have both occurred, along with every other
event that contains the outcome “2”.
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Example 3.1
An engineer has a box containing four bolts and
another box containing four nuts. The diameters
of the bolts are 4, 6, 8, and 10 mm, and the
diameters of the nuts were 6, 10, 12, and 14
mm. One bolt and one nut are chosen.
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Example 3.1 (cont.)
Let A be the event that the bolt diameter is less than 8, let B be the event that the nut diameter is greater than 10, and let C be the event that the bolt and the nut have the same diameter.
1. Find the sample space for this experiment.
2. Specify the subsets corresponding to the events A, B, and C.
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Combining Events
The union of two events A and B, denoted
𝐴 ∪ 𝐵, is the set of outcomes that belong either
to A, to B, or to both.
In words, 𝐴 ∪ 𝐵 means “A or B.” So the event
“A or B” occurs whenever either A or B (or both)
occurs.
Example: Let A = {1, 2, 3} and B = {2, 3, 4}.
What is 𝐴 ∪ 𝐵?
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Intersections
The intersection of two events A and B, denoted
by 𝐴 ∩ 𝐵, is the set of outcomes that belong to A
and to B. In words,𝐴 ∩ 𝐵 means “A and B.”
Thus the event “A and B” occurs whenever both
A and B occur.
Example: Let A = {1, 2, 3} and B = {2, 3, 4}.
What is 𝐴 ∩ 𝐵?
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Complements
The complement of an event A, denoted 𝐴𝑐, is
the set of outcomes that do not belong to A. In
words, 𝐴𝑐 means “not A.” Thus the event “not
A” occurs whenever A does not occur.
Example: Consider rolling a fair sided die. Let A be
the event: “rolling a six” = {6}.
What is 𝐴𝑐 = “not rolling a six”?
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Mutually Exclusive Events
Definition: The events A and B are said to be mutually exclusive if they have no outcomes in common.
More generally, a collection of events A1, A2, …, An
is said to be mutually exclusive if no two of them have
any outcomes in common.
Sometimes mutually exclusive events are referred to as disjoint events.
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Probabilities
Definition:Each event in the sample space has a probability of occurring. Intuitively, the probability is a quantitative measure of how likely the event is to occur.
Given any experiment and any event A:
The expression P(A) denotes the probability that the event A occurs.
P(A) is the proportion of times that the event Awould occur in the long run, if the experiment were to be repeated over and over again.
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Axioms of Probability
1. Let 𝑺 be a sample space. Then 𝑃(𝑺) = 1.
2. For any event A, 0 ≤ 𝑃(𝐴) ≤ 1.
3. If A and B are mutually exclusive events, then
𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃(𝐵). More generally, if
𝐴1, 𝐴2, … are mutually exclusive events, then
𝑃 𝐴1 ∪ 𝐴2 ∪ ⋯ = 𝑃 𝐴1 + 𝑃 𝐴2 + ⋯
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A Few Useful Things
For any event A, 𝑃(𝐴𝐶) = 1 – 𝑃(𝐴).
Let denote the empty set. Then 𝑃( ) = 0.
If 𝑆 is a sample space containing 𝑁 equally likely
outcomes, and if A is an event containing 𝑘outcomes, then 𝑃(𝐴) = 𝑘/𝑁.
Addition Rule (for when A and B are not mutually
exclusive): 𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃(𝐴 ∩ 𝐵)
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(3.1)
(3.2)
(3.3)
(3.4)
Example 3.3
A target on a test firing range consists of a bull’s-eye
with two concentric rings around it. A projectile is
fired at the target. The probability that it hits the
bull’s-eye is 0.10, the probability that it hits the inner
ring is 0.25, and the probability that it hits the outer
ring is 0.45.
1. What is the probability that the projectile hits the
target?
2. What is the probability that it misses the target?
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Example 3.4
An extrusion die is used to produce aluminum rods.
Specifications are given for the length and diameter of the
rods. For each rod, the length is classified as too short, too
long, or OK, and the diameters is classified as too thin, too
thick, or OK. In a population of 1000 rods, the number of
rods in each class are as follows:
Diameter
Length Too Thin OK Too Thick
Too Short 10 3 5
OK 38 900 4
Too Long 2 25 13
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Example 3.4 (cont.)
1. What is the probability that a randomly chosen
rod is too short?
2. If a rod is sampled at random, what is the
probability that it is neither too short or too thick?
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Section 3.2: Conditional
Probability and Independence
Definition: A probability that is based on part of the
sample space is called a conditional probability.
Let A and B be events with 𝑃(𝐵) ≠ 0. The
conditional probability of A given B is
𝑃 𝐴 𝐵 =𝑃(𝐴 ∩ 𝐵)
𝑃(𝐵)
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(3.5)
Conditional Probability
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Venn Diagram
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Example 3.6
What is the probability that a rod will have a
diameter that is OK, given that the length is
too long?
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Diameter
Length Too Thin OK Too Thick
Too Short 10 3 5
OK 38 900 4
Too Long 2 25 13
Independence
Definition: Two events A and B are independent if the probability of each event remains the same whether or not the other occurs.
If 𝑃(𝐴) ≠ 0 and 𝑃(𝐵) ≠ 0, then A and B are independent if 𝑃(𝐵|𝐴) = 𝑃(𝐵) or, equivalently, 𝑃(𝐴|𝐵) = 𝑃(𝐴).
If either 𝑃(𝐴) = 0 or 𝑃(𝐵) = 0, then A and B are independent.
These concepts can be extended to more than two events.
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Example 3.7
If an aluminum rod is sampled from the sample
space of 1000 rods, find the P(too long) and
P(too long | too thin). Are these probabilities
different? Why or why not?
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Diameter
Length Too Thin OK Too Thick
Too Short 10 3 5
OK 38 900 4
Too Long 2 25 13
The Multiplication Rule
If A and B are two events and 𝑃(𝐵) ≠ 0, then 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐵 𝑃(𝐴|𝐵)
If A and B are two events and 𝑃(𝐴) ≠ 0, then 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 𝑃(𝐵|𝐴)
If 𝑃(𝐴) ≠ 0, and 𝑃(𝐵) ≠ 0, then both of the above hold.
If A and B are two independent events, then 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 𝑃(𝐵)
If 𝐴1, 𝐴2, … , 𝐴𝑛 are independent events, then 𝑃 𝐴1 ∩ 𝐴2 ∩ ⋯ ∩ 𝐴𝑛 = 𝑃 𝐴1 𝑃 𝐴2 …𝑃(𝐴𝑛)
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(3.8)
(3.9)
(3.10)
(3.12)
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Example 3.8
It is known that 5% of the cars and 10% of the light
trucks produced by a certain manufacturer require
warranty service. If someone purchases both a car
and a light truck from this manufacturer, then,
assuming the vehicles function independently, what
is the probability that both will require warranty
service?
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Example 3.9
A system contains two components, A and B,
connected in series. The system will function only
if both components function. The probability that A
functions is 0.98 and the probability that B
functions is 0.95. Assume A and B function
independently. Find the probability that the system
functions.
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Example 3.10
A system contains two components, C and D,
connected in parallel. The system will function if
either C or D functions. The probability that C
functions is 0.90 and the probability that D
functions is 0.85. Assume C and D function
independently. Find the probability that the system
functions.
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Example 3.11
𝑃(𝐴) = 0.995; 𝑃(𝐵) = 0.99𝑃(𝐶) = 𝑃(𝐷) = 𝑃(𝐸) = 0.95𝑃(𝐹) = 0.90; 𝑃(𝐺) = 0.90, 𝑃(𝐻) = 0.98
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Section 3.3: Random Variables
Definition:A random variable assigns a numerical
value to each outcome in a sample space.
Definition:A random variable is discrete if its
possible values form a discrete set.
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Example
The number of flaws in a 1-inch length of copper wire manufactured by a certain process varies from wire to wire. Overall, 48% of the wires produced have no flaws, 39% have one flaw, 12% have two flaws, and 1% have three flaws. Let 𝑋 be the number of flaws in a randomly selected piece of wire. Write down the possible values of 𝑋 and the associated probabilities, providing a complete description of the population from which 𝑋 was drawn.
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Probability Mass Function
The description of the possible values of 𝑋 and the probabilities of each has a name: the probability mass function.
Definition:The probability mass function (pmf) of a discrete random variable 𝑋 is the function 𝑝(𝑥) = 𝑃(𝑋 = 𝑥).
The probability mass function is sometimes called the probability distribution.
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Probability Mass Function
Example
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Cumulative Distribution Function
The probability mass function specifies the probability that a random variable is equal to a given value.
A function called the cumulative distribution function (cdf) specifies the probability that a random variable is less than or equal to a given value.
The cumulative distribution function of the random variable 𝑋 is the function 𝐹(𝑥) = 𝑃(𝑋 ≤ 𝑥).
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More on a
Discrete Random Variable
Let 𝑋 be a discrete random variable. Then
The probability mass function of 𝑋 is the function 𝑝(𝑥) = 𝑃(𝑋 = 𝑥).
The cumulative distribution function of 𝑋 is the function 𝐹(𝑥) = 𝑃(𝑋 ≤ 𝑥).
𝐹 𝑥 = 𝑡≤𝑥 𝑝 𝑡 = 𝑡≤𝑥 𝑃(𝑋 = 𝑡)
𝑥 𝑝 𝑥 = 𝑥 𝑃 𝑋 = 𝑥 = 1, where the sum is over
all the possible values of 𝑋.
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Example 3.16
Recall the example of the number of flaws in a
randomly chosen piece of wire. The following is
the pmf: 𝑃(𝑋 = 0) = 0.48, 𝑃(𝑋 = 1) = 0.39, 𝑃(𝑋 =2) = 0.12, and 𝑃(𝑋 = 3) = 0.01. Compute the cdf
of the random variable 𝑋 that represents the
number of flaws in a randomly chosen wire.
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Cumulative Distribution Function
Example 3.16
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Mean and Variance for Discrete
Random Variables
The mean (or expected value) of 𝑋 is given by
𝜇𝑋 =
𝑥
𝑥𝑃(𝑋 = 𝑥)
where the sum is over all possible values of 𝑋.
The variance of 𝑋 is given by
𝜎𝑋2 =
𝑥
(𝑥 − 𝜇𝑋)2𝑃 𝑋 = 𝑥 =
𝑥
𝑥2𝑃 𝑋 = 𝑥 − 𝜇𝑋2
The standard deviation is the square root of the
variance.
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(3.13)
(3.14) (3.15)
Example 3.17
A certain industrial process is brought down for
recalibration whenever the quality of the items
produced falls below specifications. Let 𝑋represent the number of times the process is
recalibrated during a week, and assume that 𝑋 has
the following probability mass function.
Find the mean and variance of 𝑋.
x 0 1 2 3 4
p(x) 0.35 0.25 0.20 0.15 0.05
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Example 3.17
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Probability mass function
will balance if supported at
the population mean
The Probability Histogram
When the possible values of a discrete random variable are evenly spaced, the probability mass function can be represented by a histogram, with rectangles centered at the possible values of the random variable.
The area of the rectangle centered at a value 𝑥 is equal to 𝑃(𝑋 = 𝑥).
Such a histogram is called a probability histogram, because the areas represent probabilities.
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Probability Histogram for the
Number of Flaws in a Wire
The pmf is: 𝑃(𝑋 = 0) = 0.48, 𝑃(𝑋 = 1) = 0.39,
𝑃(𝑋 = 2) = 0.12, and 𝑃(𝑋 = 3) = 0.01.
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Probability Mass Function
Example
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Continuous Random Variables
A random variable is continuous if its probabilities are given by areas under a curve.
The curve is called a probability density function(pdf) for the random variable. Sometimes the pdfis called the probability distribution.
The function 𝑓(𝑥) is the probability density function of 𝑋.
Let 𝑋 be a continuous random variable with probability density function 𝑓(𝑥). Then
−∞
∞𝑓 𝑥 𝑑𝑥 = 1
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Continuous Random Variables:
Example
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Computing Probabilities
Let 𝑋 be a continuous random variable with probability
density function 𝑓(𝑥). Let 𝑎 and 𝑏 be any two numbers,
with 𝑎 < 𝑏. Then 𝑃 𝑎 ≤ 𝑋 ≤ 𝑏 = 𝑃 𝑎 ≤ 𝑋 < 𝑏 =
𝑃 𝑎 < 𝑋 ≤ 𝑏 = 𝑎𝑏𝑓 𝑥 𝑑𝑥
In addition,
𝑃 𝑋 ≤ 𝑎 = 𝑃 𝑋 < 𝑎 = −∞
𝑎
𝑓 𝑥 𝑑𝑥
𝑃 𝑋 ≥ 𝑎 = 𝑃 𝑋 > 𝑎 = 𝑎
∞
𝑓 𝑥 𝑑𝑥
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(3.16)
(3.17)
More on
Continuous Random Variables
Let 𝑋 be a continuous random variable with probability
density function 𝑓(𝑥). The cumulative distribution
function of 𝑋 is the function
𝐹 𝑥 = 𝑃 𝑋 ≤ 𝑥 = −∞
𝑥
𝑓 𝑡 𝑑𝑡
The mean of 𝑋 is given by
𝜇𝑋 = −∞
∞
𝑥𝑓 𝑥 𝑑𝑥
The variance of 𝑋 is given by
𝜎𝑋2 =
−∞
∞
(𝑥 − 𝜇𝑋)2𝑓 𝑥 𝑑𝑥 = −∞
∞
𝑥2𝑓 𝑥 𝑑𝑥 − 𝜇𝑋2
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(3.18)
(3.19)
(3.20)
(3.21)
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Example 3.21
A hole is drilled in a sheet-metal component, and then a shaft is inserted through the hole. The shaft clearance is equal to the difference between the radius of the hole and the radius of the shaft. Let the random variable 𝑋denote the clearance, in millimeters. The probability density function of X is
𝑓 𝑥 = 1.25 1 − 𝑥4 , 0 < 𝑥 < 10, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Components with clearances larger than 0.8 mm must be scraped. What proportion of components are scrapped?
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Example 3.21
46
𝑃 𝑋 > 0.8 = 0.8
∞
𝑓 𝑥 𝑑𝑥
= 0.8
1
1.25 1 − 𝑥4 𝑑𝑥
= 1.25 𝑥 −𝑥5
5
1
0.8= 0.0819
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Example 3.22
Find the cumulative distribution function 𝐹(𝑥) .
If 𝑥 < 0: 𝐹 𝑥 = −∞
𝑥𝑓 𝑡 𝑑𝑡 = −∞
𝑥0𝑑𝑡 = 0
If 0 < 𝑥 < 1:
𝐹 𝑥 = −∞
𝑥𝑓 𝑡 𝑑𝑡
= −∞
0
𝑓 𝑡 𝑑𝑡 + 0
𝑥
𝑓 𝑡 𝑑𝑡
= 0 + 0
𝑥
1.25 1 − 𝑡4 𝑑𝑡
= 1.25 𝑡 −𝑡5
5 𝑥0
= 1.25(𝑥 −𝑥5
5)
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Example 3.22
Find the cumulative distribution function F(x) .
If 𝑥 > 1:
𝐹 𝑥 = −∞
𝑥𝑓 𝑡 𝑑𝑡
= −∞
0
𝑓 𝑡 𝑑𝑡 + 0
1
𝑓 𝑡 𝑑𝑡 + 1
𝑥
𝑓 𝑡 𝑑𝑡
= 0 + 1.25 𝑡 −𝑡5
5 10
+ 0
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Example 3.24
Find the mean and variance of the clearance.
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𝜇𝑥 = −∞
∞
𝑥𝑓 𝑥 𝑑𝑥 = 0
1
𝑥 1.25 1 − 𝑥4 𝑑𝑥 = 1.25𝑥2
2−
𝑥6
6 10
= 0.4167
𝜎𝑥2 =
−∞
∞
𝑥2𝑓 𝑥 𝑑𝑥 − 𝜇𝑥2 =
0
1
𝑥2 1.25 1 − 𝑥4 𝑑𝑥 − (0.4167)2
= 1.25𝑥3
3−
𝑥7
7 10
− (0.4167)2 = 0.0645
Section 3.4: Linear Functions of
Random Variables
If X is a random variable, and 𝑎 and 𝑏 are constants,
then
–𝜇𝑎𝑋+𝑏 = 𝑎𝜇𝑋 + 𝑏
–𝜎𝑎𝑋+𝑏2 = 𝑎2𝜎𝑋
2
–𝜎𝑎𝑋+𝑏 = |𝑎|𝜎𝑋
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(3.27)
(3.28)
(3.29)
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𝑥 = 18; 𝑥 = 0.1; 𝑌 = 0.25𝑋; 𝑌 =? 𝑌 =?
𝜇𝑌 = 𝜇0.25𝑋 = 0.25𝜇𝑋 = 4.5𝜎𝑌 = 𝜎0.25𝑋 = 025𝜎𝑋 = 0.025
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Example 3.25
More Linear Functions
If 𝑋 and 𝑌 are random variables, and a and b are
constants, then
𝜇𝑎𝑋+𝑏𝑌 = 𝜇𝑎𝑋 + 𝜇𝑏𝑌 = 𝑎𝜇𝑋 + 𝑏𝜇𝑌
More generally, if 𝑋1, … , 𝑋𝑛 are random variables
and 𝑐1, … , 𝑐𝑛 are constants, then the mean of the
linear combination 𝑐1𝑋1 + ⋯+ 𝑐𝑛 𝑋𝑛 is given by
𝜇𝑐1𝑋1+𝑐2𝑋2+⋯+𝑐𝑛𝑋𝑛= 𝑐1𝜇𝑋1
+ 𝑐2𝜇𝑋2+ ⋯ + 𝑐𝑛𝜇𝑋𝑛
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(3.31)
(3.32)
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Two Independent Random
Variables
If 𝑋 and 𝑌 are independent random variables, and 𝑆and 𝑇 are sets of numbers, then
𝑃(𝑋 ∈ 𝑆 𝑎𝑛𝑑 𝑌 ∈ 𝑇) = 𝑃(𝑋 ∈ 𝑆)P(𝑌 ∈ 𝑇)
More generally, if 𝑋1, … , 𝑋𝑛 are independent random
variables, and 𝑆1, … , 𝑆𝑛 are sets, then
𝑃(𝑋1 ∈ 𝑆1, 𝑋2 ∈ 𝑆2,…, 𝑋𝑛 ∈ 𝑆𝑛)= 𝑃(𝑋1 ∈ 𝑆1)P(𝑋2 ∈ 𝑆2) …𝑃(𝑋𝑛 ∈ 𝑆𝑛)
53
(3.33)
(3.34)
Example 3.26
Cylindrical cans have specifications regarding their height and radius. Let 𝐻 be the length and 𝑅 be the radius, in mm, of a randomly sampled can. The probability mass function of 𝐻 is given by 𝑃(𝐻 = 119) = 0.2, 𝑃(𝐻 =120) = 0.7, and 𝑃(𝐻 = 121) = 0.1. The probability mass function of 𝑅 is given by 𝑃(𝑅 = 30) = 0.6 and 𝑃(𝑅 =31) = 0.4. The volume of a can is given by 𝑉 = 𝜋𝑅2𝐻. Assume 𝑅 and 𝐻 are independent. Find the probability that the volume is 108,000 mm3.𝑃(𝑉 = 108,000) = 𝑃(𝐻 = 120 𝑎𝑛𝑑 𝑅 = 30)
= 𝑃(𝐻 = 120)𝑃(𝑅 = 30)= (0.7)(0.6) = 0.42
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Variance Properties
If 𝑋1, … , 𝑋𝑛 are independent random variables,
then the variance of the sum 𝑋1 + …+ 𝑋𝑛 is given
by 𝜎𝑋1+𝑋2+⋯+𝑋𝑛2 = 𝜎𝑋1
2 + 𝜎𝑋22 + ⋯+ 𝜎𝑋𝑛
2
If 𝑋1, … , 𝑋𝑛 are independent random variables
and 𝑐1, … , 𝑐𝑛 are constants, then the variance of
the linear combination 𝑐1 𝑋1 + ⋯+ 𝑐𝑛 𝑋𝑛 is given
by𝜎𝑐1𝑋1+𝑐2𝑋2+⋯+𝑐𝑛𝑋𝑛
2 = 𝑐12𝜎𝑋1
2 + 𝑐22𝜎𝑋2
2 + ⋯+ 𝑐𝑛2𝜎𝑋𝑛
2
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(3.35)
(3.36)
More Variance Properties
If 𝑋 and 𝑌 are independent random variables
with variances 𝜎𝑋2 and 𝜎𝑌
2, then the variance of
the sum 𝑋 + 𝑌 is 𝜎𝑋+𝑌2 = 𝜎𝑋
2 + 𝜎𝑌2
The variance of the difference 𝑋–𝑌 is 𝜎𝑋−𝑌2 =
𝜎𝑋2 + 𝜎𝑌
2
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(3.37)
(3.38)
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Example 3.27
An object with initial temperature 𝑇0 is placed in an environment with ambient temperature 𝑇𝑎. According to Newton’s law of cooling, the temperature T of the object is given by 𝑇 = 𝑐𝑇0 + (1 – 𝑐)𝑇𝑎, where 𝑐 is a constant that depends on the physical properties of the object and the elapsed time. Assuming that 𝑇0 has mean 25oC and standard deviation of 2oC, and 𝑇𝑎 has mean 5oC and standard deviation of 1oC. Find the mean of 𝑇 when 𝑐 =0.25. Assuming that 𝑇0 and 𝑇𝑎 are independent, find the standard deviation of 𝑇 at that time.
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Example 3.27
58
𝜇𝑇 = 𝜇0.25𝑇0+.075𝑇𝑎
= 0.25𝜇𝑇0+ 0.75𝜇𝑇𝑎
= 6.25 + 3.75 = 10
𝜎𝑇 = 𝜎20.25𝑇0+.075𝑇𝑎
= 0.25 2𝜎2𝑇0
+ 0.75 2𝜎2𝑇𝑎
= 0.25 2(2)2 + 0.75 2(1)2
= 0.9014
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Independence and Simple
Random Samples
Definition: If 𝑋1, … , 𝑋𝑛 is a simple random
sample, then 𝑋1, … , 𝑋𝑛 may be treated as
independent random variables, all from the
same population.
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Properties of
If 𝑋1, … , 𝑋𝑛 is a simple random sample from a
population with mean and variance 2, then the
sample mean 𝑋 is a random variable with
𝜇 𝑋 = 𝜇
𝜎 𝑋2 =
𝜎2
𝑛The standard deviation of 𝑋 is
𝜎 𝑋 =𝜎
𝑛
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(3.39)
(3.40)
(3.41)
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Example 3.28
The lifetime of a light bulb in a certain application has
mean 700 hours and standard deviation 20 hours.
The light bulbs are packaged 12 to a box. Assuming
that light bulbs in a box are a simple random sample of
light bulbs, find the mean and standard deviation of the
average lifetime of the light bulbs in a box.
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𝑇= (T1 + T1 +…+ T1 )/12
𝜇 𝑇 = 𝜇 = 700
𝜎 𝑇 =𝜎
𝑛=
20
12= 5.77
Standard Deviations of
Nonlinear Functions of Random Variables
If 𝑋 is a random variable with (small) standard deviation 𝑋, and if 𝑈 is a function of 𝑋, then
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(3.42)𝜎𝑈 ≈𝑑𝑈
𝑑𝑋𝜎𝑋
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Example 3.29
The radius 𝑅 of a circle is measured to be 5.43 cm, with a standard deviation of 0.01 cm. Estimate the area of the circle, and find the standard deviation of this estimate.
Area = 𝐴 = 𝑅2 = (5.43)2
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𝜎𝐴 ≈𝑑𝐴
𝑑𝑅𝜎𝑅
= (2R)(0.01)
Standard Deviations of
Nonlinear Functions of Random Variables
If 𝑋1, … , 𝑋𝑛 are independent random variables with standard deviations 1, … ,𝑛 (small), and if
𝑈 = 𝑈(𝑋1, … , 𝑋𝑛) is a function of 𝑋1, … , 𝑋𝑛 , then
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(3.43)
𝜎𝑈 ≈𝜕𝑈
𝜕𝑋1
2
𝜎𝑥12 +
𝜕𝑈
𝜕𝑋2
2
𝜎𝑥22 + ⋯+
𝜕𝑈
𝜕𝑋𝑛
2
𝜎𝑥𝑛2
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Example 3.30
Assume the mass of a rock is measured to be m=674.01.0g, and the volume of the rock is measured to be V=261.00.1 mL. The density of the rock (D) is given by m/V. Estimate the density and find the standard deviation of the estimate.
Density = 𝐷 = 𝑚/𝑉 = 674.0/261.0 = 2.582 𝑔/𝑚𝐿
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𝜎𝐷 ≈𝜕𝐷
𝜕𝑚
2𝜎𝑚
2 +𝜕𝐷
𝜕𝑉
2𝜎𝑉
2 =
1
𝑉
2𝜎𝑚
2 +−𝑚
𝑉2
2𝜎𝑉
2
Summary
Probability and rules
Conditional probability
Independence
Random variables: discrete and continuous
Probability mass functions
Probability density functions
Cumulative distribution functions
Means and variances for random variables
Linear functions of random variables
Mean and variance of a sample mean
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