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Chapter 3:

Probability

1

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).

3

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.

5

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?

7

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)

13

(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.

35

(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

43

(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?

45

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)

47

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.

49

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

50

(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

51

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

52

(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

55

(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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