ProbabilityBasicConceptProbabilityDensityFunctionMean ApplicationsofIntegrationtoPhysicsandEngineeringWorkandImpulseFirstMomentandCenterofMassApplicationinFluidMechanics ProjectileMotionVerticalMotionProjectileMotioninTwoDimensions Outline

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Outline

1 Projectile Motion Vertical Motion

Projectile Motion in Two Dimensions

2 Applications of Integration to Physics and Engineering Work and Impulse

First Moment and Center of Mass Application in Fluid Mechanics

3 Probability Basic Concept

Probability Density Function

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Basic Concept (I)

In this section, we give a brief introduction to the use of calculus in probability theory.

We begin with a simple example involving coin-tossing.

Suppose that we toss two coins, each of which has a50%

chance of coming up heads.

If we denote heads by H and tails by T, then the four possible outcomes from tossing two coins are

HH, HT, TH and TT.

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Basic Concept (II)

Each of these four outcomes, HH, HT, TH and TT, is equally likely, so we can say that each has probability 1

4. This means that, on average, each of these events will occur in one-fourth of your tries. Said a different way, the relative frequency with which each event occurs in a large number of trials will be approximately 1

.

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Basic Concept (III)

Suppose that we are primarily interested in recording the number of heads. Based on our calculations above,

the probability of getting two heads is 1

4,

the probability of getting one head is 2

4 and

the probability of getting zero heads is 1

4.

We often summarize such information by displaying it in a histogram or bar graph (see Figure 5.51).

Figure:[5.51] Histogram for two-coin toss.

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Basic Concept (IV)

Suppose that we instead toss eight coins. The probabilities for getting a given number of heads are given in the accompanying table and the corresponding histogram is shown in Figure 5.52.

Number

of Heads Probability

1 1/256

2 8/256

3 28/256

4 56/256

5 70/256

6 56/256

7 28/256

8 1/256 Figure:[5.52] Histogram for

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Basic Concept (V)

Notice that the sum of all the probabilities is1. This is one of the defining properties of probability theory.

Number

of Heads Probability

1 1/256

2 8/256

3 28/256

4 56/256

5 70/256

6 56/256

7 28/256

8 1/256

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Basic Concept (VI)

Another basic property is called the addition principle: To

compute the probability of getting 6, 7or8heads (or any other mutually exclusive outcomes), simply add together the individual probabilities:

P(6, 7or8heads)

= 28

256+ 8 256+ 1

256

= 37

256

≈

Figure:[5.52] Histogram for eight-coin toss.

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Basic Concept (VII)

In the histogram in Figure 5.52, notice that each bar is a rectangle of width1.

Then the probability associated with each bar equals the area of the rectangle. In graphical terms,

The total area in such a histogram is1.

The probability of getting between6and8heads

(inclusive) equals the sum of the areas of the rectangles located between6and8(inclusive).

Figure:[5.52] Histogram for eight-coin toss.

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Outline

Probability Density Function

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Probability Density Function (I)

Not all probability events have the nice theoretical structure of coin-tossing. For instance, the situation is somewhat different if we want to find the probability that a randomly chosen person will have a height of5⁰9⁰⁰or5⁰10⁰⁰. There is no easy theory we can use here to compute the probabilities (since not all heights are equally likely). In this case, we would need to use the correspondence between probability and relative frequency.

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Probability Density Function (II)

If we collected information about the heights of a large number of adults (for instance, from driver’s license data), we might find the following.

Since the total number of people in the survey is1000, the relative frequency of the height5⁰9⁰⁰(69⁰⁰)is

155

1000 = 0.155and

the relative frequency of the height5⁰10⁰⁰(70⁰⁰)is 134 = 0.134.

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Probability Density Function (III)

An estimate of the probability of being 5⁰9⁰⁰or5⁰10⁰⁰is then

0.155 + 0.134 = 0.289.

A histogram is shown in Figure 5.53.

Figure:[5.53] Histogram for relative frequency of heights.

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Probability Density Function (IV)

Suppose that we want to be more specific: for example, what is the probability that a randomly chosen person is5⁰8¹₂⁰⁰or5⁰9⁰⁰? To answer this question, we would need to have our data broken down further, as in the following partial table.

The probability that a person is5⁰9⁰⁰can be estimated by the relative frequency of5⁰9⁰⁰people in our survey, which is

81

1000 = 0.081

. Similarly, the probability that a person is5⁰8¹₂⁰⁰is approximately 82 = 0.082.

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Probability Density Function (V)

A histogram for this portion of the data is shown in Figure 5.54a. Notice that since each bar of the histogram now represents a half-inch range of height, we can no longer interpret area in the histogram as the probability (we had assumed each bar has width1).

Figure:[5.54a] Histogram for relative frequency of heights.

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Probability Density Function (VI)

We will modify the histogram to make the area connection clearer. In Figure 5.54b, we have labeled the horizontal axis with the height in inches, and the vertical axis shows twice the relative frequency. The bar at69⁰⁰ has height 0.162and width ¹₂. Its area,

1

2(0.162) = 0.081,

corresponds to the relative frequency (or probability) of the height ⁰ ⁰⁰.

Figure:[5.54b] Histogram showing double the relative frequency.

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Probability Density Function (VII)

Of course, we could continue subdividing the height intervals Suppose that there arenheight intervals between5⁰8⁰⁰and 5⁰9⁰⁰. Letxrepresent height in inches andf (x)equal the height of the histogram bar for the interval containingx. Let

x₁= 68 +¹_n,x₂= 68 +²_n and so on, so that x_i = 68 + i∆x, for 1 ≤ i ≤ n where∆x = ¹_n.

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Probability Density Function (VIII)

For a randomly selected person, the probability that their height is between 5⁰8⁰⁰and5⁰9⁰⁰is estimated by the sum of the areas of the corresponding histogram rectangles, given by

P(68 ≤ x ≤ 69) ≈ f (x₁) ∆x +f (x2) ∆x + · · · + f (xn) ∆x

=

n

Xf (x_i) ∆x. ^Figure:[5.55a] Histogram for heights.

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Probability Density Function (IX)

Observe that as n increases, the histogram of Figure 5.55a will

“smooth out,” approaching a curve like the one shown in Figure 5.55b.

Figure:[5.55a] Histogram for heights.

Figure:[5.55b] Probability density function and histogram for heights.

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Probability Density Function (X)

We call this limiting functionf (x), the probability density function (pdf) for heights.

Notice that for any given

i = 1, 2, . . . , n,f (x_i)does not give the probability that a person’s height equalsxi. Instead, for small values of

∆x, the quantityf (x_i)∆xis an approximation of the probability that the height is in the range[x , x].

Figure:[5.55b] Probability density function and histogram for heights.

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Probability Density Function (XI)

Look carefully at

P(68 ≤ x ≤ 69) ≈ f (x₁) ∆x + f (x₂) ∆x + · · · + f (x_n) ∆x

=

n

X

i=1

f (xi) ∆x.

and think about what will happen asnincreases. The Riemann sum on the right should approach an integral

Z b a

f (x) dx.

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Probability Density Function (XII)

In this particular case, the limits of integration are68 (5⁰8⁰⁰)and 69 (59⁰⁰). We have

n→∞lim

n

X

i=1

f (xi) ∆x = Z 69

68

f (x) dx.

Notice that by adjusting the function values so that probability corresponds to area, we have found a familiar and direct

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Probability Density Function (XIII)

We now summarize our discussion with some definitions.

The preceding examples are of discrete probability distributions (the term discrete indicates that the quantity being measured can only assume values from a certain finite set or from a fixed sequence of values). For instance, in coin-tossing, the number of heads must be an integer.

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Probability Density Function (XIV)

By contrast, many distributions are continuous. That is, the quantity of interest (the random variable) assumes values from a continuous range of numbers (an interval). For instance, suppose you are measuring the height of people in a certain group. Although height is normally rounded off to the nearest integer number of inches, a person’s actual height can be any number.

For continuous distributions, the graph corresponding to a histogram is the graph of a probability density function (pdf).

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Probability Density Function (XV)

Definition (6.1)

Suppose thatXis a random variable that may assume any valuex witha ≤ x ≤ b. (Here, we may havea = −∞and/orb = ∞) A probability density function forXis a functionf (x)satisfying

1 f (x) ≥ 0fora ≤ x ≤ b, Probability density functions are never negative.

2

Z b a

f (x)dx = 1and The total probability is1.

the probability that the (observed) value ofXfalls betweencanddis given by the area under the graph of the pdf on that interval. That is,

P(c ≤ X ≤ d) = Z d

c

f (x)dx.

Probability corresponds to area under the curve.

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Verifying That a Function Is a pdf on an Interval

Example (6.1)

Show thatf (x) = 3x²defines a pdf on the interval[0, 1]by verifying properties (i) and (ii) of Definition 6.1

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Using a pdf to Estimate Probabilities

Example (6.2)

Suppose that f (x) = 0.4

√

2πe−0.08(x−68)²is the probability density function for the heights in inches of adult

American males. Find the probability that a randomly selected adult American male will be between5⁰8⁰⁰and5⁰9⁰⁰. Also, find the probability that a randomly selected adult American male will be between 6⁰2⁰⁰and6⁰4⁰⁰.

Figure:[5.56] Heights of adult males.

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A pdf on an Infinite Interval

Example (6.3) Given that

Z ∞

−∞

e^−u²du =√

π, confirm that f (x) = 0.4

√2πe−0.08(x−68)² defines a pdf on the interval(−∞, ∞).

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Computing Probability with an Exponential pdf

Example (6.4)

Suppose that the lifetime in years of a certain brand of lightbulb is exponentially distributed with pdff (x) = 4e^−4x. Find the probability that a given lightbulb lasts3months or less.

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Determining the Coefficient of a pdf

Example (6.5)

Suppose that the pdf for a random variable has the form f (x) = ce^−3xfor some constantc, with0 ≤ x ≤ 1. Find the value ofcthat makes this a pdf.

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Outline

Probability Density Function Mean

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Mean (I)

Given a pdf, it is possible to compute various statistics to summarize the properties of the random variable. The most common statistic is the mean, the best-known measure of average value. If you wanted to average test scores of

85, 89, 93and93,

you would probably compute the mean, given by 85 + 89 + 93 + 93

= 90.

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Mean (II)

Notice here that there were three different test scores recorded:

85, which has a relative frequency of 1 4, 89, also with a relative frequency of 1

4, and 93, with a relative frequency of 2

4.

We can also compute the mean by multiplying each value by its relative frequency and then summing:

(85)1

4 + (89)1

4 + (93)2 4 = 90.

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Mean (III)

Now, suppose we wanted to compute the mean height of the people in the following table.

It would be silly to write out the heights of all1000people, add and divide by1000. It is much easier to multiply each value by its relative frequency and add the results. Following this route, the meanmis given by

m = (63) 23

1000+ (64) 32

1000+ (65) 61

1000+ (66) 94

1000+ (67)133 1000+ · · · + (74) 26

= 68.523.

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Mean (IV)

If we denote the heights byx1, x2, . . . , xnand letf (xi)be the corresponding relative frequencies or probabilities, the mean then has the form

m = x1f (x1) + x2f (x2) + x3f (x3) + · · · + x12f (x12).

If the heights in our data set were given for every half-inch or tenth-of-an-inch, we would compute the mean by multiplying eachxi

by the corresponding probabilityf (x_i) ∆x, where∆xis the fraction of an inch between data points. The mean now has the form

m = [x1f (x1) + x2f (x2) + x3f (x3) + · · · + xnf (xn)] ∆x =

n

X

i=1

xif (xi) ∆x,

wherenis the number of data points.

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Mean (V)

Notice that, asnincreases and∆xapproaches0, the Riemann

sum n

X

i=1

xif (xi) ∆x approaches the integral

Z b a

xf (x) dx.

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Mean (VI)

Definition (6.2)

The meanµof a random variable with pdff (x)on the interval [a, b]is given by

µ = Z b

a

xf (x) dx.

Notice that the formula for the mean is analogous to the formula for the first moment of a one-dimensional object with density f (x). For this reason, the mean is sometimes called the first moment.

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Mean (VII)

Although the mean is commonly used to report the average value of a random variable, it is important to realize that it is not the only measure of average used by statisticians.

An alternative measurement of average is the median, the x-value that divides the probability in half. (That is, half of all values of the random variable lie at or below the median and half lie at or above the median.) In example 6.6 and in the exercises, we will explore situations in which each measure provides a better or worse indication about the average of a

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Finding the Mean Age and Median Age of a Group of Cells

Example (6.6)

Suppose that the age in days of a type of single-celled organism has pdff (x) = (ln2)e^−kx, where k =¹₂ln 2. The domain is

0 ≤ x ≤ 2(the assumption here is that upon reaching an age of2 days, each cell divides into two daughter cells). Find

1 the mean age of the cells,

2 the proportion of cells that are younger than the mean and

3 the median age of the cells.

Figure:[5.57]y = ln 2e^{−(ln 2)x/2}