PROB ABILITY Calculus plays a role in the analysis of random behavior. Suppose we consider the choles-

8.5 PROB ABILITY Calculus plays a role in the analysis of random behavior. Suppose we consider the choles-

terol level of a person chosen at random from a certain age group, or the height of an adult female chosen at random, or the lifetime of a randomly chosen battery of a certain type. Such quantities are called continuous random variables because their values actually range over an interval of real numbers, although they might be measured or recorded only to the nearest integer. We might want to know the probability that a blood cholesterol level is greater than 250, or the probability that the height of an adult female is between 60 and

70 inches, or the probability that the battery we are buying lasts between 100 and 200 hours. If X represents the lifetime of that type of battery, we denote this last probability as follows:

P 共100 艋 X 艋 200兲

According to the frequency interpretation of probability, this number is the long-run pro- portion of all batteries of the specified type whose lifetimes are between 100 and 200 hours. Since it represents a proportion, the probability naturally falls between 0 and 1.

Every continuous random variable X has a probability density function . This means f that the probability that X lies between a and b is found by integrating from a to b: f

1 b P 共a 艋 X 艋 b兲 苷 y

a f 共x兲 dx

For example, Figure 1 shows the graph of a model for the probability density function

f for a random variable X defined to be the height in inches of an adult female in the United States (according to data from the National Health Survey). The probability that the height of a woman chosen at random from this population is between 60 and 70 inches is equal to the area under the graph of from 60 to 70. f

area =probability that the height of a woman

y=ƒ

is between 60 and

70 inches

FIGURE 1

Probability density function

for the height of an adult female

In general, the probability density function of a random variable X satisfies the con- f dition for f 共x兲 艌 0 all x . Because probabilities are measured on a scale from 0 to 1, it fol- lows that

共x兲 dx 苷 1 f

EXAMPLE 1 Let f 共x兲 苷 0.006x共10 ⫺ x兲 for 0 艋 x 艋 10 and f 共x兲 苷 0 for all other

values of . x (a) Verify that is a probability density function. f (b) Find . P 共4 艋 X 艋 8兲

CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION

SOLUTION

(a) For 0 艋 x 艋 10 we have 0.006x 共10 ⫺ x兲 艌 0 , so f 共x兲 艌 0 for all . We also need to x

check that Equation 2 is satisfied:

10 y 10

共x兲 dx 苷 y 0 共10 ⫺ x兲 dx 苷 0.006 y 共10x ⫺ x 0 兲 dx

f 0.006x

5x 2 1 x 3 ⫺ 10

Therefore is a probability density function. f (b) The probability that X lies between 4 and 8 is

8 P 8 共4 艋 X 艋 8兲 苷 y

f 共x兲 dx 苷 0.006 y 共10x ⫺ x 2 兲 dx

[ 5x 2 1 3 苷 0.006 8 ⫺ 3 x ] 4 苷 0.544

V EXAMPLE 2 Phenomena such as waiting times and equipment failure times are com- monly modeled by exponentially decreasing probability density functions. Find the exact

form of such a function. SOLUTION Think of the random variable as being the time you wait on hold before an

agent of a company you’re telephoning answers your call. So instead of x, let’s use t to represent time, in minutes. If is the probability density function and you call at time f

, then, from Definition 1, t苷0 2 x 0 f 共t兲 dt represents the probability that an agent answers within the first two minutes and 5 x 4 f 共t兲 dt is the probability that your call is answered

during the fifth minute.

It’s clear that f 共t兲 苷 0 for t ⬍0 (the agent can’t answer before you place the call). For t ⬎ 0 we are told to use an exponentially decreasing function, that is, a function of the form f 共t兲 苷 Ae ⫺ct , where A and c are positive constants. Thus

再 Ae ⫺ct if t 艌 0

0 if t ⬍ 0

f 共t兲 苷

We use Equation 2 to determine the value of A:

0 1苷 ⬁ y

⫺⬁ 共t兲 dt 苷 y ⫺⬁

f 共t兲 dt ⫹ f f 共t兲 dt

Ae ⫺ct dt 苷 lim x y Ae ⫺ct dt

f(t)= 0ce if

t<0 if t˘0

_ct

Therefore A 兾c 苷 1 and so A苷c . Thus every exponential density function has the form

再 ce ⫺ct if t 艌 0

An exponential density function

A typical graph is shown in Figure 2.

SECTION 8.5 PROBABILITY

AVERAGE VALUES

Suppose you’re waiting for a company to answer your phone call and you wonder how long, on average, you can expect to wait. Let f y=f(t) be the corresponding density function, 共t兲 where t is measured in minutes, and think of a sample of N people who have called this Ît company. Most likely, none of them had to wait more than an hour, so let’s restrict our attention to the interval 0 艋 t 艋 60 . Let’s divide that interval into n intervals of length ⌬t and endpoints

1 0, t ,t 2 , ... t 60 . (Think of ⌬t as lasting a minute, or half a minute, or 10 sec- onds, or even a second.) The probability that somebody’s call gets answered during the time period from t i ⫺ 1 to is the area under the curve t i y 苷 f 共t兲 from t i ⫺ 1 to , which is t i

0 t t i-1 t i

approximately equal to f 共t i 兲 ⌬t . (This is the area of the approximating rectangle in Fig-

ure 3, where is the midpoint of the interval.) t i

FIGURE 3

Since the long-run proportion of calls that get answered in the time period from t i ⫺ 1 to t i is f 共t i 兲 ⌬t , we expect that, out of our sample of N callers, the number whose call was answered in that time period is approximately N f 共t i 兲 ⌬t and the time that each waited is about . Therefore the total time they waited is the product of these numbers: approxi- t i mately t i 关N f 共t i 兲 ⌬t兴 . Adding over all such intervals, we get the approximate total of every- body’s waiting times:

i苷1 兺

Nt i f 共t i 兲 ⌬t

If we now divide by the number of callers N, we get the approximate average waiting time:

兺 t i f 共t i i苷1 兲 ⌬t

We recognize this as a Riemann sum for the function t f 共t兲 . As the time interval shrinks (that is, ⌬t l 0 and nl⬁ ), this Riemann sum approaches the integral

y 60

tf

0 共t兲 dt

This integral is called the mean waiting time.

N It is traditional to denote the mean by the

In general, the mean of any probability density function is defined to be f

Greek letter (mu). ␮

␮苷 ⬁ y

共x兲 dx

The mean can be interpreted as the long-run average value of the random variable X. It can also be interpreted as a measure of centrality of the probability density function.

The expression for the mean resembles an integral we have seen before. If ᏾ is the region that lies under the graph of , we know from Formula 8.3.8 that the x-coordinate of f

the centroid of ᏾ is

y=ƒ

x=m ⬁ y

苷 y x f ⫺⬁ 共x兲 dx 苷 ␮

y ⫺⬁ 共x兲 dx 0 f

FIGURE 4

because of Equation 2. So a thin plate in the shape of ᏾ balances at a point on the vertical T balances at a point on the line x=m

line x苷␮ . (See Figure 4.)

CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION

EXAMPLE 3 Find the mean of the exponential distribution of Example 2:

ce 再 ⫺ct if t 艌 0

0 if t ⬍ 0

f 共t兲 苷

SOLUTION According to the definition of a mean, we have

␮苷 ⬁ y

To evaluate this integral we use integration by parts, with u苷t and d v 苷 ce ⫺ct dt :

x y tce dt 苷 lim x ⫺te ] 0 ⫹ y e dt 0 l ⬁ 0 l ⬁ 冉 0 冊

x y ⫺ct

⬁ tce ⫺ct

dt 苷 lim x ⫺ct ⫺ct x

x ⫺xe ⫺cx ⫹ ⬁ ⫺ c c l’Hospital’s Rule. 苷 l 冉 冊 c

N The limit of the first term is by 0 1 e 苷 lim 1

⫺cx

The mean is ␮ 苷 1兾c , so we can rewrite the probability density function as

再 ␮ ⫺ 1 e ⫺t

V EXAMPLE 4 Suppose the average waiting time for a customer’s call to be answered by a company representative is five minutes.

(a) Find the probability that a call is answered during the first minute. (b) Find the probability that a customer waits more than five minutes to be answered.

SOLUTION

(a) We are given that the mean of the exponential distribution is ␮苷5 min and so, from the result of Example 3, we know that the probability density function is

再 0.2e ⫺t if t 艌 0

Thus the probability that a call is answered during the first minute is

P 1 共0 艋 T 艋 1兲 苷 y

f 共t兲 dt

1 0.2e ⫺t 兾5

dt

0 苷 0.2共⫺5兲e ] 0

So about 18% of customers’ calls are answered during the first minute. (b) The probability that a customer waits more than five minutes is

共T ⬎ 5兲 苷 兾5 y f

5 共t兲 dt 苷 y

0.2e ⫺t dt

⫺ 1 ⫺x 苷 lim 兾5

⫺t 兾5

x l y ⬁ 0.2e 5 dt 苷 lim x 共e ⫺e l 兲 ⬁

1 苷 ⬇ 0.368 e

About 37% of customers wait more than five minutes before their calls are answered. M

SECTION 8.5 PROBABILITY

Notice the result of Example 4(b): Even though the mean waiting time is 5 minutes, only 37% of callers wait more than 5 minutes. The reason is that some callers have to wait much longer (maybe 10 or 15 minutes), and this brings up the average.

Another measure of centrality of a probability density function is the median. That is a number m such that half the callers have a waiting time less than m and the other callers have a waiting time longer than m. In general, the median of a probability density func- tion is the number m such that

m f 共x兲 dx 苷 2

This means that half the area under the graph of lies to the right of m. In Exercise 9 you f are asked to show that the median waiting time for the company described in Example 4 is approximately 3.5 minutes.

NORMAL DISTRIBUTIONS Many important random phenomena—such as test scores on aptitude tests, heights and

weights of individuals from a homogeneous population, annual rainfall in a given loca- tion—are modeled by a normal distribution. This means that the probability density function of the random variable X is a member of the family of functions

3 1 ⫺ 共x⫺␮兲 2 f 2 兾共2␴ 共x兲 苷 兲 e ␴ s2␲

You can verify that the mean for this function is . The positive constant is called the ␮ ␴

N The standard deviation is denoted by the

standard deviation ; it measures how spread out the values of X are. From the bell-shaped

lowercase Greek letter (sigma). ␴

graphs of members of the family in Figure 5, we see that for small values of the values ␴ of X are clustered about the mean, whereas for larger values of the values of X are more ␴ spread out. Statisticians have methods for using sets of data to estimate and . ␮ ␴

s =1 s =2

FIGURE 5

Normal distributions

The factor 1 兾 ( ␴ s2␲ ) is needed to make a probability density function. In fact, it can f

be verified using the methods of multivariable calculus that

e ⫺ 共x⫺␮兲 兾共2␴ 兲

V EXAMPLE 5 Intelligence Quotient (IQ) scores are distributed normally with mean

0 60 80 100 120 140 x

100 and standard deviation 15. (Figure 6 shows the corresponding probability density function.)

FIGURE 6

(a) What percentage of the population has an IQ score between 85 and 115? Distribution of IQ scores

(b) What percentage of the population has an IQ above 140?

CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION

SOLUTION

(a) Since IQ scores are normally distributed, we use the probability density function given by Equation 3 with ␮ 苷 100 and ␴ 苷 15 :

1 共x⫺100兲 2 2

P 共85 艋 X 艋 115兲 苷

y 兾共2ⴢ15 兲

85 15 s2␲ e dx

Recall from Section 7.5 that the function 2 y苷e ⫺x doesn’t have an elementary anti- derivative, so we can’t evaluate the integral exactly. But we can use the numerical

integration capability of a calculator or computer (or the Midpoint Rule or Simpson’s Rule) to estimate the integral. Doing so, we find that

P 共85 艋 X 艋 115兲 ⬇ 0.68

So about 68% of the population has an IQ between 85 and 115, that is, within one stan- dard deviation of the mean.

(b) The probability that the IQ score of a person chosen at random is more than 140 is

P 共X ⬎ 140兲 苷 y e ⫺ 共x⫺100兲 兾450 dx

140 15 s2␲

To avoid the improper integral we could approximate it by the integral from 140 to 200. (It’s quite safe to say that people with an IQ over 200 are extremely rare.) Then

1 ⫺ 共x⫺100兲 2

P 共X ⬎ 140兲 ⬇ y e 兾450 dx ⬇ 0.0038

140 15 s2␲

Therefore about 0.4% of the population has an IQ over 140.

8.5 EXERCISES 1. Let f 共x兲

be the probability density function for the lifetime of a

5. Let . f 共x兲 苷 c兾共1 ⫹ x 2 兲

manufacturer’s highest quality car tire, where is measured in x (a) For what value of is a probability density function? c f miles. Explain the meaning of each integral.

(b) For that value of , find c P 共⫺1 ⬍ X ⬍ 1兲 .

(a) ⬁

6. y 2 f 共x兲 dx (b) y 共x兲 dx f Let if f

共x兲 苷 kx 30,000 共1 ⫺ x兲 0艋x艋1 and f 共x兲 苷 0 if x ⬍0

or . x ⬎ 1

2. Let f 共t兲 be the probability density function for the time it takes (a) For what value of is a probability density function? k f 1 you to drive to school in the morning, where is measured in t

(b) For that value of , find k P ( X 艌 2 ) .

minutes. Express the following probabilities as integrals.

(c) Find the mean.

(a) The probability that you drive to school in less than 7. A spinner from a board game randomly indicates a real number 15 minutes

between 0 and 10. The spinner is fair in the sense that it indi- (b) The probability that it takes you more than half an hour to cates a number in a given interval with the same probability as get to school it indicates a number in any other interval of the same length.

3. Let for 3 f 共x兲 苷 64 x s16 ⫺ x 2 0艋x艋4 and f 共x兲 苷 0 for all (a) Explain why the function

other values of . x

再 0 if x ⬍ 0 or x ⬎ 10

(a) Verify that is a probability density function. f 0.1 f if 0 艋 x 艋 10 共x兲 苷

(b) Find . P (X⬍2 兲

is a probability density function for the spinner’s values. (a) Verify that is a probability density function. f (b) What does your intuition tell you about the value of the (b) Find . P 共1 艋 X 艋 2兲

4. Let if f 共x兲 苷 xe ⫺x x 艌 0 and f 共x兲 苷 0 if x ⬍0 .

mean? Check your guess by evaluating an integral.

SECTION 8.5 PROBABILITY

8. (a) Explain why the function whose graph is shown is a proba- of 500 g. At what target weight should the manufacturer set its bility density function.

filling machine?

(b) Use the graph to find the following probabilities: (i) P 共X ⬍ 3兲

15. The speeds of vehicles on a highway with speed limit 100 km 兾h (c) Calculate the mean.

(ii) P 共3 艋 X 艋 8兲

are normally distributed with mean 112 km 兾h and standard deviation . 8 km 兾h

y (a) What is the probability that a randomly chosen vehicle is 0.2

y=ƒ traveling at a legal speed? (b) If police are instructed to ticket motorists driving 125 km 0.1 兾h or more, what percentage of motorists are targeted?

0 2 4 6 8 10 x

16. Show that the probability density function for a normally dis-

tributed random variable has inflection points at x苷␮⫾␴ . Show that the median waiting time for a phone call to the com-

pany described in Example 4 is about 3.5 minutes. 17. For any normal distribution, find the probability that the 10. (a) A type of lightbulb is labeled as having an average lifetime

random variable lies within two standard deviations of the of 1000 hours. It’s reasonable to model the probability of

mean.

failure of these bulbs by an exponential density function with mean

. Use this model to find the probability ␮ 苷 1000 18. The standard deviation for a random variable with probability that a bulb

density function and mean is defined by f ␮ (i) fails within the first 200 hours,

1 (ii) burns for more than 800 hours. 兾2

冋 y ⫺⬁

(b) What is the median lifetime of these lightbulbs?

11. The manager of a fast-food restaurant determines that the average time that her customers wait for service is 2.5 min-

Find the standard deviation for an exponential density function utes.

with mean . ␮

(a) Find the probability that a customer has to wait more than 19. The hydrogen atom is composed of one proton in the nucleus 4 minutes.

and one electron, which moves about the nucleus. In the quan- (b) Find the probability that a customer is served within the tum theory of atomic structure, it is assumed that the electron first 2 minutes. does not move in a well-defined orbit. Instead, it occupies a (c) The manager wants to advertise that anybody who isn’t state known as an orbital, which may be thought of as a served within a certain number of minutes gets a free ham- “cloud” of negative charge surrounding the nucleus. At the burger. But she doesn’t want to give away free hamburgers state of lowest energy, called the ground state, or 1s-orbital, to more than 2% of her customers. What should the adver- the shape of this cloud is assumed to be a sphere centered at tisement say? the nucleus. This sphere is described in terms of the probability

12. According to the National Health Survey, the heights of adult

density function

males in the United States are normally distributed with mean 69.0 inches and standard deviation 2.8 inches.

r 2 ⫺ 共r兲 苷 2r 3 e 兾a 0 r 艌 0 (a) What is the probability that an adult male chosen at random

is between 65 inches and 73 inches tall? a 0 Bohr radius 共a 0 ⬇ 5.59 ⫻ 10 ⫺ 11 兲 (b) What percentage of the adult male population is more than

m . The 6 feet tall?

where is the

integral

The “Garbage Project” at the University of Arizona reports r P 4 2 ⫺ 2s

共r兲 苷 3 s e 兾a y 0 0 ds

that the amount of paper discarded by households per week is

normally distributed with mean 9.4 lb and standard deviation gives the probability that the electron will be found within the 4.2 lb. What percentage of households throw out at least 10 lb

sphere of radius meters centered at the nucleus. r of paper a week?

(a) Verify that p 共r兲 is a probability density function. 14. Boxes are labeled as containing 500 g of cereal. The machine

(b) Find lim r l ⬁ p 共r兲 . For what value of does r p 共r兲 have its filling the boxes produces weights that are normally distributed

maximum value?

with standard deviation 12 g.

; (c) Graph the density function.

(a) If the target weight is 500 g, what is the probability that the (d) Find the probability that the electron will be within the machine produces a box with less than 480 g of cereal?

sphere of radius 4a 0 centered at the nucleus. (b) Suppose a law states that no more than 5% of a manufac-

(e) Calculate the mean distance of the electron from the turer’s cereal boxes can contain less than the stated weight

nucleus in the ground state of the hydrogen atom.

CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION

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