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Chapter 7

Additional

Integration

Topics

Section 2

Applications in

Business and

Economics

2Barnett/Ziegler/Byleen Business Calculus 12e

Learning Objectives for Section 7.2

Applications in Business/Economics

The student will be able to:

1. Construct and interpret probability

density functions.

2. Evaluate a continuous income

stream.

3. Evaluate the future value of a

continuous income stream.

4. Evaluate consumers’ and producers’

surplus.

3

Random Variables

Random variables come in two varieties:

Discrete

• Values are distinct or separate and can be counted

and listed

Continuous

• Infinite number of values that are within an interval

Barnett/Ziegler/Byleen Business Calculus 12e

4

Continuous vs Discrete

Discrete Random Variable

• Suppose we roll a single die. How many possible

outcomes are there?

• There are 6 discrete possible outcomes.

Continuous Random Variable

• Suppose we randomly choose a real number (x) in the

interval [1, 6]. How many possible outcomes are there?

• There are an infinite number of possible outcomes.


5

Continuous Random Variables

Suppose an experiment is designed in such a way that any

real number x on the interval [c, d] is a possible outcome.

Examples of what x could represent:

• Inches of rain in one day

• Height of a person between 5 ft and 7 ft

• Life of a lightbulb between 40 hours and 100 hours

These are all examples of continuous random variables

because the possible outcomes are not discrete. Rather,

there is an infinite number of possible outcomes over a

specified interval.


6

Probability Density Function

In certain situations, we can find a function 𝑓 that can be

used to model the probability that a continuous random

variable will take on a value over a specified interval.

Such functions are called probability density functions.



Probability Density Functions

A probability density function must satisfy 3

conditions:

1. f (x) 0 for all real x

2. The area under the graph of f (x) over the interval

(-, ) is 1

3. If [c, d] is a subinterval of (-, ) then

the probability that x falls in the interval [c, d]

is equal to:

d

cdxxf )(


Graph Examples

𝐴𝑟𝑒𝑎 = 1

𝑐 𝑑


Example 1

In a certain city, the daily use of water in hundreds of gallons

per household is a continuous random variable with

probability density function

Find the probability that a household chosen at random will

use between 300 and 600 gallons. (Use graphing calculator.)

𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 3 ≤ 𝑥 ≤ 6 = 3

6

.15𝑒−.15𝑥𝑑𝑥

≈ 0.23

There is a 23% probability that a household chosen at

random uses between 300-600 gallons of water.

f x = .15𝑒−.15𝑥 𝑥 ≥ 00 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒


Conceptual Insight

The probability that a household in the previous example uses

exactly 300 gallons is given by:

In fact, for any continuous random variable x with

probability density function f (x), the probability that x is

exactly equal to a constant c is equal to 0.

𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 3 ≤ 𝑥 ≤ 3 = 3

3

.15𝑒−.15𝑥𝑑𝑥

= 0

See khanAcademy.org “probability density functions” for

an additional explanation.

11

Example 2

Suppose that the length of a phone call (in minutes) is a

continuous random variable with the probability density

function:

Find the probability that a call selected at random will last

4 minutes or less. (Use graphing calculator.)

Solve for b so that the probability of a call selected at

random lasting b minutes or less is 90%.


𝑓 𝑡 = 0.25𝑒−.25𝑡 𝑖𝑓 𝑡≥0

0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

12

Example 2 (continued)


𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 0 ≤ 𝑡 ≤ 4 = 0

4

.25𝑒−.25𝑡𝑑𝑡

≈ 0.63 (𝑢𝑠𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑜𝑟)

There is a 63% probability that a phone call chosen at

random will last 4 minutes or less.

Find the probability that a call selected at random will

last 4 minutes or less.

13



Solve for b so that the probability of a call selected at

random lasting b minutes or less is 90%.

0

𝑏

.25𝑒−.25𝑡𝑑𝑡 = 0.9

𝑢 = −.25𝑡

𝑑𝑢 = −.25𝑑𝑡

0

−.25𝑏

−𝑒𝑢𝑑𝑢 = 0.9

−𝑒𝑢 −.25𝑏

0= 0.9

−𝑒−.25𝑏 − −𝑒0 = 0.9

−𝑒−.25𝑏 + 1 = 0.9

−𝑒−.25𝑏 = −0.1

𝑒−.25𝑏 = 0.1

ln 𝑒−.25𝑏 = ln 0.1

−.25𝑏 = ln 0.1

𝑏 ≈ 9.21

There is a 90% probability of a call lasting 9.21 minutes or less.

14

Application: Continuous Income

A function that models the flow of money represents a

continuous income stream.

Let f(t) represent the rate of flow of a continuous income

stream where t is time.

We can use calculus to find the total income produced over

a specified time interval.



Continuous Income Stream

Total Income for a Continuous Income Stream:

If f (t) is the rate of flow of a continuous income stream,

the total income produced during the time period from t = a

to t = b is

b

adttf )( income Total

a Total Income b

16

Continuous Income Stream

This makes sense if you recall what we have been saying

about definite integrals.

If you integrate a rate of change of a quantity on an

interval then you get the total change of the quantity on

that interval.

Since the rate of flow represents the rate of change of

income produced then the definite integral from a to b

represents the total income produced on that interval.



Example 3

Find the total income produced by a continuous income

stream in the first 2 years if the rate of flow is

f (t) = 600 e 0.06t

𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑐𝑜𝑚𝑒 = 0

2

600𝑒 .06𝑡𝑑𝑡

𝑢 = .06𝑡

𝑑𝑢 = .06𝑑𝑡 = 0

.12

10000𝑒𝑢𝑑𝑢

= 10000 0

.12

𝑒𝑢𝑑𝑢10000𝑑𝑢 = 600𝑑𝑡

18



= 10000 ∙ 𝑒𝑢 .12

0

= 10000 ∙ (𝑒 .12 − 𝑒0)

= 10000 ∙ (𝑒 .12 − 1)

≈ 1,274.97

= 10000 0

.12

𝑒𝑢𝑑𝑢

The total income after the first 2 years is $1,274.97

19


What would be the total income produced during the

second two years? (Use graphing calculator.)

Interval will be [2, 4] because it represents the end of the

2nd year to the end of the 4th year.


𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑐𝑜𝑚𝑒 = 2

4

600𝑒 .06𝑡𝑑𝑡

≈ 1,437.52

The total income produced during the next

two years is $1437.52

20

Homework

#7-2A

Pg 430

(13-19 odd, 21, 25)


khanAcademy.org

“discrete and continuous random variables”

“probability density functions”

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