Chapter 1 Linear Equations and Graphspmcnamara.weebly.com/uploads/1/3/3/8/13389841/bcb12e_ppt... ·...
Transcript of Chapter 1 Linear Equations and Graphspmcnamara.weebly.com/uploads/1/3/3/8/13389841/bcb12e_ppt... ·...
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Chapter 7
Additional
Integration
Topics
Section 2
Applications in
Business and
Economics
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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.
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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
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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.
Barnett/Ziegler/Byleen Business Calculus 12e
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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.
Barnett/Ziegler/Byleen Business Calculus 12e
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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.
Barnett/Ziegler/Byleen Business Calculus 12e
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7Barnett/Ziegler/Byleen Business Calculus 12e
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 )(
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8Barnett/Ziegler/Byleen Business Calculus 12e
Graph Examples
𝐴𝑟𝑒𝑎 = 1
𝑐 𝑑
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9Barnett/Ziegler/Byleen Business Calculus 12e
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
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10Barnett/Ziegler/Byleen Business Calculus 12e
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.
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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%.
Barnett/Ziegler/Byleen Business Calculus 12e
𝑓 𝑡 = 0.25𝑒−.25𝑡 𝑖𝑓 𝑡≥0
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
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12
Example 2 (continued)
Barnett/Ziegler/Byleen Business Calculus 12e
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.
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13
Example 2 (continued)
Barnett/Ziegler/Byleen Business Calculus 12e
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.
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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.
Barnett/Ziegler/Byleen Business Calculus 12e
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15Barnett/Ziegler/Byleen Business Calculus 12e
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
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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.
Barnett/Ziegler/Byleen Business Calculus 12e
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17Barnett/Ziegler/Byleen Business Calculus 12e
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
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18
Example 3 (continued)
Barnett/Ziegler/Byleen Business Calculus 12e
= 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
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19
Example 3 (continued)
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.
Barnett/Ziegler/Byleen Business Calculus 12e
𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑐𝑜𝑚𝑒 = 2
4
600𝑒 .06𝑡𝑑𝑡
≈ 1,437.52
The total income produced during the next
two years is $1437.52
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20
Homework
#7-2A
Pg 430
(13-19 odd, 21, 25)
Barnett/Ziegler/Byleen Business Calculus 12e
khanAcademy.org
“discrete and continuous random variables”
“probability density functions”