Download Chapter 7 Random Variables

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n  7.1
Discrete and Continuous Random Variables
n  6.2
Transforming and Combining Random Variables
n  Example:
El Dorado Community College
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Chapter 7
Random Variables
El Dorado Community College considers a student to be full-time if he
or she is taking between 12 and 18 credits. The number of credits X
that a randomly-selected EDCC full-time student is taking in the fall
semester has the following distribution.
Credits
Prob pi
12
13
0.25 0.10
14
15
16
17
18
0.05
0.30
0.10
0.05
0.15
Find the mean and standard deviation.
EDCC charges $50 per credit. Let T = tuition charge for a randomly-selected fulltime student. T = 50X.
Tuit. T
$600
$650
$700
$750
$800 $850
$900
Prob pi
0.25
0.10
0.05
0.30
0.10
0.15
0.05
Find the mean and standard deviation
Compare the shape, center, and spread of the two probability distributions.
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Skee Ball
Probability pi 0.32 0.27 0.19 0.15 0.07
Find the mean and standard deviation. Describe the
distribution.
A player receives one ticket from the game for every 10 points scored. Let the
random variable T = number of tickets a player gets on a randomly-selected throw.
Find and interpret the mean and standard deviation of the random variable T.
Tickets T
Probability pi
1
2
3
4
5
0.32
0.27
0.19
0.15
0.07
Find the mean – state in words the meaning. Find the standard
deviation – start in words what this means.
n  Linear
Transformations
Effect on a Random Variable of Adding (or Subtracting) a Constant
Adding the same number a (which could be negative) to
each value of a random variable:
•  Adds a to measures of center and location (mean,
median, quartiles, percentiles).
•  Does not change measures of spread (range, IQR,
standard deviation).
•  Does not change the shape of the distribution.
Transforming and Combining Random Variables
How does adding or subtracting a constant affect a random variable?
Transforming and Combining Random Variables
Good skee ball players always roll for the 50-point slot. The probability
distribution of a randomly-selected score at the WLPCS skee ball
championship is shown below. Let X be the score. Find the mean
and the standard deviation of the random variable X. Describe the
s
shape of the distribution.
Score xi
10
20
30
40
50
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n  Example:
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Transformations
Collected ci
300
450
600
750
900
Probability pi
0.15
0.25
0.35
0.20
0.05
Find the mean and standard deviation
It costs Pete $100 per trip to buy permits, gas, and a ferry pass. The random
variable V describes the profit Pete makes on a randomly selected day.
Profit vi
200
350
500
650
800
Probability pi
0.15
0.25
0.35
0.20
0.05
Find the mean and standard deviation.
Compare the shape, center, and spread of the two probability distributions.
n  Example:
El Dorado Community College
Transforming and Combining Random Variables
Consider Pete’s Jeep Tours again. We defined C as the amount of
money Pete collects on a randomly selected day.
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n  Linear
In addition to tuition charges, each full-time student is assessed student
fees of $100 per semester. If C = overall cost for a randomly selected
full-time student, C = 100 + T
Overall
Cost
$700
$750
$800
$850
$900 $950 $1000
Prob pi
0.25
0.10
0.05
0.30
0.10
0.05
0.15
Find the mean and standard deviation.
Tuit. T
$600
$650
$700
$750
$800 $850
$900
Prob pi
0.25
0.10
0.05
0.30
0.10
0.15
0.05
Find the mean and standard deviation.
Compare the shape, center, and spread of the two probability distributions.
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YOUR UNDERSTANDING
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n  CHECK
Cars Sold
Prob pi
0
1
2
3
0.3
0.4
0.2
0.1
The random variable X has mean 1.1 and standard deviation is 0.943.
1)  Suppose the dealership’s manager receives a $500 bonus from the
company for each car sold. Let Y = the bonus received from car
sales during the first hour on a randomly-selected Friday. Find the
mean and standard deviation of Y.
2)  To encourage customers to buy cars on Friday mornings, the
manager spends $75 to provide coffee and doughnuts. The
manager’s net profit T on a randomly-selected Friday is the bonus
earned minus the $75. Find the mean and standard deviation of T.
YOUR UNDERSTANDING
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n  CHECK
Prob pi
0
1
2
3
0.3
0.4
0.2
0.1
The random variable X has mean 1.1 and standard deviation is 0.943.
1)  Suppose the dealership’s manager receives a $500 bonus from the
company for each car sold. Let Y = the bonus received from car
sales during the first hour on a randomly-selected Friday. Find the
mean and standard deviation of Y.
2)  To encourage customers to buy cars on Friday mornings, the
manager spends $75 to provide coffee and doughnuts. The
manager’s net profit T on a randomly-selected Friday is the bonus
earned minus the $75. Find the mean and standard deviation of T.
Transforming and Combining Random Variables
A large auto dealership keeps track of sales made during each hour of the
day. Let X = the number of cars sold during the first hour of business on
a randomly-selected Friday. The probability distribution is:
Cars Sold
Transforming and Combining Random Variables
A large auto dealership keeps track of sales made during each hour of the
day. Let X = the number of cars sold during the first hour of business on
a randomly-selected Friday. The probability distribution is:
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ERROR ALERT!
n  Linear
Transformations
Effect on a Linear Transformation on the Mean and Standard Deviation
If Y = a + bX is a linear transformation of the random
variable X, then
•  The probability distribution of Y has the same shape
as the probability distribution of X.
•  µY = a + bµX.
•  σY = |b|σX (since b could be a negative number).
Transforming and Combining Random Variables
Whether we are dealing with data or random variables, the
effects of a linear transformation are the same.
Transforming and Combining Random Variables
When you are asked about the effect of a linear transformation on
summary statistics, it is easy to forget that adding (subtracting) a
constant to (from) every value in the distribution has no effect on
measures of spread, including the standard deviation, variance,
range, and IQR.
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n  AP
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The Baby and the Bathwater
Define the random variable Y to be the water temperature in Fahrenheit (F =
(9/5)C + 32) when the dial is set to “babysafe”. Find the mean and standard
deviation of Y.
According to Babies R Us, the temperature of a baby’s bathwater should be
between 90°F and 100°F. Find the probability that the water temperature on a
randomly-selected day when the “babysafe” setting is used meets the Babies R Us
recommendations.
n  Example:
Scaling a Test
Define the random variable Y to be scaled score of a randomly-selected
student from the class. Find the mean and standard deviation of Y.
What is the probability that a randomly-selected student has a scaled score of at
least 90?
Transforming and Combining Random Variables
In a large introductory statistics class, the distribution of X = raw
scores on a test was approximately Normally distributed with a mean
of 17.2 and a standard deviation of 3.8. The professor decides to
scale the scores by multiplying the raw scores by 4 and adding 10.
Transforming and Combining Random Variables
One brand of bathtub comes with a dial to set the water
temperature. When the “babysafe” setting is selected and the tub is
filled, the temperature X of the water follows a Normal distribution
with a mean of 34°C and a standard deviation of 2°C.
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n  Example:
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Cereal
Define the random variable Y to be the excess amount of cereal beyond
what’s advertised in a randomly-selected box, measured in grams (1oz =
28.35g). Find the mean and standard deviation of Y.
What is the probability that a randomly-selected box at least 3 grams more cereal
than advertised?
n  Combining
Random Variables
Transforming and Combining Random Variables
If we buy a large drink at a fast-food restaurant and grab a largesized lid, we assume that the lid will fit snugly on the cup. Usually,
it does. Sometimes, though, the lid is a bit too small or too large to
fit securely. Why does this happen? Even though the large cups
are made to be identical in size, the width actually varies slightly
from cup to cup. In fact, the diameter of a randomly-selected large
drink cup is a random variable (C) that has a particular probability
distribution. Likewise, the diameter of a randomly-selected large lid
(L) is a random variable with its own probability distribution. For a
lid to fit on a cup, the value of L has to be bigger than the value of
C, but not too much bigger. To find the probability that the lid fits,
we need to know more about what happens when we combine
these two random variables.
Transforming and Combining Random Variables
A company’s cereal boxes advertise 9.63 ounces of cereal. In fact, the
amount of cereal X in a randomly-selected box follows a Normal
distribution with a mean of 9.70 ounces and a standard deviation of
0.03 ounces.
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n  Example:
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Random Variables
Let’s investigate the result of adding and subtracting random variables.
Let X = the number of passengers on a randomly selected trip with
Pete’s Jeep Tours. Y = the number of passengers on a randomly
selected trip with Erin’s Adventures. Define T = X + Y. What are the
mean and variance of T?
Passengers xi
2
3
4
5
6
Probability pi
0.15
0.25
0.35
0.20
0.05
Mean µX = 3.75 Standard Deviation σX = 1.090
Passengers yi
2
3
4
5
Probability pi
0.3
0.4
0.2
0.1
Mean µY = 3.10 Standard Deviation σY = 0.943
n  Combining
Random Variables
Since Pete expects µX = 3.75 and Erin expects µY = 3.10 , they
will average a total of 3.75 + 3.10 = 6.85 passengers per trip.
We can generalize this result as follows:
Mean of the Sum of Random Variables
For any two random variables X and Y, if T = X + Y, then the
expected value of T is
E(T) = µT = µX + µY
In general, the mean of the sum of several random variables is the
sum of their means and the range is the sum of all ranges.
How much variability is there in the total number of passengers who
go on Pete’s and Erin’s tours on a randomly selected day? To
determine this, we need to find the probability distribution of T.
Transforming and Combining Random Variables
How many total passengers can Pete and Erin expect on a
randomly selected day?
Transforming and Combining Random Variables
So far, we have looked at settings that involve a single random variable.
Many interesting statistics problems require us to examine two or
more random variables.
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n  Combining
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Random Variables
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n  Combining
Definition:
If knowing whether any event involving X alone has occurred tells us
nothing about the occurrence of any event involving Y alone, and vice
versa, then X and Y are independent random variables.
Probability models often assume independence when the random variables
describe outcomes that appear unrelated to each other.
You should always ask whether the assumption of independence seems
reasonable.
In our investigation, it is reasonable to assume X and Y are independent
since the siblings operate their tours in different parts of the country.
Random Variables
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n  Combining
Transforming and Combining Random Variables
The only way to determine the probability for any value of T is if X and Y
are independent random variables.
Let T = X + Y. Consider all possible combinations of the values of X and Y.
Recall: µT = µX + µY = 6.85
σT2 = ∑ (t i − µT ) 2 pi
= (4 – 6.85)2(0.045) + … +
(11 – 6.85)2(0.005) = 2.0775
Note: σX2 = 1.1875 and σY2 = 0.89
What do you notice about the
variance of T?
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Random Variables
Variance of the Sum of Random Variables
For any two independent random variables X and Y, if T = X + Y, then the
variance of T is
σT2 = σX2 + σY2
Is that the Pythagorean
Theorem I see? Why,
In general, the variance of the sum of several independent random
yesvariables
it is!
is the sum of their variances.
Remember that you can add variances only if the two random variables are
independent, and that you can NEVER add standard deviations!
n  Example:
El Dorado Community College
Number of
Credits (Y)
12
15
18
Probability
0.3
0.4
0.3
Transforming and Combining Random Variables
EDCC also has a campus downtown, specializing in just a few
fields of study. Full-time students at the downtown campus
take only 3-credit classes. Let Y = the number of credits taken
in the fall semester by a randomly-selected full-time student at
the downtown campus. Here is the probability distribution:
Transforming and Combining Random Variables
As the preceding example illustrates, when we add two
independent random variables, their variances add. Standard
deviations do not add.
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n  Combining
Mean µX = 15 credits; Variance σ2X = 5.4 credits; Standard Deviation σX = 2.3 credits
If you were to randomly select one full-time student from the
main campus and one full-time student from the downtown
campus and add their number of units, the expected value of
the sum (S = X + Y) would be µS = µX + µY = 14.65 + 15 =
29.65.
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n  Example:
El Dorado Community College
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n  Example:
Tuition, Fees, & Books
Transforming and Combining Random Variables
Let B = the amount spent on books in the fall semester for a
randomly-selected full-time student at EDCC. Suppose that
µB = 153 and σB = 32. Recall that C = overall cost for tuition
and fees for a randomly-selected full-time student at EDCC
and that µC = $832.50 and σC = $103. Find the mean and
standard deviation of the cost of tuition, fees, & books (C + B)
for a randomly-selected full-time student at EDCC.
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Here are all possible combinations of X and Y:
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SAT Scores & the Role of
Independence
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n  Example:
Mean
Standard Deviation
SAT Math Score X
519
115
SAT CR Score Y
507
111
What are the mean and standard deviation of the total score
Math & CR among students applying to this college?
YOUR UNDERSTANDING
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n  CHECK
Cars sold xi 0
1
2
3
Prob. pi
0.4
0.2
0.1
µx = 1.1
σx = 0.943
Cars leased yi
0
1
2
Prob. pi
0.4
0.5
0.1
µy = 0.7
σy = 0.64
Define
Define TT == XX ++ YY
1) 
1) FFind
ind and
and interpret
interpret µµTT..
2) 
2) C
Compute
ompute σσTT assuming
assuming that
that XX and
and YY are
are independent.
independent.
3) 
T
he
dealer’s
manager
receives
a
$500
3) The dealer’s manager receives a $500 bonus
bonus for
for each
each car
car sold
sold and
and
aa $300
$300 bonus
bonus for
for each
each car
car leased.
leased. Find
Find the
the µµ && σσ for
for the
the total
total bonus.
bonus.
Transforming and Combining Random Variables
A large auto dealership keeps track of sales and lease agreements
made during each hour of the day. Let X = the number of cars
sold and Y = the number of cars leased during the first hour of
business on a randomly-selected Friday. Based on previous
records, the probability distributions of X and Y are:
0.3
Transforming and Combining Random Variables
A college uses SAT scores as one criterion for admission.
Experience has shown that the distribution of SAT scores
among its entire population of applicants is such that:
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Random Variables
Mean of the Difference of Random Variables
For any two random variables X and Y, if D = X - Y, then the expected value
of D is
E(D) = µD = µX - µY
In general, the mean of the difference of several random variables is the
difference of their means. The order of subtraction is important!
Variance of the Difference of Random Variables
For any two independent random variables X and Y, if D = X - Y, then the
variance of D is
σD2 = σX2 + σY2
In general, the variance of the difference of two independent random
variables is the sum of their variances.
n  Example:
We have defined several random variables related to Pete’s and Erin’s
tour businesses. For a randomly-selected day:
X = number of passengers on Pete’s trip; C = money Pete collects
µx = 3.75 σx = 1.090
µC = 562.50
σC = 163.50
µY = 3.10 σY = 0.943
µG = 542.50
σG = 165.03
Y = number of passengers on Erin’s trip; G = money Erin collects
Calculate the mean and standard deviation of the difference D = C – G in the
amounts that Pete & Erin collect on a randomly-chosen day. Interpret each value
in context.
Transforming and Combining Random Variables
Pete’s Jeep Tours & Erin’s
Adventures
Transforming and Combining Random Variables
We can perform a similar investigation to determine what happens
when we define a random variable as the difference of two random
variables. In summary, we find the following:
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n  Combining
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n  Example:
El Dorado Community College
T = average tuition at downtown campus: mean = 732.50; stdev = 102.8
U = average tuition at main campus: mean = 825; stdev = 126.50
YOUR UNDERSTANDING
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n  CHECK
Prob. pi
1
0.3
0.4
µx = 1.1
Cars leased yi
0
Prob. pi
0.4
µy = 0.7
2
3
0.2
0.1
σx = 0.943
1
2
0.5
0.1
σy = 0.64
Define D = X – Y. Find and interpret the mean and standard deviation.
The dealer’s manager receives a $500 bonus for each car sold and a
$300 bonus for each car leased. Find the µ & σ for the total bonus.
Transforming and Combining Random Variables
A large auto dealership keeps track of sales and lease agreements made
during each hour of the day. Let X = the number of cars sold and Y = the
number of cars leased during the first hour of business on a randomly
selected Friday. Based on previous records, the probability distribution of
X and Y are as follows.
Cars sold xi 0
Transforming and Combining Random Variables
Suppose we randomly select one full-time student from each of the two
campuses. What are the mean and standard deviation of the
difference in tuition charges?
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Normal Random Variables
An important fact about Normal random variables is that any sum or
difference of independent Normal random variables is also Normally
distributed.
Example
Mr. Boyd likes between 8.5 and 9 grams of sugar in his coffee. Suppose
the amount of sugar in a randomly selected packet follows a Normal distribution
with mean 2.17 g and standard deviation 0.08 g. If Mr. Boyd selects 4 packets at
random, what is the probability his tea will taste right?
µT = µX1 + µX2 + µX3 + µX4 = 2.17 + 2.17 + 2.17 +2.17 = 8.68
σT2 = σX2 1 + σX2 2 + σX2 3 + σX2 4 = (0.08) 2 + (0.08) 2 + (0.08) 2 + (0.08) 2 = 0.0256
σT = 0.0256 = 0.16
n  Example:
Apples
What is the probability that a random sample of 12 apples has a total weight less
than 100 ounces?
Transforming and Combining Random Variables
Suppose that a certain variety of apples have weights that are
approximately Normally distributed with a mean of 9 ounces and a
standard deviation of 1.5oz. If bags of apples are filled by randomly
selecting 12 apples, what is the probability that the sum of the
weights of the 12 apples is less than 100 ounces?
Transforming and Combining Random Variables
So far, we have concentrated on finding rules for means and variances
of random variables. If a random variable is Normally distributed, we
can use its mean and standard deviation to compute probabilities.
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n  Combining
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n  Example:
Put a Lid on It!
What is the probability that a randomly-selected large lid will fit on a randomly
chosen large drink cup?
Transforming and Combining Random Variables
The diameter C of a randomly selected large drink cup at a fast-food restaurant
follows a Normal distribution with a mean of 3.96 inches and a standard
deviation of 0.01 inches. The diameter L of a randomly-selected large lid at
this restaurant follows a Normal distribution with mean 3.98 inches and
standard deviation 0.02 inches. For a lid to fit on a cup, the value of L has to
be bigger than the value of C, but not by more than 0.06 inches.
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Transforming and Combining Random Variables
Summary
In this section, we learned that…
ü 
Adding a constant a (which could be negative) to a random variable
increases (or decreases) the mean of the random variable by a but does not
affect its standard deviation or the shape of its probability distribution.
ü 
Multiplying a random variable by a constant b (which could be negative)
multiplies the mean of the random variable by b and the standard deviation
by |b| but does not change the shape of its probability distribution.
ü 
A linear transformation of a random variable involves adding a constant a,
multiplying by a constant b, or both. If we write the linear transformation of X
in the form Y = a + bX, the following about are true about Y:
ü 
Shape: same as the probability distribution of X.
Center: µY = a + bµX
ü 
Spread: σY = |b|σX
ü 
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Transforming and Combining Random Variables
Summary
In this section, we learned that…
ü 
If X and Y are any two random variables,
µ X ±Y = µ X ± µY
ü 
If X and Y are independent random variables
σX2 ±Y = σX2 + σY2
ü 
The sum or difference of independent Normal random variables follows a
Normal distribution.
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