Download 1.6 Solving Linear Inequalities

Lesson Objectives
At the end of the lesson, students can:
•
Perform linear transformations on Random
Variables.
•
Determine the mean and standard deviation of
transformed Random Variables.
•
Determine the mean and standard deviation of
sums and differences of Random Variables.
Linear Transformation Review
In Unit 1, we studied the effects of transformations on the
shape, center and spread of a distribution of data.
𝒙𝒏𝒆𝒘 = 𝒂 + 𝒃𝒙𝒐𝒍𝒅
Adding (or subtracting) a constant: Adding the same
number, 𝑎, to each observation
•
Adds 𝑎 to measures of center
•
Does not change shape or measures of spread
Multiplying (or dividing) each observation by constant, 𝑏 :
•
Multiplies (divides) measures of center by 𝑏
•
Multiplies (divides) measures of spread
•
Does not change shape of distribution
Linear Transformation of
Random Variables
How are the probability distributions of random variables
affected by similar transformations?
Let’s follow an example through this…
Pete’s Jeep Tours
Pete’s Jeep Tours offers a popular half-day trip in a tourist
area. There must be at least 2 passengers for the trip to run,
an the vehicle will hold up to 6 passengers. Here’s the
probability distribution:
Number of Passengers (X)
Probability
2
.15
3
.25
4
.35
5
.20
Draw the probability histogram.
What is the mean, 𝜇𝑋 ?
What is the standard deviation, 𝜎𝑋 , and variance, 𝜎𝑋 2 ?
6
.05
Pete’s Jeep Tours
Pete’s charges $150 per passenger. Let C = the amount of
money Pete collects on a randomly selected trip. Show the
probability distribution of C: C = 150 * X
Total collected (C)
Probability
Draw the probability histogram.
What is the mean, 𝜇𝐶 ?
What is the standard deviation, 𝜎𝐶 , and variance, 𝜎𝐶 2 ?
How do these relate to the numerical summaries for X?
(shape, center, spread)
Pete’s Jeep Tours
It costs Pete $100 to buy permits, gas, and a ferry pass for
each half-day trip. The amount of profit that Pete makes from
the trip, V, is the total amount of money, C, minus $100. What
is the probability distribution of V. V = C - 100
Total collected (V)
Probability
Draw the probability histogram.
What is the mean, 𝜇𝑉 ?
What is the standard deviation, 𝜎𝑉 , and variance, 𝜎𝑉 2 ?
How do these relate to the numerical summaries for C? for X?
(shape, center, spread)
Rules for MEANS
RULES FOR MEANS Means behave like averages!
Rule 1: If X is a random variable and a and b are fixed
numbers, then
μa+bx = a + bμx
μ a–bx = a – bμx
Rule 2: If X and Y are random variables, then
μ x + y = μx + μy
μ x – y = μx – μy
In other words, the mean of the sum = sum of the means
and the mean of the difference = difference of the means.
Linear Transforms of Random Variables
EXAMPLE: Consider the data on the distribution of the
number of communications units sold by the military division
and the civilian division.
Military Division:
# of Units Sold (X)
Probability
1000
.1
3000
.3
300
.4
500
.5
5000 10,000
.4
.2
Civilian Division:
# of Units Sold (Y)
Probability
750
.1
(a) Find the mean # of units sold collectively by both divisions.
Let X =
Let Y =
μx =
μy =
μX+Y =
Linear Transforms of Random Variables
EXAMPLE: Consider the data on the distribution of the
number of communications units sold by the military division
and the civilian division.
(b) The company makes a profit of $2,000 per military units
sold and $3500 on each civilian unit sold.
What will next year’s mean profit from military sales be?
What will next year’s mean profit from civilian sales be?
(c) Suppose we multiplied each value of X by 2 and added 10.
(2X + 10). How would this affect the mean?
Rules for Variances
RULES FOR VARIANCE
Rule 1: If X is a random variable and a and b are fixed
numbers, then
σ2a+bx = b2σ2x
σ2a-bx = b2σ2x
NOTE: Multiplying X by a constant “b” multiplies the
variance of X by the “b2”. The variance of X + a is the
same as the variance of X.
Rules for Variance
Rule 2: If X and Y are independent random variables, then
σ2x+y = σ2x + σ2y
σ2x-y = σ2x + σ2y
s.dev. = σ2x+y =
(σ2x + σ2y )
In other words: Add the variances, then take square root
(This is called the “Addition rule for variances of independent random variables.”)
Pete’s Sister Erin
Pete’s sister, Erin, who lives in another part of the country,
decided to join the business. She has a slightly smaller
vehicle. The probability distribution for her business is:
Number of Passengers (Y)
Probability
2
.3
3
.4
4
.2
5
.1
Draw the probability histogram.
What is the mean, 𝜇𝑌 ?
What is the standard deviation, 𝜎𝑌 , and variance, 𝜎𝑌 2 ?
What is the average number of passengers Pete and Erin
expect to have on their tours on a randomly selected day?
Independent Random Variables
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.
Are X, the number of Pete’s passengers on a random day,
and Y, the number of Erin’s passengers on a random day,
independent?
Pete’s and Erin’s Combined Business
Pete and Erin looked at their businesses together, where
T = X + Y. Assuming their businesses are independent, the
probability distribution for their combined businesses is:
Number of Passengers (T)
Probability
4
5
6
.045 .135 .235
7
.265
8
9
.190 .095
10
11
.030
.005
What is the mean, 𝜇 𝑇 ?
What is the standard deviation, 𝜎𝑇 , and variance, 𝜎𝑇 2 ?
NOTE: You have the know-how to compute the probabilities for
the above table yourself!
(See page 366 in book)
Rules for Means and Variances
Rules for Variance
NOTE: When random variables are not independent, the
variance of their sum depends on the relationship between
them as well as on their individual variances. We use ρ
(Greek letter “rho”) for the correlation between two random
variables. The correlation ρ is a number between -1 and 1 that
measures the strength and direction of the linear relationship
between the two variables. The correlation between two
independent random variables is zero.
We will not be looking into detail with variance of not
independent random variables.
EXAMPLE: Earlier, we defined X = the number of passengers
on Pete’s trip, Y = the number of passengers on Erin’s trip, and
C = the amount of money that Pete collects on a randomly
selected day. We also found the means and standard
deviations of these variables:
μx = 3.75
μy = 3.10
μC = 562.50
σx = 1.090
σy = 0.943
σC = 163.50
(a) Erin charges $175 per passenger for her trip. Let G = the
amount of money that she collects on a randomly selected day.
Find the mean and standard deviation of G.
EXAMPLE: Earlier, we defined X = the number of passengers
on Pete’s trip, Y = the number of passengers on Erin’s trip, and
C = the amount of money that Pete collects on a randomly
selected day. We also found the means and standard
deviations of these variables:
μx = 3.75
μy = 3.10
μC = 562.50
σx = 1.090
σy = 0.943
σC = 163.50
(b) Calculate the mean and the standard deviation of the total
amount that Pete and Erin collect on a randomly chosen day.
Linear Transformations of
Random Variables
EXAMPLE: 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 of X is:
# of Cars Sold (X)
Probability
0
.3
1
.4
2
.2
3
.1
The mean is 𝜇𝑋 = 1.1 ; the standard deviation is 𝜎𝑋 = 0.943 .
Suppose the dealership’s manager receives a $500 bonus for
each car sold. Let Y = the bonus received. Find 𝜇𝑌 and 𝜎𝑌 .
𝜇𝑌 = (500)(1.1) = $550
𝜎𝑌 = (500)(0.943) = $471.50
Linear Transformations of
Random Variables
EXAMPLE: 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 of X is:
# of Cars Sold (X)
Probability
0
.3
1
.4
2
.2
3
.1
The mean is 𝜇𝑋 = 1.1 ; the standard deviation is 𝜎𝑋 = 0.943 .
To encourage customers to buy cars, the manager spends $75
to provide coffee and doughnuts. The manager’s net profit, T,
on a random Friday is the bonus earned minus $75. Find 𝜇 𝑇
and 𝜎𝑇 .
𝜇 𝑇 = (500)(1.1) – 75 = $475; 𝜎𝑇 = (500)(0.943) = $471.50
Combining Random Variables
Example: A large auto dealership keeps track of
sales and lease agreements made during each
hour of the day. Let X = # of cars sold, and Y =
# 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 as follows:
(continued on next slide . . .)
Example continued. . .
Cars Sold, xi
0
1
2
3
Probability, pi
0.3
0.4
0.2
0.1
Cars Leased, yi
0
1
2
Probability, pi
0.4
0.5
0.1
Example continued. . .
µx = 1.1; σx = 0.943 and µY = 0.7; σY = 0.64
Define T = X + Y.
1) Find and interpret µT.
µT = µx + µY = 1.1 + 0.7 = 1.8. On average, this
dealership sells or leases 1.8 cars in the first hour of
business on Fridays.
2) Compute σT assuming that X and Y are independent.
σT = (0.943)2 +(0.64)2 = 1.14
Example continued . . .
3) The dealership’s manager receives a $500 bonus for
each car sold and a $300 bonus for each car leased.
Find the mean and standard deviation of the manager’s
total bonus, B. Show work!
µB = 500(1.1) + 300(0.7) = $760
σB =
𝟓𝟎𝟎 𝟐 (𝟎. 𝟗𝟒𝟑)𝟐 +(𝟑𝟎𝟎)𝟐 (𝟎. 𝟔𝟒)𝟐 = $𝟓𝟎𝟗. 𝟎𝟗
Combining Normal Random Variables
If a random variable is Normally distributed, we can use its
mean and variance to compute probabilities. What if we
combine two normal random variables?
Any linear combination of independent Normal random
variables is also Normally distributed!
If X and Y are independent Normal random variables and a
and b are any fixed numbers, then aX + bY is also Normally
distributed. You must communicate this fact about the
distribution’s shape!
Combining Normal Random Variables
EXAMPLE: Tom and George are avid golf players. Their
scores vary as they play the course repeatedly according to
the following distributions:
Tom’s score X: N(110, 10)
George’s score Y: N(100, 8)
If they play independently, what is the probability that Tom will
score lower than George (and thus do better in the
tournament)?
APPLES!
Suppose that a certain variety of apples
have weights that are approximately
Normally distributed with a mean of 9 oz
and a standard deviation of 1.5 oz. 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 oz?
P(X<100) = 0.0620
Speed Dating
• To save time and money, many single people have decided to
try speed dating. At a speed-dating event, women sit in a
circle, and each man spends about 5 minutes getting to know a
woman before moving on to the next one.
• Suppose that the height M of male speed daters follows a
Normal distribution, with a mean of 70” and a standard
deviation of 3.5”, and suppose that the height F of female
speed daters follows a Normal distribution, with a mean of 65”
and a standard deviation of 3”.
• What is the probability that the man is taller than the woman in
a randomly selected speed-dating couple? Use 4-step process!
• P(M>F) = P(D>0), where D = M-F. P(D>0) = 0.8610
Lesson Objectives
At the end of the lesson, students can:
•
Perform linear transformations on Random
Variables.
•
Determine the mean and standard deviation of
transformed Random Variables.
•
Determine the mean and standard deviation of
sums and differences of Random Variables.