### 7.2, part 2 - TeacherWeb

```7.2 Part 3 Rules for Means and
Variances

Students will use the rules for means and variances
to solve problems.
Linear Transformations Review


Multiplying (or dividing) each observation by a constant b (positive,
negative, or zero):

Multiplies (or divides) measures of center and location (mean, median,
quartiles, percentiles) by b

Multiplies (or divides) measures of spread (range, IQR, standard deviation)
by |b|

Does not change the shape of the distribution
If you are multiplying or dividing, you are basically stretching or
Linear Transformations Review


(positive, negative or zero) to each observation:

Adds a to measures of center and location (mean, median, quartiles,
percentiles)

Does not change shape or measures of spread (range, IQR, standard
deviation)
the left or to the right
Rules for Means and Variances
1. Suppose μX = 5 and μY = 10.
According to the rules for means, what
is μX+Y ?

Rule 2 for means states that μX+Y = μX + μY
 So
μX+Y = 5 + 10 = 15
2. Suppose μX = 2. According to the rules for
means, what is μ3+4X ?
• Rule 1 for means states that μa+bX = a + bμX
– So μ3+4X = 3 + 4(2) = 11
3. Suppose σX2 = 2 and σY2 = 3 and X and Y are
independent random variables. According to the
rules for variances, what is σX2+ Y ? What is σX + Y ?

Rule 2 for variances states that

2
X Y
  
2
 So 𝜎𝑋+𝑌
=


2
X
2
Y

2
X Y
  
2
X
2
Y
2 + 3 = 5 and 𝜎𝑋+𝑌 = 5 = 2.23
This rule only applies if X and Y are independent
 we add the variances whether
Keep in mind that
we are finding the sum or difference because in
both situations we are combining distributions,
which gives us more variability
4. Suppose σ2= 4. According to the rules for
variances, what is σ23+4x? What is σ3+4x?

Rule 1 for variances states that
 So

2
a bX
b 
2
2
X
σ23+4x = (42)(4) = 64 and σ3+4x = 8.
5. What is the best
way to combine standard
deviations?
• Standard deviations are most easily combined by
using the rules for variances.
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, and the vehicle will hold up to 6
passengers. The number of passengers X on a randomly selected day has the
following probability distribution:
# of passengers X
2
3
4
5
6
Probability
0.15
0.25
0.35
0.20
0.05

Find the mean and standard deviation of X.
1.
L1: number of passengers X
2.
L2: probability
3.
L3: L1 x L2 (The mean is the sum of L3.)
4.
L4: (L1-mean)2
5.
L5: L4 x L2 (The variance is the sum of L5.)
6.
The Standard deviation is the square root of the sum of L5.
Pete’s Jeep Tours
# of passengers X
2
3
4
5
6
Probability
0.15
0.25
0.35
0.20
0.05

Pete charges \$150 per passenger. Let C = the total amount of money
that Pete collects on a randomly selected trip.

C = 150X

The probability distribution of C looks like this:

Amount of money
collected C
\$300
\$450
\$600
\$750
\$900
Probability
0.15
0.25
0.35
0.20
0.05
Note that the probability doesn’t change
Pete’s Jeep Tours
# of passengers X
2
3
4
5
6
Probability
0.15
0.25
0.35
0.20
0.05
Amount of money
collected C
\$300
\$450
\$600
\$750
\$900
Probability
0.15
0.25
0.35
0.20
0.05
μX = 3.75
σX = 1.090
σ2X = 1.188
μC = \$562.50
σC = \$163.50
σ2C = \$26,718.75
Using the rules, we know that since C = 150X:
μC = 150(μX) = 150(3.75) = 562.5
σ2C = 1502(σ2X) = (22,500)(1.188) = 26,730
Or
σC = 1.090(150) = 163.5
Pete’s Jeep Tours
Amount of money
collected C
\$300
\$450
\$600
\$750
\$900
Probability
0.15
0.25
0.35
0.20
0.05

It costs Pete \$100 to buy permits, gas, and a ferry pass for each half-day
trip. The amount of profit V that Pete makes from the trip is the total
amount of money C that he collects from passengers minus \$100. That is,
V = C – 100. If Pete has only two passengers on the trip (X = 2), then C =
300 and V = 200.

V = C – 100 or V = 150X – 100

The probability distribution of V looks like this:
Amount of profit V
\$200
\$350
\$500
\$650
\$800
Probability
0.15
0.25
0.35
0.20
0.05
Pete’s Jeep Tours
μX = 3.75
σX = 1.090
μC = \$562.50
σ2X = 1.188
σC = \$163.50
σ2C = \$26,718.75
Amount of profit V
\$200
\$350
\$500
\$650
\$800
Probability
0.15
0.25
0.35
0.20
0.05
Using the rules, we know that
since V = C – 100:
μV = μC – 100 = 562.5 – 100 = 462.5
σ2V and σV stay exactly the same
as σ2C and σC because the
amount of variability hasn’t
increased or decreased
μV = \$462.50
σV = \$163.50
σ2V = \$26,718.75
Using the rules, we know that since V =
150X – 100
μV = 150(μX) – 100 = 150(3.75) – 100 =
562.5 – 100 = 462.5
σ2V = 1502(σ2X) = (22,500)(1.188) = 26,730
Or
σV = 1.090(150) = 163.5

Pete’s sister Erin, who lives near a tourist area in another part of the
country, is impressed by the success of Pete’s business and decides to
join the business, running tours on the same days as Pete in her slightly
smaller vehicle, under the name “Erin’s Adventures.” After a year of
steady bookings, Erin discovers that the number of passengers Y on her
half-day tours has the following probability distribution:
# of passengers Y
2
3
4
5
Probability
0.3
0.4
0.2
0.1
μY = 3.10
σY = 0.943
σ2Y = .889
# of passengers Y
2
3
4
5
Probability
0.3
0.4
0.2
0.1
μY = 3.10

σY = 0.943
σ2Y = .889
How many total passengers T can Pete and Erin expect to have
on their tours on a randomly selected day?
Using the rules, we know that since T = X + Y:
μT = μX + μY = 3.75 + 3.10 = 6.85
σ2T = σ2X + σ2Y = 1.188 + .889 = 2.077
σT = 2.077 = 1.441
• The probability distribution for T can be found:
xi
pi
yi
pi
ti = xi + yi
pi
2
0.15
2
0.3
4
(0.15)(0.3) = 0.045
2
0.15
3
0.4
5
(0.15)(0.4) = 0.060
2
0.15
4
0.2
6
(0.15)(0.2) = 0.030
2
0.15
5
0.1
7
(0.15)(0.1) = 0.015
3
0.25
2
0.3
5
(0.25)(0.3) = 0.075
3
0.25
3
0.4
6
(0.25)(0.4) = 0.100
3
0.25
4
0.2
7
(0.25)(0.2) = 0.050
3
0.25
5
0.1
8
(0.25)(0.1) = 0.025
4
0.35
2
0.3
6
(0.35)(0.3) = 0.105
4
0.35
3
0.4
7
(0.35)(0.4) = 0.140
4
0.35
4
0.2
8
(0.35)(0.2) = 0.070
4
0.35
5
0.1
9
(0.35)(0.1) = 0.035
5
0.2
2
0.3
7
(0.2)(0.3) = 0.060
5
0.2
3
0.4
8
(0.2)(0.4) = 0.080
5
0.2
4
0.2
9
(0.2)(0.2) = 0.040
5
0.2
5
0.1
10
(0.2)(0.1) = 0.020
6
0.05
2
0.3
8
(0.05)(0.3) = 0.015
6
0.05
3
0.4
9
(0.05)(0.4) = 0.020
6
0.05
4
0.2
10
(0.05)(0.2) = 0.010
6
0.05
5
0.1
11
(0.05)(0.1) = 0.005

The probability distribution for T can be found:
Combined # of
passengers
4
Probability
0.045
μT = 6.85
5
6
7
8
9
10
11
0.135 0.235 0.265 0.190 0.095 0.030 0.005
σT = 1.441
σ2T = 2.0775

Let D = X – Y, or the difference in the number of
passengers that Pete and Erin have on their tours on a
randomly selected day.
μD = μX – μY = 3.75 – 3.10 = 0.65
σ2D = σ2X + σ2Y = 1.188 + .889 = 2.077
σD = 2.0477 = 1.441
7.41 Time and Motion (pg. 500)

(a)
A time and motion study measures the time required for an
assembly-line worker to perform a repetitive task. The data show
that the time required to bring a part from a bin to its position on
an automobile chassis varies from car to car with mean 11
seconds and standard deviation 2 seconds. The time required to
attach the part to the chassis varies with mean 20 seconds and
standard deviation 4 seconds.
What is the mean time required for the entire operation of
positioning and attaching the part?
7.41 Time and Motion (pg. 500)
(b) If the variation in the worker’s performance is reduced by better
training, the standard deviations will decrease. Will this decrease
change the mean you found in (a) if the mean times for the two
steps remain as before?
(c) The study finds that the times required for the two steps are
independent. A part that takes a long time to position, for
example, does not take more or less time to attach than other
parts. Find the standard deviation of the time required for the
two-step assembly operation.
7.42 Electronic circuit (pg. 500)

The design of an electronic circuit calls for a 100-ohm resistor and
a 250-ohm resistor connected in series so that their resistances
add. The components used are not perfectly uniform, so that the
actual resistances vary independently according to Normal
distributions. The resistance of 100-ohm resistors has mean 100
ohms and standard deviation 2.5 ohms, while that of 250-ohm
resistors has mean 250 ohms and standard deviation 2.8 ohms.
(a)
What is the distribution of the total resistance of the two
components in series?
(b)
What is the probability that the total resistance lies between 345
and 355 ohms?
HOMEWORK

7.38-40, 42, 45, 46 (show work for b, you
can use your calculator to calculate
expected value for a), 47, 49-52
```