Combining Two Random Variables
1. Suppose X and Y are random variables with  X =35,  X =8,  Y =72,  Y =4. Given that X and Y are independent
variables, calculate the following:
 X Y  107
 X Y 
 2X 
-37
64
 2 X Y 
 X Y 
 X Y 
64+16=80
80  8.94
80  8.94
2. Now that the new models are here, a car dealership has lowered prices on last year’s models. An aggressive
salesperson estimates the following probability distribution of X, the number of cars that she’ll sell next week.
X
0 1 2
3 4
P( X ) .05 .15 .35 .25 .20
a) Determine the expected value, variance and standard deviation (from homework).
E(X) or mean = (0)(.05)+(1)(.15)+(2)(.35)+(3)(.25)+(4)(.2) = 2.4
 2 =1.24
  1.24  1.114
b) Suppose that this salesperson earns a $200 commission for each car sold. What are her expected wages for
next week? What is the standard deviation of those wages?
All this means is what now happens to the answers in part a if the numbers are multiplied by 200?
200 X = (200)(2.4) = $480
 200 X = (200)(1.114) = $222.80
c) Suppose the employee’s complain to the owner about their pay. They want some type of guaranteed salary
and more commission. The owner relents and decides they get $150 each plus $250 commission for each
car sold. What are her expected wages for next week? What is the standard deviation of those wages?
150250 X
= 150+(250)2.4) = $750
150250 X = (250)(1.114) = 278.50
3. The probability distribution below represents (Y) the length of long distance calls in minutes.
Y
5 10 15 20 25
P(Y ) .1 .2 .3 .3 .1
a) What is the expected length of a long distance call? The variance and standard deviation.
Mean = 15.5 mins
Variance = 32.25
Standard deviation = 5.678 mins
b) Suppose that there is an initial connection charge of $.60 and a further charge of $.015 for each minute.
What is the expected cost of a long distance call? What is the standard deviation of the cost of the call?
Mean = 0.60 + (0.015)(15.5) = $0.8325 = approx. 83 cents per call
Standard deviation = (0.015)(5.678) = $0.08517 = approx. 8.5 cents per call
4. For a given high school basketball team, the number of baskets (X) for the leading scorer is E(X) = 8.3 with (X) =
1.25 and the number of baskets (Y) for the second leading scorer is E(Y) = 6.6 with (Y) = 2.31.
a) Together, how many baskets would we expect each game?
8.3 + 6.6 = 14.9 baskets
b) What is the standard deviation of this number?
  1.252  2.312  2.63
c) How many more baskets would we expect the leading scorer to have?
 X Y = 8.3 – 6.6 = 1.7 more baskets
d) What is the standard deviation of this number?
  1.252  2.312  2.63 (variances can only add)
e) Hypothetically, let’s say the leading scorer shoots only 3-point baskets and the second leading scorer shoots
2-point baskets, how many points would we expect each game from the two players?
3 X  2Y = (3)(8.3)+(2)(6.6) = 38.1 points
f)
What is the standard deviation of this number?
 3 X  2Y  3.752  4.622  5.95
5. A department supervisor is considering purchasing one of two comparable photocopy machines, A or B.
Machine A costs $10,000 and machine B costs $10,500. This department replaces photocopy machines
every three years. The repair contract for machine A costs $50 per month and covers an unlimited number
of repairs. The repair contact for machine B costs $200 per repair. Based on past performance, the
distribution of the number of repairs needed over any one-year period for machine B is shown below.
Number of Repairs
0
1
2
3
Probability
0.50 0.25 0.15 0.10
You are asked to give a recommendation based on overall cost as to which machine, A or B, along with its
repair contact, should be purchased. What would your recommendation be? Give a statistical justification
to support your recommendation.
Machine A cost over 3 years (36 months):
$10,000 + 36 × $50 = $11,800
Machine B cost over 3 years (36 months):
$10,500 + $200 × 3 × ( 0×.5 + 1×.25 + 2×.15 + 3×.10) = $11,010
Go with Machine B
6.
Suppose X and Y are random variables with  X =100,  X =10,  Y =72,  Y =18. Given that X and Y are
independent variables, calculate the following:
 2 X Y 
 X Y 
 X Y 
 X Y 
 X Y 
 2X 
μx+Y = 172
μx-Y = 28
σ2x = 100
σ2X+Y = 424
σX+Y = 20.59
σX-Y = 20.59
7. The following are the distributions for the number of goals scored per soccer game for the three top forwards from
Vareebull High School in Random, Texas.
Xavier
Yousuf
X
0
1
2
P( X ) .10 .55 .35
a)
b)
c)
d)
e)
a)
b)
c)
d)
e)
Zoué
Y
0
1
2
P(Y ) .20 .75 .05
Z
0
1
2
3
P( Z ) .80 .10 .05 .05
Calculate the mean, variance and standard deviation for the number of goals that Xavier scores.
Calculate the mean, variance and standard deviation for the number of goals that Yousuf scores.
Calculate the mean, variance and standard deviation for the number of goals that Zoué scores.
Calculate the mean and standard deviation for the number of goals that all three players score together.
Calculate the mean and standard deviation for the difference in goals between Xavier and Zoué.
Xavier:
Yousuf:
Zoue:
All 3:
Xavier-Zoue:
μ=
1.25
μ=
0.85
μ=
0.35
μx+y+z = 2.45
μx-z = 0.9
σ2 = 0.3875
σ2 = 0.2275
σ2 = 0.6275
σ2x+y+z = 1.2425
σ2x-z = 1.015
σ = 0.6225
σ = 0.477
σ = 0.792
σx+y+z = 1.1147
σx-z = 1.0075
8. Chen, Hope, Marco and Panchali love to visit the local bakery that specializes in muffins. Listed below are mean and
standard deviation of the number of muffins each buys per visit.
Chen:  = 2.3 muffins,  = 1.1 muffins
Hope:  = 3.9 muffins,  = 2.2 muffins
Marco:  = 5.2 muffins,  = 0.7 muffins
Panchali:  = 4.2 muffins,  = 1.4 muffins
a) If all four friends visit the bakery, what is the expected number of muffins purchased? The standard
deviation?
b) How many more muffins would we expect Marco to purchase than Chen? The standard deviation?
c) Hope promised to buy 4 extra muffins for her colleagues. How many muffins would we expect her to
purchase? The standard deviation?
d) The two ladies love going Thursday mornings because all muffins are $2.50 each. How much would we
expect them to spend? The standard deviation?
e) Marco buys the $4 mega muffins and Panchali buys the regular $3.25 muffins. How much higher will
Marco’s cost than Panchali’s? The standard deviation?
a)
b)
c)
d)
e)
μC+H+M+P = 15.6
σ2C+H+M+P = 8.5
σC+H+M+P = 2.915
2
μM-C = 2.9
σ M-C
= 1.7
σM-C
= 1.304
μH+4 = 7.9
σ2H+4
= 4.84
σH+4
= 2.2
($2.50)μ (C+H) = $15.50
σ2C+H
= 6.05
($2.50)σC+H = 6.15
The cost: 4(5.2) – 3.25(4.2) = $7.15
The standard deviation: 5.34
You get the standard deviation by taking the square root of {(4*0.7)2 + (3.25*1.4)2}