251solnJ1 3/4/08
(Open this document in 'Page Layout' view!)
J. Random Variables.
1. Definitions.
2. Probability Distribution of a Discrete Random Variable.
3. Expected Value (Expectation) of a Discrete Random Variable.
D&C pg. 140 means for 1 - 4. J1, J3. (J0A, J1)
4. Variance of a Discrete Random Variable.
D&C pg. 140 standard deviations for 1 - 4. J2 (J0B). Text 5.3, 5.4, 5.6. J4 –J7 (J2 – J5).
5. Summary
J8, J9 (J6, J7)
6. Continuous Random Variables.
a. Normal Distribution (Overview).
b. The Continuous Uniform Distribution.
Text 6.23, 6.25[6.24*, 6.26*], J10, J12 (J8. J10)
c. Cumulative Distributions, Means and Variances for Continuous Distributions.
d. Chebyshef’s Inequality Again.
J11 (J9).
7. Skewness and Kurtosis (Short Summary).
This document includes solutions for sections 1-5
D&C pg 130, Application 1: (Mean only) Find
the expected value from the distribution below.
x
Px 
x P x 
1
.25
0.25
2
.25
0.50
3
.25
0.75
4
.25
1.00
1.00
2.50
 Px   1.00 (valid
distribution!),   E x    xPx   2.5 .
From this table
D&C pg 130, Application 2: (Mean only) Find the expected value from the distribution below.
x
P x 
x Px 
Px   1.00 (valid
From this table
10
.500
5.00
20
.250
5.00
xPx   18 .75 .
distribution!),   E x  
30
.125
3.75
40
.125
5.00
1.000
18.75

D&C pg 130, Application 3: (Mean only)   0 . For computations see below.
D&C pg 130, Application 4: (Mean only)   3 For computations see below.

251solnJ1 3/4/08
(Open this document in 'Page Layout' view!)
Problem J1 (Formerly J0A): Flip a coin 4 times. Let x be the number of heads. Find the following:
a. The distribution of x .
b. Find the probability of (i) at least one head, (ii) an even number of heads, (iii) at least 2 heads.
c. Find E x 
Solution: a) First, using the multiplication rule for independent events, we know that the probability of any
one pattern, like HTHT is  1 2  1 2  1 2  1 2   116 . For this particular pattern there are two heads, so x  2 .
But there are other ways to get 2. We have to find out how many. There are two ways to do this.
The easy way: We have to replace two locations in TTTT with heads. The number of ways we can do this
4!
43
is C 24 

 6 , so the probability is 6  116  616 .
2! 2! 2 1
The hard way: List all the possible patterns and add their probabilities. There are 16 equally likely
patterns.
Pattern
HHHH
x Probabilit y
1
4
16
HHHT
HHTH
HHTT
HTHH
3
3
2
3
HTHT
HTTH
2
2
HTTT
THHH
1
3
THHT
THTH
2
2
THTT
TTHH
1
2
TTHT
TTTH
1
1
TTTT
0
Regardless of how we do it, we get the
following:
x Number of ways to
Probability
get this value
P0  1 116 
4!
4
0 C 0  4! 0!  1
 .0625
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
C 34 
4! 4
 4
1! 3! 1
2
C 24 
4!
43

6
2! 2! 2 1
3
4! 4
C 34 
 4
1! 3! 1
P1  4  116  416
 .2500
P2  6  116  616
 .3750
P3  4  116  416
 .2500
P4  1  116 
4!
1
0! 4!
 .0625
Note that these probabilities add to one.
4
C 44 
1
16
1
16
b. (i) At least one head. Px  1  1  P0  1  .0625  .9375
(ii) An even number of heads P2  P4  .3750  .0625  .4375
(iii) At least 2 heads. Px  2  P2  P3  P4  .3750  .2500  .0625  .6875
c.
x
0
1
2
3
4
P x 
.0625
.2500
.3750
.2500
.0625
1.0000
x Px 
0.0000
0.2500
0.7500
0.7500
0.2500
2.0000
x 2 P x 
0.00
0.25
1.50
2.25
1.00
5.00
 Px   1.00 (valid
distribution!),   E x    xPx   2.00 . .
From this table
251solnJ1 3/4/08
(Open this document in 'Page Layout' view!)
D&C pg 130, Application 1: (Variance) Find the variance of the distribution below. The first two columns
only were given. The solution is given using both the computational and definitional method. Please use
only one method on an exam.
x
Px 
x P x 
x 2 P x 
 x    P x   x   2 P x 
x   
1
.25
0.25
0.25
-1.5
-0.375
0.5625
2
.25
0.50
1.00
-0.5
-0.125
0.0625
3
.25
0.75
2.25
0.5
0.125
0.0625
4
.25
1.00
4.00
1.5
0.375
0.5625
1.00
2.50
7.50
0.000
1.2500
2
xPx   2.5 ,
Px   1.00 (valid distribution!),   E x  
x Px   7.50 ,
From this table

 x   Px   0.00 (a check for accuracy!) and  x   


2
Px   1.25 . Remember! You do not need
the last two columns if you use the computational formula.
If you use the computational formula (the easy way!): Var x    x2  E x 2   2 
 
x
2
Px    2
 7.50  2.52  1.25 . If you use the definitional formula (the hard way – 2 extra columns):

  x    Px  1.25 .
Varx    x2  E x   2 
2
D&C pg 140, Application 2: (Variance) Find the variance of the distribution below. The first two columns
only were given. The solution is given using both the computational and definitional method
x
P x 
x Px 
x 2 P x 
 x    P x 
 x   2 P x 
x   
10
.500
5.00
50.00
-8.75
-4.37500
38.2812500
20
.250
5.00
100.00
1.25
0.31250
0.3906250
30
.125
3.75
112.50
11.25
1.40625
15.8203125
40
.125
5.00
200.00
21.25
2.65625
56.4453125
1.000
18.75
462.50
0.00000
110.937500
2
x Px   462 .50 ,
Px   1.00 (valid distribution!),   E x  
xPx   18 .75 ,
From this table



 x   Px   0.00 and  x    Px  110.7735 . If you use the computational formula:
Var x     E x      x Px     462 .50  18 .75   110 .9375 . If you use the definitional
formula: Varx     E x       x   2 Px  110.9375 .
2
2
x
2
2
2
2
2
2
2
x
D&C pg 140, Application 3: (Variance)
Find the variance of the distribution below. The first two columns only were given. The solution is given
using both the computational and definitional method.
x
P x 
x Px 
x 2 P x 
-1
.5
-.5
.5
1
.5
.5
.5
1.0
0.0
1.0
xPx   0 and
Px   1.00 (valid distribution!),   E x  
x 2 P  x   1 .0 .
From this table
Var x   
2
x

 E x      x
2
2

2

Px     1.0  0  1 . Because the mean is zero and the standard
2
2
deviation is 1, this can be called a standardized variable.
251solnJ1 3/4/08
(Open this document in 'Page Layout' view!)
D&C pg 140, Application 4: (Variance)
Find the variance of the distribution below. The first two columns only were given. The solution is given
using both the computational and definitional method
Px   1.00 (valid
From this table
x
P x 
x Px 
x 2 P x 
1
.1
0.1
0.1
xPx   3.0 and
distribution!),   E x  
2
.2
0.4
0.8
2
3
.3
0.9
2.7
x Px   10 .0 .
4
.4
1.6
6.4
Var x    x2  E x 2   2 
x 2 Px    2
1.0
3.0
10.0


 
 10 .0  9.0  1.0 .


2
Problem J2 (Formerly J0B): In problem J1 find the standard deviation of the number of heads.
P x 
.0625
.2500
.3750
.2500
.0625
1.0000
x
0
1
2
3
4
x Px 
0.0000
0.2500
0.7500
0.7500
0.2500
2.0000

Px   1.00 (valid
x 2From
Px  this table
0.00
xPx   2.0 and
distribution!),   E x  
0.25
1.50
x 2 Px   5.00 .
2.25
Var x    x2  E x 2   2 
x 2 Px    2
1.00
5.00
 5.0  1.02  1.0 .   Var x   1.0  1.0.

 


Exercise 5.3: A dealership has the frequency of occurrence for sales over the last 200 days as shown below
(for example on there were 100 days out of the 200 on which one car was sold).
Find the probability distribution, the expected value and standard deviation and find the probability that on
a given day d) fewer than 4 cars will be sold, e) at most 4 cars will be sold, f) at least 4 cars will be sold, g)
exactly four cars will be sold and g) more than 4 cars will be sold.
a) – c) To find the probabilities; divide the frequencies by the total size of the sample.
P x 
.080
.200
.284
.132
.072
.060
.052
.040
.032
.028
.016
.004
1.000
f
x
0
1
2
3
4
5
6
7
8
9
10
11
40
100
142
66
36
30
26
20
16
14
8
2
500
x Px 
0.000
0.200
0.568
0.396
0.288
0.300
0.312
0.280
0.256
0.252
0.160
0.044
3.056
x 2 P x 
0.000
0.200
1.136
1.188
1.152
1.500
1.872
1.960
2.048
2.268
1.600
0.484
15.408
 Px   1.00 (valid distribution!),   E x    xPx   3.056 and  x
 E x      x Px     15 .408  3.056   6.068864 .
From this table
Var x    x2
2
2
2
2
2
Px   15 .408 .
2
  6.068864  2.4635064. Please note that you should be able to find the probabilities below even
if they are not in your edition!
According to the Instructor’s Solutions Manual
(d)
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= (0.080 + 0.200 + 0.284 + 0.132) = 0.696
(e)
P(X  4) = P(X < 4) + P(X = 4) = 0.696 + 0.072 = 0.768
251solnJ1 3/4/08
(Open this document in 'Page Layout' view!)
P(X  4) = 1 – P(X < 4) = 1 – 0.696 = 0.304
P(X = 4) = 0.072
P(X > 4) = 1 – P(X  4) = 1 – 0.768 = 0.232
(f)
(g)
(h)
Exercise 5.4: If the number of traffic accidents in a small city is as below, find the mean and standard
deviation.
x P x 
Row
1
2
3
4
5
6
0
1
2
3
4
5
0.10
0.20
0.45
0.15
0.05
0.05
(a)-(b) According to the Instructor’s Solutions Manual (a) Mean = 2.00, (b) Variance = 1.40 and )
Stdev = 1.18321596. If you can’t get the same answers, go back to application 1!
Exercise 5.6: The game is called Under-or-Over-Seven and there are 3 ways to play. If y is the sum of the
number of dots on the two faces up, Game a is to win $1 if y  7 and to lose $1 if y  7 , Game b is to win
$1 if y  7 and to lose $1 if y  7 and Game c is to win $4 if y  4 and to lose $1 4 if y  4 . Find the
probability distribution function for the outcomes of games a, b, and c. Show that the expected profit or loss
is the same regardless of game played.
y
1
2
3
4
5
6
7
1
2
Diagram for dice problems.
x 3
4
5
6
2
3
4
5
6
7
8
3
4
5
6
4 5 6 7
5 6 7  8
6 7 8 9
7  8  9  10 
8  9  10  11 
9  10  11  12 
From this you should be able to count points to get the probabilities.
Game a: $1 if sum under 7, $-1 if sum not under 7.
Game c: $4 if sum is 7, $-1 if not 7.
Game b: $1 if sum over 7, $-1 if sum not over 7.
Px 
x
-1
+1
Sum
21
36
15
36
Variance
x 2 P x 
21
36
15
36
6

36
21
36
15
36

1
2 
x P x 
x
2
Px  
1
-1
+4
Sum
 xPx
2
Px 
x
30
36
6
36
x 2 P x 
30
36
24
36
6

36
30
36
96
36
126
36

1
2 
x P x 
x
2
Px  
 xPx
2
2
2
Variance
126  6 
 6 
 1     0.97222

    3.47222
36  36 
 36 
  .9860
  1.8634
a, b and c are answered above and, according to the Instructor’s Solutions Manual,
(d)
Ex   6 36  $ – 0.167 for each method of play.
But the most interesting part of this problem is that, if you compute variances, you can see that game c is
much riskier than game a or game b.
251solnJ1 3/4/08
(Open this document in 'Page Layout' view!)
Problem J3 (Formerly J1): Assume that values of x as shown below represent sales of your product.
Sales
Probability
Except for part a do the parts below in two ways - first,
x
p(x)
by figuring out each value of the function of x and its
0
.50
probability, and, second, by using a formula for a function
1
.10
of x .
2
.10
a. What is Ex ,E x 2 ?
3
.10
b. If costs are C1  10 x  10 , what is E C1  ?
4
.20
c. If costs are C2  0.5x 2  10x  10 , What is E C 2  ?
 
d. If the product sells for $25 per unit, and R represents revenue, what is E R  ?
e. If Costs are described by C1 , and  represents profit, what is E   ?
f. If Costs are described by C 2 , and  represents profit, what is E   ?
Solution: To do these directly, make a table by first computing the values of the functions of x for each
value of x and then multiplying each value by the probability of the corresponding value of x.
x
0
1
2
3
4
px 
xpx 
.50
.10
.10
.10
.20
1.00
0
.10
.20
.30
.80
1.40
x2
x 2 px 
0
1
4
9
16
0
0.10
0.40
0.90
3.20
4.60
E x2
C1 px 
C1
10
20
30
40
50
5
2
3
4
10
24
E C1 
 
x
 
a. From the table: Ex  1.40,E x 2  4.60 .
b. Using the formula Eax  b  aEx   b,
 10 Ex   10  101.4  10  24 .
C 2 px 
C2
10.0
19.5
28.0
35.5
42.0
5.00
1.95
2.80
3.55
8.40
21.70
E C 2 
R
0
25
50
75
100
EC1   E10 x  10 
   
add and subtract means, EC   EC  .5x   EC   E.5x   EC   .5Ex 
c. Using the formula E ax  aEx  , E .5x 2  .5E x 2 . Then, using the fact that you can
2
2
 24  .54.60   21.70 .
d. The formula for revenue is
1
2
2
1
1
R  25x , so using the formula Eax  aEx  ,
ER   E 25 x   25 E x   251.4  35 .
e. Since we can add or subtract means, and   R  C , E    E R   E C1   35  24  11 .
f. E    E R   E C2   35  21 .70  13.30 .
Problem J4 (Formerly J2): If y  200x  50, x  3 and  x2  10 , what are 
y
and  y2 ?
Solution: The formula says that if y  ax  b, y  a x  b and  y2  a 2 x2 , so, if
a  200 and b  50 , then  y  200 x  50  2003  50  650 , and
 y2  2002  x2  2002 10  400000 . Thus  y  400000  632.46 .
Rpx 
0
2.5
5.0
7.5
20.0
35.0
E R 
251solnJ1 3/4/08
(Open this document in 'Page Layout' view!)
Problem J5 (Formerly J3):.I have measured the temperature at which successful outcomes of a
biochemical process occur. It turns out that the sample mean is 72 degrees Fahrenheit and the sample
standard deviation is 10 degrees Fahrenheit. My European affiliate needs this data. What are the mean and
standard deviation in Celsius? Remember that if x f is the temperature in Fahrenheit and xc is the
temperature in Celsius, x f  9 5 xc  32
 
Solution: x f  72 ,Var x f  10 2  100 . If we solve the equation above for Celsius temperature, we get
xc 
5
9
xf
 160
9
.
Using the formula y  ax  b , and remembering E  y   aEx   b and
Var y   a 2Varx  , let y  xc , x  x f , a  5 9 , and b   1609 .
Also, remember that we can use these formulas for sample means and variances, so that
2500
2
2
.
xc  ax f  b  5 9 72  1609  22.22 and Varxc   5 9  Var x f  5 9  10 2 
81
 
5 9 2 10 2  5 9 10   5.556 .
So that the standard deviation of x c is
Problem J6 (Formerly J4):.A package of a product sells for $21.00/lb. plus a $1.00 packing cost, so that
A 2 lb. package would sell for $21.00(2)+$1.00 = $43.00. If the mean package size is 37 oz. and the
standard deviation of package size is 3 oz., what are the mean and standard deviation of package price?
Solution: If we choose to work in pounds, let x be the weight in pounds,  x  3716 and  x  316 . The
cost is y  21x  1 . Using the formula y  ax  b , and letting a  21 and b  1 , we can use our formulas
 37 
for the means and variances of y .  y  E  y   aEx   b  a x  b  21   1  49 .56 and
 16 
2
 3
  15 .5039 . So that  y  15.5039  3.9375 .
 16 
 y2  Var  y   a 2Var x   a 2 x2  212 
We could also take the square root of the variance equation and say  y  a 2 x2  a  x
 3
 21   15 .5039 .
 16 
Continued on next page.
251solnJ1 3/4/08
(Open this document in 'Page Layout' view!)
Problem J7 (Formerly J5): You are investing in a project with the following distribution of profits (in
millions of dollars):
x
P(x)
a. What is the mean and variance of the profits?
10
.500
b. Assume that you must pay me 50% of the profits plus a
20
.250
finder's fee of $1(million). What is the mean and variance of
30
.125
your profit?
40
.125
c. Use your answer in part b. to find the mean and variance
of your profit if I raise my finder's fee to $2(million)?
Solution: a)
x
10
20
30
40
Px  xPx  x 2 Px 
.500
5.00
50 .0
.250
5.00 100 .0
.125
3.75 112 .5
.125
5.00 200 .0
1.000 18 .75 462 .5
From the table at left
 xPx  18.75
Var x     E x      x Px   
x 
2
x
2
2
x
2
2
x
 462 .5  18 .75 2  110 .9375
 x  110 .9375  10 .53
b) Since x represents the profits, the formula for my share is 0.5x  1 . So your share is
y  x  0.5x  1  0.5x  1 . Using the formula y  ax  b , and letting a  0.5 and b  1 , we can use
our formulas for the means and variances of y .  y  E y   aEx  b  a x  b  0.518.75  1  8.375
and  y2  Var y   a 2Varx  a 2 x2  0.52 110.9375  27.7344 . (or  y  4.892 )
c) Just as if y  ax  b ,  y  E y   aEx  b  a x  b and  y2  Var  y   a 2Var x   a 2 x2 , so if
w  cy  d , w  Ew  cE y   d  c x  d and  w2  Var w  c 2Var  y   c 2 y2 . Our new return is
w  1y  1 , so that c  1 and d  1 . Therefore w  Ew  c x  d  18.375   1  7.375 and
 w2  Varw  c2 y2  12 27.7344  27.7344.
Problem J8 (Formerly J6): Assume that x is a standardized random variable. If y  5x  3 , find the
mean and variance of y .
Solution: By definition a standardized random variable has a mean of zero and a standard deviation of 1.
 x  0 and  x  1 . y  5x  3 . Using the formula y  ax  b , and letting a  5 and b  3 , we can use
our formulas for the means and variances of y .  y  E y   aEx  b  a x  b  50  3  3 and
 y2  Var y   a 2Varx  a 2 x2  52 12  25 .
Continued on next page.
251solnJ1 3/4/08
(Open this document in 'Page Layout' view!)
Problem J9 (Formerly J7): Consider the random variable x in the table below. Make a probability table
for a standardized random variable z , created by
subtracting the mean (which is 4) and dividing
x
P x 
by the standard deviation (which is 2). Show
0
1/16
that the mean and standard deviation of z are 0
2
1/4
and 1.
4
3/8
6
8
Solution: We standardize a variable by using the formula z 
z
2
x


24

 1 . First we check the mean and variance of x .
2
1/4
1/16
x4
, so that x  2 is replaced by
2

x P x  xPx  x 2 P x 
From the table at left
0 1
0.00
0.00
16
x 
xPx   4.00
1
2
0.50
1.00
4
Var x    x2  E x 2   x2 
x 2 Px    x2
3
4
1.50
6.00
8
 20 .00  4.00 2  4.00
1
6
1.50
9.00
4
 x  4.00  2.00
8 1
0.50
4.00
16
1.00 4.00 20 .00
Now we write out the variable z with its probabilities and compute its mean and standard deviation.
z Pz 
zP z  z 2 Pz 
From the table at left
2 1
 0.125
0.25
16
 z  zP z   0.00
1 1
 0.250
0.25
4
Var z    z2  E z 2   z2 
z 2 Pz    z2
3
0
0.000
0.00
8
 1.00  0.00 2  1.00
1 1
0.250
0.25
4
 x  1.00  1.00
2 1
0.125
0.25
16

 

 
1.00
0.000
1.00
