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1.
The probability of you not having a beer tonight is 35% and the probability of me not having a beer is 60%. If I have a
beer, the probability is that I’ll drink a Bud is 45%, that I drink a Coors is 25% and that I’ll drink something else is
30%. On the other hand, if you drink a beer, the probability is that 50% that you will drink a Coors and 50% that you
will drink something else. Assuming statistical independence, what is the probability of you having a Coors tonight
and me having a Bud tonight ?
a. 16%
b. 21%
c. 26.5%
d. 50.5 %
e. None of the above
Let MB  I have a beer , YB  You have a beer , B  It' s a Bud, C  It' s Coors . The following are given:
 
 
P MB  .60 , P YB  .35 , PB MB  .45 , PC YB  .50 . So PMB   .40 PYB   .65 . What is requested is




PB  MB  C  YB  = PB  MB   PC  YB  = P B MB PMB  P C YB PYB =.45(.40)(.50)(.65) =.0585
2.
A survey of top executives revealed that 35% of them regularly read Time magazine and 40% read U.S. News and
World Report. 20% read them both. What is the probability that an executive who reads US News and World Report
will also read Time?
a. 20%
b. 35%
c. 50%
d. 60%
e. None of the above
PT  U  .20
Given: PT   .35 , PU   .40 , PT U   .20 . Requested: P T U 

 .50
PU 
.40
 
3.
The ages of a sample of typewriters used in a typing pool were organized into the following table:
Ages (in years)
2 up to 5
5 up to 8
8 up to 11
11 up to 14
14 up to 17
Number
2
5
10
4
2
Class
x
f
2 - 5
5 - 8
8 - 11
11-14
14-17
2
5
10
4
2
23
3.5
6.5
9.5
12.5
15.5
What is the sample variance?
a. About 10.2
b. About 6.1
c. About 14.0
d. About 3.2
e. None of the above
x
 fx  215 .5  9.3696
n
23
fx
s2 
 fx
2
 nx 2
n 1

7.0
32.5
95.0
50.0
31.0
215.5
2243 .75  23 9.3696 2
 10 .2
22
fx2
24.50
211.25
902.50
625.00
480.50
2243.75
4.
The main computer used by a company for inventory and payroll crashes, on average, once a week. What is the
probability that the time between the next two crashes will be less than two weeks?
a. 13.57%
b. 23.33%
c. 77.67%
d. 86.47%
e. None of the above
Exponential Distribution Problems
From the outline
F x   1  cx
, when the mean time to a success is
1
.
c
  
1
c
Note that this is only for x  0 . There is no probability below zero.
1
For this problem c   1

Px  2  F 2  1  e
12
 1  e 2  1  .1353  .8647
5.
A cola dispensing machine is set to dispense a mean of 2.02 liters into a container labeled 2 liters. Actual quantities
dispensed vary and the amounts are normally distributed with a standard deviation of 0.015 liters. What is the
probability a container will have less than 2 liters?
a. 0.0918
b. 0.3413
c. 0.1926
d. 0.8741
e. None of the above
Make a diagram: Show a Normal curve with a mean at 2.02 and shade the area below 2.
2.00  2.02 

Px  2.00   P  z 
 Pz  1.33  Pz  0  P 1.33  z  0  .5  .4082  ..0918
0.015 

6.
A sample of assistant professors on the business faculty at state supported institutions in Ohio revealed the mean
income to be $62,000 for 9 months with a standard deviation of $4,500. With no assumptions regarding whether the
sample data follows a Normal distribution, what is the proportion of faculty that earn more than $53,000 but less than
$71,000 for 9 months?
a. At least 50%
b. At least 25%
c. At least 75%
d. At least 94%
e. None of the above
1
1
Chebyshef's Inequality P  k  x    k   1  2 or P x    k  2
k
k
You're right. I don't have time to do the solutions. Fortunately, these are both easy applications of Chubby Checker's
inequality. In the formula above 53000 = 62000 - 9000 = 62000 - 2(4500) and 71000 = 62000 + 9000 = 62000 + 2(4500).
1
1
So k  2 . P  k  x    k   1  2 or P x    k  2 . Check your text to correct the inequalities. If you
k
k
fill in the blanks here, You should find out that at least 75% are in the box.




7.
A company is in the process of hiring new employees for its sales force. The Human Resource manager estimates that
the following are the probabilities of how many sales people will be hired during the next three months:
No. To Be Hired
Probability
0
5%
1
10%
2
20%
3
20%
4
25%
5
10%
6
10%
What is the expected range of new hires over the next three months with approximately 75% accuracy?
a. 0 to 6.4 persons
b. .4 to 4.2 persons
c. 1.2 to 4.4 persons
d. 1.6 to 5.4 persons
e. None of the above
Looks like the same stuff to me. k  2 . You need a mean and standard deviation to start. A worksheet from another
problem is below. Use it to get the mean and standard deviation.
x
Px 
x P x 
x 2 P x 
 x    P x   x   2 P x 
x   
1
.2
0.2
0.2
-2.8
-0.56
1.568
2
.4
0.8
1.6
-1.8
-0.72
1.296
4
.2
0.8
3.2
0.2
0.04
0.008
10
.2
2.0
20.0
6.2
1.24
7.688
1.0
3.8
25.0
0.00
10.560
xPx   3.8 ,
Px   1.00 (valid distribution!),   E x  
x 2 Px   25 .0 ,
From this table


 x   Px   0.00 (a check for accuracy!) and  x   
columns if you use the computational formula.
xPx   3.8
a)   E x  

2
Px  10.560 . Remember! You do not need the last two
 
b) Computational Formula (the easy way!): Var x    x2  E x 2   2 
 25 .0  3.82  10 .56

x
2
Px    2
  x    Px  10.560 .
Definitional Formula (the hard way): Varx    x2  E x   2 
c)   10.56  3.2496

2