Probability, Decision Analysis and Business Statistics
BNAL 206

Fall 2016
1.
=100 =10 n=25 x =
a.
P ( x <95 )=P ( Z=2.5 ) =0.0062
i.
b.
2.
95100
=2.5
2
Z=
97.5100
=1.25
2
P ( x >102.2 )=P ( Z>1.1 ) =10.8643=0.1357
i.
d.
Z=
P ( 95< x < 97.5 )=P (2.5<Z <1.25 )=0.10560.0062=0.0994
i.
c.
10
=2
25
Z=
102.2100
=1.1
2
P ( x >99.22 )
Probability, Decision Analysis and Business Statistics
BNAL 206

Fall 2016
5. Additional cell phone
a. X = 135, N = 500
i. Point estimate  0.27
ii. Max Error 
0.080.22=0.058
iii. Confidence 99%
Sample X N Sample p
1
99% CI
135 500 0.270000 (0.218858, 0.321142)
b. The manger could have 99% confidence that the 22% to 32% of hous
Probability, Decision Analysis and Business Statistics
BNAL 206

Fall 2016
Summarizing and Describing Numerical Data
Problem #1
a) Chicken
a. Mean
7 +9+15+16+16 +18+22+25+27+ 33+ 39
20.6364
11
b. Median = 18
c. Mode = 16
d. Q1
Burger
a. Mean
19+ 31+34 +35+39+39+ 43
34.2857
7
b. Median = 35
c. Mode = 39
c) Both are skewed. The
Probability, Decision Analysis and Business Statistics
BNAL 206

Fall 2016
1. Columbia Construction Company
a. Houses  $70,000, Shopping Center  $105,000, Leasing  $40,000. Therefore, the
shopping center would be the Maximax decision.
b. Houses  $30,000, Shopping Center  $20,000, Leasing  $40,000. Therefore, leasing
would
Probability, Decision Analysis and Business Statistics
BNAL 206

Fall 2016
1.
2.
3.
=0 =1
a.
P ( Z >+1.08 )=10.8599=0.1401
b.
P ( Z <0.21 )=0.4168
c.
P (1.96, Z ,0.21 ) =0.41680.0250=0.3918
d.
P (1.96< Z <+1.08 )=0.85990.0250=0.8349
e.
P (+1.08< Z <+1.96 )=0.97500.8599=0.1151
=240 =40
a.
P ( X <180 )=P ( Z<1.5 )=0.0668
b.
P (180
Probability, Decision Analysis and Business Statistics
BNAL 206

Fall 2016
5. Miramar Company
P(positive/favorable) = .60
P(positive/stable) = .30
P(positive/unfavorable) = .10
State of conditions
Favorable
Stable
Unfavorable
a. EMV
i.
ii.
iii.
P(negative/favorable) = .40
P(negative/stable) = .70
P(negative/unfavorable) = .90
Po
Probability, Decision Analysis and Business Statistics
BNAL 206

Fall 2016
6. At the 0.01 level of significance, is there evidence that the proportion of dataprocessing clerks
who have gone through the new training and are no longer employed is less than 0.25?
a.
=.25 n=150 X=29 Ps =
29
=.29
100
1)Null & Alternative Hypothesis
Probability, Decision Analysis and Business Statistics
BNAL 206

Fall 2016
1.
100
n=64 X =350 =100 X =
=12.5
64
a.
3501.96 ( 12.5 ) 350+1.96 ( 12.5 )=35024.5 350+24.5=325.5 374.5
i. We can be 95% confident that the mean life is between 325.5 and 374.5 hours.
ii. Point Estimate  350
iii. Max error is  24.5
iv. Confidence 95%
b
Probability, Decision Analysis and Business Statistics
BNAL 206

Fall 2016
3.
n=100, X =315.40, S=43.20
a. Using the 0.10 level of significance, is there evidence that the population mean is above
$300?
i.
H 0 : 300 , H 1 : >300
ii.
=0. 10 , n=100 , df =99, critical value=1.2902
iii.
n=100, X =315.40, S=43.20
iv. Test Statistic
Probability, Decision Analysis and Business Statistics
BNAL 206

Fall 2016
3. Miramar Company
b. EMV
i. Widgets  .2(120,000) + .7(70,000) + .1(30,000) = $70,000
ii. Hummer  .2(60,000) + .7(40,000) + .1(20,000) = $42,000
iii. Nimnot  .2(35,000) + .7(30,000) + .1(30,000) = $31,000
iv. Therefore, Widgets would be the expected m
Probability, Decision Analysis and Business Statistics
BNAL 206

Fall 2016
2. Nashville p=.60 n=20
a.
P( X <5)=0.0003+0.000+0+ 0+0=0.0003
b.
P( X 10)=1 (0.0710+ 0.0355+0.0146+0.0049+0.0013+ 0.0003)=.8724
c.
P( x=20)=0.0000
3. Ticket fees p=.40 n=10
a.
=( 10 )( .40 )=4
b.
probabilities
= (10 ) (.40)(1.40)=1.549
a)
P( X =0)=0.006
Probability, Decision Analysis and Business Statistics
BNAL 206

Fall 2016
Bayes Theorem
7. TV viewing habits
a) P (Husband is watching/Wife is watching) = 67%
b) P (Wife is watching) = .36
Event
Prior Probability
Joint Probability
.6
Conditional
Probability
.4
.24
Posterior
Probability
.24/.36 = .67
Husband watches
TV
Husband d
Probability, Decision Analysis and Business Statistics
BNAL 206

Fall 2016
Multiplication Rule
5. Deck of cards
a) P (Q1 and Q2) = 4/52 * 3/51 = 12/2652 =0.0045
b) P (10(1) and 5 or 6 (2) = 4/52 * 8/51 = 32/2652 = 0.0121
c) P (Q1 and Q2) = 4/52 * 4/52 = 16/2704 = 0.0059
6. Gloves
a) P (R1 and R2) = 7/9 * 6/8 = 42/72 = 0.5833
b)
Probability, Decision Analysis and Business Statistics
BNAL 206

Fall 2016
Conditional probability and independence
2. Restaurant servings
e) P (No Dessert/Female) = 240/280 = 0.8571
f) No gender and ordering dessert are not independent
i. P (Female/Ordered Desert) = 40/136 = 0.2941
ii. P (Male/Ordered Dessert) = 96/136 = 0.7059
Probability, Decision Analysis and Business Statistics
BNAL 206

Fall 2016
5.
i.
Z=
.50.50
=0
0 .035 4
ii.
Z=
.60.50
=2.83
0 .035 4
P ( 0.4419< X <0.5581 )=P (1.65<Z <1.65 )=.90
b.
i.
ps=.501.650.035 4=0.4419
ii.
ps=.50+ 1.650 .035 4=0.5581
P ( X >.65 )=P ( Z > 4.24 ) =1.9999=.0001
c.
i.
ps=
Z=
.65.50
=.9999
0.035 4
.375 (1. 375
Probability, Decision Analysis and Business Statistics
BNAL 206

Fall 2016
1.
=70, =3.5,n=49, X =69.1, =.05
a. Is there evidence that the machine is not meeting the manufacturers specifications for
mean breaking strength?
1) H 0 : =70, H 1 : 70
2) 2 =0.025, critical value= 1.96
3) X =69.1,n=49, =3.5
4)Test Statistic:
5)Decision:
QUESTION 1
RsQ_001 The process of estimating the population average family
expenditure on food based on the sample average expenditure of
1,000 families is an example of
a statistic.
a
.
descriptive
b statistics.
.
a parameter.
c
.
inferential
d statistic
Durham, Jonathan
BNAL 610
1.
Mod 2 HW
Durham, Jonathan
BNAL 610
Mod 2 HW
2. QM for 1, 14, 22
1.
The current optimum solution (X1 = 4, X2= 0) holds as long as the unit profit of product X1
remains at or between $9 and infinity. The solution will change if
Durham, Jonathan
Chapter 1
BNAL 610
Module 7
1.
Using business analytics would help them apply the data, certain times and days of the week
have longer lines than others, to make decisions in regards to staffing the registers. Using
prescriptive analytics
Durham, Jonathan
Module 9
2.
BNAL 610
11/29/2016
A data warehouse is a pool of data produced to help in decision making processes. More
specifically it is subject oriented, integrated, timevariant, nonvolatile collection of data that
supports the decisio
Durham, Jonathan
Chapter 2
BNAL 610
Module 8
Chapter Application Case
1.
Their market share stagnated at 56% and total number of customers leveled of
Competitors emerged
Cost of government compliance increased
Revenue per customer lagged
2.
Vodafone forme
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
B
C
D
E
F
Assembling and testing computers
Cost per labor hour assembling
Cost per labor hour testing
$11
$15
Inputs for assembling and testing a computer
Basic
5
1
$150
$300
$80
Labor ho
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
B
C
Cost per labor hour assembling
Cost per labor hour testing, line 1
Cost per labor hour testing, line 2
E
F
G
H
I
J
$11
$19
$17
Inputs for assembling and testing a computer
Model 1
Labor hours for assembly
4
Lab
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
B
C
Cost per labor hour assembling
Cost per labor hour testing, line 1
Cost per labor hour testing, line 2
E
F
G
H
I
J
$11
$19
$17
Inputs for assembling and testing a computer
Model 1
Labor hours for assembly
4
Lab