MGT 201 - Quantitative Analysis
Fall 2015
Homework 6
Due Thursday November 12, 2015
1. In a study of 500 executives, 315 stated that their company informally monitored social
networking sites to stay on top of information related to their company.
(a) Con

MGT 201 - Quantitative Analysis
Fall 2015
Homework 5
Due Friday November 6, 2015
1. The owner of a sh market determined that the mean weight for salmon is 12.3 pounds,
with a standard deviation of 2 pounds. Assuming that the weights of salmon are normally

CHAPTER
2
CASES
Managing Ashland MultiComm
Services
Type and Cost of Wrong Billing Errors
Type of Wrong Billing Errors
Cost ($thousands)
Recently, Ashland MultiComm Services has been criticized for its
inadequate customer service in responding to question

CHAPTER
CASES
Managing Ashland MultiComm
Services
For what variable in the Chapter 2 Managing Ashland MultiComm Services case (see page 104) are numerical descriptive
measures needed?
1. For the variable you identify, compute the appropriate
numerical des

CardioGood Fitness
The Choice Is Yours Follow-Up
The market research team at AdRight is assigned the task to identify the profile of the typical customer for each treadmill product
offered by CardioGood Fitness. The market research team decides
to investi

MGT 201 QUANTITATIVE ANALYSIS
Fall Quarter 2015
Reading 51
STATISTICAL NOISE IN ELECTION POLLS
Some headlines on Tuesday suggested that Barack Obama has a problem among his African-American
base. The warrant for

MGT 201 - Quantitative Analysis
Winter 2015
Homework 2
Due Saturday, October 10, 2015
1. Textbook problem 103, page 120. Assume that the data provided is the entire population
under consideration, and use the formulas or Excel functions presented in class

MGT 201 - Quantitative Analysis
Fall 2015
Homework 7
Due Friday November 20, 2015
1. The Federal Communications Commissionreported that in 2014, the mean wait for repairs
for Verizon customers was 36.5 hours. In an eort to improve this service, suppose th

861108877
Junjie Duan
1
a Phat=315/500=0.63
P(z<1.96)=97.5% c=1.96
L=c*(p(1-p)/n)=1.96*(0.63*0.37/500)=0.04
(0.63-L,0.63+L)=(0.59, 0.67)
b The interval (0.59, 0.67)is called a 95% confidence interval for the proportion of companies
that informally monitor

861108877
Junjie Duan
1
a P(12<x<15)=P(Z<(15-12.3)/2)-P(Z<(12-12.3)/2)=.47111
b P(x<10)=P(z<(10-12.3)/2)=.125072
c z=norm.s.inv(2.5%)=-1.96 x=-1.96*2+12.3=8.38
z=norm.s.inv(97.5%)=1.96 x=1.96*2+12.3=16.22
2
a P(x>170)=1-P(x<170)=1-P(z<(170-150)/35)=0.2843

Junjie Duan
861108877
1a Covariance(C,T)=540
b Correlation=0.95
c Yes, because the correlation is 0.95. The higher temperature is related to higher crime rate
d No, Correlation is not causual.
2b
3b
4b
5a
6a
7f
8f
9a P(S)=128/328=0.39
b P(B and do not D)=

861108877
Junjie Duan
1.
A P(3<x<5)=1/6
B no
C yes
D P(C)=(1/6)(1/6)=1/36
E P(C|D)=1/6
F no
G P(E|D)=P(E and D)/P(D)=(1/6)(1/6)/(1/6)=
2.
P(yes)=.65 P(no)=.30 P(undecided)=.05
P(last|yes)=.50 P(last|no)=.60 P(last|undecided)=.15
P(last and yes)=.65*.5=.32