Florida Institute of
Technology
Instructions: Read the questions very carefully because the requirements for each are specific. Each
group should turn in one copy of this WORD document (email to me) with the names of the participating
group members typed
Florida Institute of Technology
Instructions: Read the questions very carefully because the requirements for each are
specific. Each group should turn in one copy of this WORD document (email to me) with the
names of the participating group members typed
MGT 5006
1/12/2016
Chapter 1
1. The type of beverage sold is an example of a categorical variable because it does not
represent a quantity, but a designation.
b. The type of beverage sold is a nominal scale because it does not portray a ranking system
lik
Event
1
2
3
4
5
6
Probability
P(F)
Given that
0.5
0.3
0.2
0.01
0.03
0.06
Posterior Probabililty
Given P(D), the P it came
from F1 =
Conditional
Joint
Posterior Probability
Probability (Given Probability P(F
given that D) = P(D
That)
P(D (And)
P(D
and F) /
a. State scores changed minimally between the years of 2011 and 2010.
b. Between 2001 and 2011, overall scores decreased but there was a
significant increase in the amount of students in the 625 Bin.
c. The same results are interpeted from b.
d. Michigan
9f4bfa5dc8c13bea99dbb8e4e8466c7be583614e.xls
Problem 3.1, p.
a)
b)
c)
P(E1) =
P(E2) =
P(E4) =
P(E5) =
P(E3) =
P(E1) =
P(E2) =
P(E4) =
P(E5) =
P(E3) =
The probabilities must sum to 1.0.
P(E3) + (0.1+0.2+0.1+0.1) = 1.0
P(E3) = 1.0 - (0.1+0.2+0.1+0.1)
0.1
0.
min
max
Average
Median
5.2
13.5
8.492
8.05
To find minimum number: =min( "click & drag" )
To find maximum number: =max( "click & drag" )
Starting bin should be below your min and above your highest value.
Company RDPct
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
6.3 p 307
A random sample of n measurements was selected from a population with unknown mean and
know std.dev. Calculate a 95% confidence interval for mean for each of the following situations:
a.
b.
c.
d.
n
75
200
100
100
x bar
28
102
45
4.05
s^2
12
22
0
Binomial Probability Example
Probability of Success, p
Number of Trials, n
mean=np
std dev=sqrt(npq)
Number of
successes
0.2 <- Probability must be between 0 and 1
30 <- Enter value up to 100
6.00000
2.19089
Binomial
Prob.
0 0.00124
1 0.00928
2 0.03366
3
7.11 p 366
Student Default Rate. The national student loan default rate has flucuated over the last several years. A few
years ago, the Dept of Education reported the default rate (i.e. the proportion of college students that default
on their loans) at 0.
Probability of Success, p
0.2 <- Probability must be between 0 and 1
Number of Trials, n
30 <- Enter value up to 100
mean=np
6.00000
std dev=sqrt(npq)
2.19089023
Binomial
Prob.
Number of successes
0
1
2
3
4
5
6
0.00124
0.00928
0.03366
0.07853
0.13252
0.17
We are given that:
Bayes Theorem
Three factories(F1, F2 and F3) produce parts. F1 produces half the total parts, F2 produces 30%, and F3 produces the
remaining 20%. The set D represents defective parts. 1% of the parts produced by F1 are defective, 3% of
Chapter Three
Question 4
a)
The mean
7 +5+8+7 +9
X =
5
The median
5+ 1
Median =
2
=
10
5
=2
= 3rd ranked value
Arrange values from the lowest to the highest: -5 -8 7 7 9
The median is 7 because the data set is an odd number so the median is the
middle-ran
Chapter One
Question One
a) The type of beverage sold is a type is an example of categorical variable
because these values (soft drink, tea, coffee, bottled water) can only be
sorted/placed into categories. These values cannot me measured or counted.
For
Event
1
2
3
4
5
6
Probability
P(F)
Given that
0.5
0.3
0.2
0.01
0.03
0.06
Posterior Probabililty
Given P(D), the P it came
from F1 =
Conditional
Joint
Posterior Probability
Probability (Given Probability P(F
given that D) = P(D
That)
P(D (And)
P(D
and F) /