Business Statistic Assignment 2
Essay by Hupan Hupan • January 16, 2016 • Course Note • 1,799 Words (8 Pages) • 1,085 Views
BUSINESS STATISTIC ASSIGNMENT 2
PAN HU ZID: Z1785373
CHAPTER 4
4.45
A is not a valid number for a probability. B is a valid number C is a valid number
D is a valid number E is not a valid number F is not a valid number
G is not a valid number H is a valid number I is a valid number
4.46:P=13/52=0.25
4.51
A 13/90=14.4% B 33/90=36.7% C 37/90=41% D 18/90=20%
E This is an example of empirical probability.
4.55
A 0.43*0.64=27.5% B 40%+27.5%=67.5%
4.56
A 239/968=24.7% B 707/968=73%
C 130/261=49.8% D 68/307=22%
4.65
4*4*3*2*4=384 384 choices
Chapter 5
5.4: A: 0.1+0.1+0.2+0.4+0.3+0.1=1.2 invalid
B: 0.0+0.1+0.2+0.3+0.3+0.1=1.0 valid
C: 0.1+0.1+0.2+0.3+0.2+0.1=1.0 valid
D: 0.1+0.1+0.2+0.4+1.3+0.1=2.2 invalid
E: 0.1+0.1+0.2+0.2+0.2+0.1=0.9 valid
5.6
(1) Station City
rating | station city | city ranking | probability | rating sq | prob*rating sq |
1 | 4 | 4 | 0.05 | 1 | 0.05 |
2 | 12 | 24 | 0.15 | 4 | 0.6 |
3 | 8 | 24 | 0.1 | 9 | 0.9 |
4 | 36 | 144 | 0.45 | 16 | 7.2 |
5 | 20 | 100 | 0.25 | 25 | 6.25 |
Total | 80 | 296 | 15 |
Mean=3.7
Variance=1.31
Std=1.144552314
(2) Newark
rating | Newark | City*Rating | Probability | Rating SQ | Prob*Rating Sq |
1 | 12 | 12 | 0.16 | 1 | 0.16 |
2 | 12 | 24 | 0.16 | 4 | 0.64 |
3 | 18 | 54 | 0.24 | 9 | 2.16 |
4 | 15 | 60 | 0.2 | 16 | 3.2 |
5 | 18 | 90 | 0.24 | 25 | 6 |
Total | 75 | 240 | 12.16 |
Mean=3.2
Variance=1.92
Std=1.385640646
Station has higher average rate and contingency.
5.15 a. =BINOM.DIST(3,10,0.28,FALSE)=0.26
b. =BINOM.DIST(3,10,0.28,TRUE)=0.70
c. =1-BINOM.DIST(4,10,0.28,TRUE)=0.12
d. mean=0.28*10=2.8 std=10*0.28*0.72=2.016
Number of Success P(x) | |
1 | 0.145596354 |
2 | 0.254793619 |
3 | 0.26423042 |
4 | 0.17982348 |
5 | 0.083917624 |
6 | 0.027195526 |
7 | 0.00604345 |
8 | 0.000881337 |
9 | 7.61649E-05 |
10 | 2.96197E-06 |
[pic 1]
5.20
5.20 a. =BINOM.DIST(0,15,0.06,FALSE)=0.40
b. =BINOM.DIST(2,15,0.06,TRUE)=0.94
c. =1-BINOM.DIST(3,15,0.06,TRUE)=0.23
d.
[pic 2]
e. mean=0.06*15=0.9
5.22 a. =POISSON.DIST(3,3,FALSE)=0.2240
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