MGMT2262 midterm solutions
VERSION A
VERSION
B(click here)
Question 1
a) With raw values:
|
|
bald |
not bald |
total |
|
higher |
880 |
410 |
1290 |
|
not higher |
220 |
640 |
860 |
|
total |
1100 |
1050 |
2150 |
with percentages:
|
|
bald |
not bald |
total |
|
higher |
0.4093 |
0.1907 |
0.6 |
|
not higher |
0.1023 |
0.2977 |
0.4 |
|
total |
0.5116 |
0.4884 |
1.0 |
b) 7,917,940 x 80% = 6,334,352
c) P(higher | not balding) = 410/1050 = 0.3905
d) P(not bald and not higher) = 640/2150 = 0.2977
e) P(not bald | not higher) = 640/860 = 0.7442
f) P(bald or not higher) = (1100 + 860 220)/2150 = 1740/2150 = 0.8093
Question 2
a) (income < 50K) = (240 + 554)/1000 = 794/1000 = 0.794
b) P(female and income ³ 50K) = (342 + 114)/1000 = 456/1000 = 0.456
c) P(male | income between 20K and 50K) = 212/554 = 0.3827
d) P(income ³ 50K | male) = 92/400 = 0.23
P(income ³ 50K | female) = 114/600 = 0.19
Males are more likely
e) P(I < 25K and male) = 96/1000 = 0.096
P(I < 25K)P(male) = (240/1000)(400/1000) = 0.096
Therefore, income under $25,000 does not depend on gender.
P(I ³ 50K and male) = 92/1000 = 0.092
P(I ³ 50K)P(male) = (206/1000)(400/1000) = 0.0824
Therefore, income of $50,000 or more depends on gender.
f) None since each gender is represented in each income category.
Question 3
a) P(5) = 0.103
b) P(X £ 5) = 0.939
c) P(X > 5) = 1 0.939 = 0.061
d) P(3) + P(4) + P(5) + P(6) = 0.25 + 0.188 + 0.103 + 0.043 = 0.584
Or P(X £ 6) P(X £ 2) = 0.982 0.398 = 0.584
e) P(X £ 8) = 0.99
f) m = 10(0.2) = 2; s = sqrt(2*0.8) = 1.2649
Question 4
a) 38C5 = 501,942
b) 33P3 = 32,736
c) 6 x 3 x 2 = 36
Question 5
a) P(X < 5) = P(Z < -1) = 0.5 0.3413 = 0.1587
b) P(X < 10) = P(Z < 1.5) = 0.5 + 0.4332 = 0.9332
c) P(X < 6) + P(X > 9) = P(Z < -0.5) + P(Z > 1) = (0.5 0.1915) + 0.1587 = 0.4672
d) P(4 < X < 11) = P(-1.5 < X < 2) = 0.4332 + 0.4772 = 0.9104
e) 1.282 = (X 7)/2; X = 7 1.282(2) = 4.436 seconds
f) P(Xbar > 8) = P(Z > 2.96) = 0.5 0.4985 = 0.0015
g) Lower limit = 7 2(2) = 3
Upper limit = 7 + 2(2) = 11
h) 1 1/(2.52) = 1 0.16 = 0.84
Question 6
a) mean = 5; P(X > 3) = 1 P(X £ 3) = 1 0.265 = 0.735
b) mean = 5; P(3) + P(4) + P(5) + P(6) = 0.6376
c) mean = 2.5; P(X < 3) = 0.5438
d) mean = 7.5; P(0) = 0.0006
e) standard deviation = sqrt(5) = 2.2361
Question 7
a) Here is the comparison:
|
|
Cable |
DSL |
|
mean |
13.375 |
11.0167 |
|
median |
13.2 |
10.6 |
|
stdev |
1.1802 |
0.9778 |
|
variance |
1.393 |
0.9561 |
|
range |
3.6 |
2.7 |
|
cv |
8.82% |
8.88% |
b) The DSLs cv is higher than that of cable. Thus, it has the highest relative variability.
c) Using DSL:
Q1 = (10.2 + 10.3)/2 = 10.25
Q3 = (11.7 + 12.1)/2 = 11.9
IQR = 11.9 10.25 = 1.65
d) 50% of Canadians using DSL spend between 10.25 and 11.9 hours per week online.
e) Lower limit = 11.0167 2(0.9778) = 9.0611
Upper limit = 11.0167 + 2(0.9778) = 12.9723
100% of the values lie between these limits which exceeds the 75% threshold under Chebyshevs theorem.
VERSION B
Question 1
a) With raw values:
|
|
bald |
not bald |
total |
|
higher |
99 |
73 |
172 |
|
not higher |
11 |
32 |
43 |
|
total |
110 |
105 |
215 |
with percentages:
|
|
bald |
not bald |
total |
|
higher |
0.4604 |
0.3396 |
0.8 |
|
not higher |
0.0512 |
0.1488 |
0.2 |
|
total |
0.5116 |
0.4884 |
1.0 |
b) 7,917,940 x 90% = 7,126,146
c) P(higher | not balding) = 73/105 = 0.6952
d) P(not bald and not higher) = 32/215 = 0.1488
e) P(not bald | not higher) = 32/43 = 0.7442
f) P(bald or not higher) = (110 + 42 - 11)/215 = 142/215 = 0.6605
Question 2
a) (income < 50K) = (220 + 611)/1000 = 831/1000 = 0.831
b) P(female and income ³ 50K) = (434 + 112)/1000 = 546/1000 = 0.546
c) P(male | income between 20K and 50K) = 177/611 = 0.2897
d) P(income ³ 50K | male) = 57/300 = 0.19
P(income ³ 50K | female) = 112/700 = 0.16
Males are more likely
e) P(I < 25K and male) = 66/1000 = 0.066
P(I < 25K)P(male) = (220/1000)(300/1000) = 0.066
Therefore, income under $25,000 does not depend on gender.
P(I ³ 50K and male) = 57/1000 = 0.092
P(I ³ 50K)P(male) = (169/1000)(300/1000) = 0.0507
Therefore, income of $50,000 or more depends on gender.
f) None since each gender is represented in each income category.
Question 3
a) P(5) = 0.162
b) P(X £ 5) = 0.329
c) P(X > 5) = 1 0.329 = 0.671
d) P(3) + P(4) + P(5) + P(6) = 0.047 + 0.101 + 0.162 + 0.198 = 0.508
Or P(X £ 6) P(X £ 2) = 0.527 0.018 = 0.509
The discrepancy is due to rounding errors adding the values from Table A.1
e) P(X £ 8) = 0.596
f) m = 10(0.4) = 4; s = sqrt(4*0.6) = 1.5492
Question 4
a) 37C6 = 2,324,784
b) 31P4 = 755,160
c) 5 x 4 x 3 = 60
Question 5
a) P(X < 6) = P(Z < -0.8) = 0.5 0.2881 = 0.2119
b) P(X < 11) = P(Z < 1.2) = 0.5 + 0.3849 = 0.8849
c) P(X < 7) + P(X > 10) = P(Z < -0.4) + P(Z > 0.8) = (0.5 0.1554) + 0.2119 = 0.5565
d) P(5 < X < 12) = P(-1.2 < X < 1.6) = 0.3849 + 0.4452 = 0.8301
e) 1.282 = (X 8)/2.5; X = 8 1.282(2.5) = 4.795 seconds
f) P(Xbar > 9) = P(Z > 2.53) = 0.5 0.4943 = 0.0057
g) Lower limit = 8 2(2.5) = 3
Upper limit = 8 + 2(2.5) = 13
h) 1 1/(2.52) = 1 0.16 = 0.84
Question 6
a) mean = 6; P(X > 3) = 1 P(X £ 3) = 1 0.1512 = 0.8488
b) mean = 6; P(3) + P(4) + P(5) + P(6) = 0.5443
c) mean = 3; P(X < 3) = 0.4232
d) mean = 9; P(0) = 0.0001
e) standard deviation = sqrt(6) = 2.4495
Question 7
a) Here is the comparison:
|
|
Cable |
DSL |
|
mean |
11.0167 |
13.375 |
|
median |
10.6 |
13.2 |
|
stdev |
0.9778 |
1.1802 |
|
variance |
0.9561 |
1.393 |
|
range |
2.7 |
3.6 |
|
cv |
8.88% |
8.82% |
b) The cables cv is higher than that of DSL Thus, it has the highest relative variability.
c) Using cable:
Q1 = (10.2 + 10.3)/2 = 10.25
Q3 = (11.7 + 12.1)/2 = 11.9
IQR = 11.9 10.25 = 1.65
d) 50% of Canadians using cable spend between 10.25 and 11.9 hours per week online.
e) Lower limit = 11.0167 2(0.9778) = 9.0611
Upper limit = 11.0167 + 2(0.9778) = 12.9723
100% of the values lie between these limits which exceeds the 75% threshold under Chebyshevs theorem.