MGMT2262
Worksheet #2 Solutions
First,
we’ll make up a crosstab:
|
Bay |
Not Bay |
Total |
|
|
Walmart |
7 |
33 |
40 |
|
Not Walmart |
8 |
52 |
60 |
|
Total |
15 |
85 |
100 |
1)
P(Bay
or Walmart) = P(Bay) + P(Walmart)
- P(Bay and Walmart) = 15% + 40% - 7% = 48%
2)
P(Walmart | Bay) = P(Bay and Walmart)/P(Bay)
= 7/15 = 46.67%
3)
P(Walmart and not Bay) = 33% from the crosstab
4)
P(not
Bay | Walmart) = P(Walmart
and not Bay)/P(Walmart) = 33/40 = 82.5%
5)
P(not
Walmart | not Bay) = P(not Bay and not Walmart)/P(not Bay) = 52/85 = 61.18%
6)
P(neither Bay nor Walmart)
= 52% from crosstab. The other solution is to use DeMorgan’s
Law: P(neither Bay nor Walmart)
= 100% - P(Bay or Walmart) = 100% - 48% = 52%.
7)
P(Bay | Walmart) = 7/40
= 17.5%; P(Bay | not Walmart) = 8/60 = 13.33%. So,
someone who shops at Walmart is more likely to shop
at The Bay than someone who doesn’t shop at Walmart.
8)
We
have P(Bay and Walmart) =
7%. P(Bay)*P(Walmart) =
(0.15)(0.4) = 0.06 = 6%. Since these are not equal, the events are not
independent and so shopping at one store depends on shopping at the other. One
other solution is to note that P(Bay | Walmart) = 17.5% from question 7. However, P(Bay) = 15%. Since P(Bay) doesn’t
equal P(Bay | Walmart), we reach the same conclusion.
9)
Since
P(Bay and Walmart) = 7% is
not zero, the events of shopping at the 2 stores are not mutually exclusive.
10) P(revenue ³ 500K) = (438 + 365)/1000 = 803/1000 = 80.3%
11) P(profit ³ 100K | revenue ³ 500K) = (81 + 98 + 11 + 67)/(438 + 365) = 257/803 = 32%
12) P(profit < 100K) = (268 + 473)/1000 = 741/1000 = 74.1%
13) P(revenue < 500K | profit < 100K) = (141 + 54)/(268 + 473) = 195/741 = 26.32%
14) Revenue under $500,000 and net profit of $500,000 or more are mutually exclusive since the crosstab value is zero. These categories must be mutually exclusive since you can’t have a net profit of $500,000 or more if your revenue is less than that.
15) P(revenue ³ 500K or profit ³ 100K) = P(revenue ³ 500K) + P(profit ³ 100K) – P(revenue ³ 500K and profit ³ 100K) = (438 + 365)/1000 + (181 + 78)/1000 – (81 + 98 + 11 + 67)/1000 = 805/1000 = 80.5%
16) This table summarizes the results:
|
Profit|Revenue |
< $500K |
$500K - $1000K |
$1000K + |
|
$100K - $500K |
2 |
81 |
98 |
|
$500K + |
0 |
11 |
67 |
|
Total |
197 |
438 |
365 |
|
|
1.02% |
21.00% |
45.21% |
For example, for revenue under $500,000, the percentage of these companies with net profit of $100,000 or more is (2 + 0)/197 = 1.02%. As revenues increase, the percentage of companies with net profit of $100,000 or more increases.
17) P(revenue ³ 500K and profit ³ 100K) = (81 + 98 + 11 + 67)/1000 = 0.257
P(revenue ³ 500K)*P(profit ³ 100K) = [(438 + 365)/1000]* [(181 + 78)/1000] = (0.803)(0.259) = 0.208. Since the two sides of the equation are not equal, having profit of $100,000 or more depends on having revenue of $500,000 or more.
18)
We
first need to compute the overall market share for the under
25 age group. This Excel output summarizes the results:
|
Brand |
% market share |
% used by under 25 age group |
Weighted average |
|
Brand A |
0.25 |
0.7 |
0.175 |
|
Brand B |
0.15 |
0.55 |
0.0825 |
|
Brand C |
0.6 |
0.4 |
0.24 |
|
|
|
Total |
0.4975 |
We see the overall market share for the under 25 age group is 49.75%. We see that Brand C comprises
the largest percentage at 24%. Therefore, the percentage of the under 25 age
group that uses Brand C is 24/49.75 = 48.24%.
19)
12P4
= 11,880. The other method is 12x11x10x9 = 11,880
20) 12C3 = 220
21) 36P4 = 1,413,720. The other method is 36x35x34x33 = 1,413,720
22) 26x35x34x33 = 1,021,020. The other method is 26 x 35P3 = 26 x 39,270 = 1,021,020.
From the textbook
3.11 (page
99)
Part a
P(M) =
0.25, P(V) = 0.4, P(M and V) = 0.1
Part b
|
|
M |
Not M |
Total |
|
V |
0.1 |
0.3 |
0.4 |
|
Not V |
0.15 |
0.45 |
0.6 |
|
Total |
0.25 |
0.75 |
1.0 |
Part c
P(M or V)
= 0.25 + 0.4 – 0.1 = 0.55
P(not M
and not V) = 0.45
P(exactly
one) = 0.15 + 0.3 = 0.45
3.24 (page
106)
Part a
P(M) =
0.15, P(MBA) = 0.25, P(MBA | M) = 0.6
P(M and
MBA) = P(M)*P(MBA | M) = (0.15)(0.6) = 0.09
Setting up
the crosstab:
|
|
M |
Not M |
Total |
|
MBA |
0.09 |
0.16 |
0.25 |
|
Not MBA |
0.06 |
0.69 |
0.75 |
|
Total |
0.15 |
0.85 |
1.0 |
P(M |
MBA) = 0.09/0.25 = 0.36
Part b
Approach 1
LS = P(M and MBA) = 0.09
RS = P(M)*P(MBA) = (0.15)(0.25) = 0.0375
LS does not equal RS. Therefore, these are not independent
events
Approach 2
LS = P(MBA | M) = 0.6
RS = P(MBA) = 0.25
Again, same
result.
3.38 (page
111)
Part a
P(S) = 0.6 P(pass | S) = 0.85 P(S and pass) = (0.6)(0.85) = 0.51
P(not S)
= 0.4 P(pass | not S) = 0.1 P(not S and pass) = (0.4)(0.1) = 0.04
P(pass) =
0.51 + 0.04 = 0.55
P(S | pass
test) = 0.51/0.55 = 0.9273
Part b
Based on
the results from part a, the test appears to be, by and large, a valuable way
to screen applicants.
I also
looked at the problem from the perspective of seeing the percentage who succeed given that they fail the test:
P(S) = 0.6 P(fail | S) = 0.15 P(S and fail) = (0.6)(0.15) = 0.09
P(not S)
= 0.4 P(fail | not S) = 0.9 P(not S and fail) = (0.4)(0.9) = 0.36
P(fail) =
0.09 + 0.36 = 0.45
P(S | fail)
= 0.09/0.45 = 0.2
Of those
who fail the test, only 20% are successful managers. Again, based on this
perspective, the test appears to be a valuable way to screen applicants.