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Lauren Frye
Peyton Riddle
Brittanie Corn
Excel Assignment #2
1.
a. We were asked to calculate and report the expected number of devices in use, the
standard deviation of the number of devices in use and the probability that a
randomly selected customer will use more than 6 devices.
b. First, we created a graph of the probability distribution by creating a Pivot Table
of the number of devices and the percentage of the grand total. We made a second
graph by creating a Pivot Table of the number of devices and the percentage of
the running total in. Last, we created a chart that computed the expected number
of devices in use, the standard deviation of the number of devices in use, and the
probability that a randomly selected customer will use more than six devices. The
expected number of 5.605 was calculated by multiplying each outcome by their
corresponding probability, then calculating the sum of each of these numbers. The
standard deviation of 2.39 was computed by taking the square root of the
variance, which is a sum of all weighted square deviations. The probability,
36.5%, that a randomly selected customer will use more than six devices was
calculated by finding the sum of the probabilities in relation to more than six
devices.
# of Devices
1
2
3
4
5
6
7
8
9
10
11
12
13
Grand Total
Count of Devices
2.50%
7.00%
14.00%
12.00%
12.00%
16.00%
12.00%
14.50%
6.00%
1.50%
1.50%
0.50%
0.50%
100.00%
Row Labels
1
2
3
4
5
6
7
8
9
10
11
12
13
Grand Total
Outcome
x
1
2
3
4
5
6
7
8
9
10
11
12
13
P(# of devices)
0.45%
2.94%
10.44%
19.00%
29.71%
46.83%
61.82%
82.52%
92.15%
94.83%
97.77%
98.84%
100.00%
Probability
P(X=x)
0.025
0.07
0.14
0.12
0.12
0.16
0.12
0.145
0.06
0.015
0.015
0.005
0.005
Outcome * Prob Deviation Squared Deviation Weighted Square Deviation
x * P(X=x)
x - E(X)
(x - E(X))^2
(x - E(X))^2 * P(X=x)
0.025
-4.605
21.206025
0.530150625
0.14
-3.605
12.996025
0.90972175
0.42
-2.605
6.786025
0.9500435
0.48
-1.605
2.576025
0.309123
0.6
-0.605
0.366025
0.043923
0.96
0.395
0.156025
0.024964
0.84
1.395
1.946025
0.233523
1.16
2.395
5.736025
0.831723625
0.54
3.395
11.526025
0.6915615
0.15
4.395
19.316025
0.289740375
0.165
5.395
29.106025
0.436590375
0.06
6.395
40.896025
0.204480125
0.065
7.395
54.686025
0.273430125
5.605
E(x)
5.728975
Var(x)
2.393527731
StDev(x)
36.50%
P(X>6)
c. The majority of customers use between 3-8 devices. The standard deviation for
the sample of 200 customers was 2.3935 units. This is relatively low; leading us
to believe that our data is useful. The probability that a random customer will use
more than 6 devices is 36.50%. That means that a majority of customers use 6 or
less devices in their residence.
2.
a. The goal was to utilize Excel functions and the information for Highway Miles
per Gallon (HwyMPG) to calculate mean, median, mode, range, variance,
standard deviation, interquartile range, and whether there would be any mild or
extreme outliers. Then, we were asked to calculate the statistics for City Miles
per Gallon (CityMPG).
b.
First, we made a table to calculate the values of mean, median, mode, range,
variance, standard deviation, interquartile range, and whether there would be any
mild or extreme outliers. To make these calculations, we used the excel
statements below. We made them functions by adding the “=” symbol before each
equation. Then, we used the Descriptive Analysis tool under the Data Analysis tab
to generate summary statistics for CityMPG. This automatically generated the
chart containing the information in regards to the mean, standard error, median,
mode, standard deviation, sample variance, kurtosis, skewness, range, minimum,
maximum, sum, and count based off of the information in cells Y2:Y76. For
HwyMPG, the mean was 29.25, the median was 29, the mode and the range were
30, the variance was 31.22, the standard deviation was 5.59, and the interquartile
range was 6.5. After analyzing the interquartile range value, there were 2 mild
outliers and 1 extreme outliers. For City MPG, the mean was 22.76, the median
was 22, the mode was 18, the range was 31, the variance was 34.48, and the
standard deviation was 5.87.
CityMPG
HwyMPG
Item
Mean
Median
Mode
Range
Variance
Standard Deviation
Interquartile Range
Outliers
Mild
Extreme
Value
Excel Statement
29.25333333
AVERAGE(Z2:Z76)
29
MEDIAN(Z2:Z76)
30
MODE(Z2:Z76)
30
MAX(Z2:Z76)-MIN(Z2:Z76)
31.21873874
VAR(Z2:Z76)
5.587373152
STDEV(Z2:Z76)
6.5
QUARTILE(Z2:Z76,3)-QUARTILE(Z2:Z76,1)
no mild outliers, some extreme outliers
1.656115151
ABS((MIN(Z2:Z76)-AD4)/AD9)
3.713134259
ABS((MAX(Z2:Z76)-AD4)/AD9)
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
22.76
0.678057639
22
18
5.872151408
34.48216216
3.731247432
1.671679587
31
15
46
1707
75
c. We concluded that, on average, you have more HwyMPG than CityMPG. You
can achieve more miles to the gallon when driving on the highway versus in the
city, thus making it cheaper for the driver.
3.
a. We were asked to make a decision on suppliers from Buenos Aires, Callao,
Montevideo, Valparaiso, and Santos based on differing future conditions by
determining the Maximax, Maximin, Minimax regret, Hurwicz, and Equal
Likelihood.
b. We had to calculate the Maximax by using =MAX of the corresponding profits
or loss of each individual supplier. We calculated the Maximin by using =MIN of
the corresponding profits or loss of each individual supplier. We created a chart of
the Maximum Regrets by calculating which suppliers had the highest profits
within each state of nature. We used this number to generate the amount of regret
for each supplier in each state of nature by subtracting their profit/loss from the
maximum of each state of nature. To complete that table, we found the Minimax
Regret by finding the maximum profit from each supplier in all conditions of the
regret table. To calculate Hurwicz, we found the maximum profit from each
supplier, multiplied that by the coefficient of optimization, 0.2, then added that
number to the minimum profit/loss from each supplier multiplied by 1 minus the
coefficient of optimization. An example of the equation previously stated is:
=MAX(C5:E5)*$K$3+MIN(C5:E5)*(1-$K$3). We found the Equal Likelihood
by finding the average of the corresponding profits/losses for each individual
supplier, =AVERAGE. Last, we calculated the best of each computation. We used
=MAX to find the maximum numbers for the Maximax, Maximin, Hurwicz, and
Equal Likelihood. We used =MIN to find the option with the least amount of
regret in the Minimax Regret section.
0.2
0.5
Alternative Maximax Maximin Minimax Regret Hurwicz Equal Likelihood
Buenos Aires
56
-11
12
2.4
21.333
Callao
60
-13
8
1.6
24.333
Montevideo
44
-17
16
-4.8
17.667
Valparaiso
39
-7
21
2.2
15.667
Santos
48
-5
12
5.6
24.667
BEST
60
-5
8
5.6
24.667
c. After looking at the excel results, we said that Callao would be the best alternative
to get the supplies from. We chose Callao because it was the supplier with the
least amount of regret, which we believe, would make it the smartest business
decision.
4.
a. The objective was to determine which of the four bids should be selected to
purchase the owner’s internet company, based off of the combination of
probabilities of a bonus and expected returns.
b. First, we constructed a Payoff table in Excel. It displays the 4 decision
alternatives, the bonus or no bonus it offered, and the expected value. We entered
all the data from the problem into the payoff table. The probability of a bonus
state of nature was 25%, and the probability of a no bonus state of nature was
75%. We calculated 75% by subtracting the 25% from 100%. Next, we used the
expected values from the payoff table and created a sensitivity analysis. We used
the “what-if analysis” data table to create this sensitivity analysis. We referenced
the bonus state of nature probability of 25% as the column part of the table in
order to generate the entire data. From this data, we created a chart that clearly
stated each individual bid’s probability and bonus in millions.
Probability
(Bonus)
Payoff Table:
0.25
0.75
States of Nature
Decision Alternatives Bonus No Bonus Expected Value
1
10
10
10
2
12
9
9.75
3
14
8
9.5
4
14
7
8.75
Best
10
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Bid 1
10
10
10
10
10
10
10
10
10
10
10
10
Decision Alt.
Bid 2
Bid 3
9.75
9.5
9
8
9.3
8.6
9.6
9.2
9.9
9.8
10.2
10.4
10.5
11
10.8
11.6
11.1
12.2
11.4
12.8
11.7
13.4
12
14
Bid 4
8.75
7
7.7
8.4
9.1
9.8
10.5
11.2
11.9
12.6
13.3
14
c. From the payoff table, the best decision alternative was Bid 1 with an expected
value of 10 million. We chose this because it was the highest expected value in
regards to the other bids. Looking at the sensitivity analysis graph, we believe that
the best decision alternative was Bid 3 because it had the highest return overall in
relation to the other bids.
5.
a. A restaurant chain has been working on a new line of seasonal deserts. We were
asked to help make a decision between three alternatives; scrap, launch, and
conduct a market test. If we were to advise a market test, we had to recommend
options based on if the results were positive or negative.
b. To complete this problem, we made a decision tree in Excel. We made our first
decision with three options; scrap, launch, or conduct market research. If the
company decided to scrap, the payoff was automatically zero. If the company
launched, we entered in the starting cost of $2 million. If they launched, the event
that could take place was high or low acceptance. We entered in the
corresponding profits that they expect from high and low acceptance. If they
chose to conduct market research, then the event that could take place would be
positive or negative results with a 50/50 chance. After the results came back, the
company would have to make a decision to launch or scrap. If the company
decided to launch, a new event would take place with 70% high profit or 30%
mild profit. Then, the decision tree calculated the appropriate expected returns
based on the probabilities of profit and loss.
Scrap
0
0
0
65%
High
3000000
Launch
-2000000 1600000
5000000 3000000
35%
Low
-1000000
1000000 -1000000
70%
Launch
2
1600000
2500000
Launch
5000000 2500000
-2000000 1300000
30%
Launch
50%
positive
-1500000
1
1000000 -1500000
0 1300000
Scrap
-500000
0 -500000
Conduct
30%
Launch
-500000 500000
2500000
Launch
5000000 2500000
-2000000 -300000
70%
Launch
50%
negative
-1500000
1
1000000 -1500000
0 -300000
Scrap
-500000
0 -500000
c. If the company was to decide to conduct a market test and the results were
positive, we believe that the best decision would be to launch the new product
system. This is because the chance of success with a return of $2.5 million is
70%. If the company was to decide to conduct a market test and the results were
negative, we believe the best option would be to scrap the project. This is
because, the loss incurred from scrapping the project is only $500,000. In
contrast, if the company launched the project, they hold a 70% chance to lose $1.5
million. Ultimately, we believe that it would be best for the company to launch
without conducting a market test because they would have the highest outcome
with this option.