WGU C207 Data Driven Decision Making
Module 2 Exam (Answered)
Which of the following is NOT an application of statistics in business?
a Determine the internet advertiser that will reach the most people from extensive
page view data.
b Determine a target market based on information on household incomes
throughout the country.
c Predict trends in certain investments of big companies from previous results of
Fortune 100 companies.
d Forecast likelihood of board of director decisions from previous voting habits.
d Forecast likelihood of board of director decisions from previous voting habits.
Although this might be in a business context, this is the use of statistics in politics. The
other three examples are applications of statistics in business.
Bethany notices that her husband is wearing a blue sweater on Tuesday. She
cannot remember what he has worn previous Tuesdays. The next Tuesday she
notices he is wearing another blue sweater. She concludes that if it is Tuesday,
he will wear a blue sweater having data from her experiment to support this. What
is the flaw with this experiment?
a Small Sample Size
b Operationalization
c Missing Data
d Assumptions
a Small Sample Size
This is an experiment with too few data entries to form a statistically relevant
conclusion.
Doctor Andrews has been trying to measure the likelihood of heart attack risk.
Doctor Andrews decides to monitor hair length in people to determine those at
high risk of heart attack. What is the flaw in this experiment?
a Assumptions
b Association vs. Causation
c Response Bias
d Operationalization
d Operationalization
Monitoring hair growth does not measure the risk of heart attack. There is a flaw in the
experiment because the experiment is not measuring what the objective is trying to
determine.
Mr. Wonka notices that the last twenty times he invented a new chocolate candy,
his major competitors, Count Chocula, and the Easter Bunny, have big sales in
,late October. Mr. Wonka feels directly responsible for the profit of his
competitors. What is the flaw in this experiment?
a Small Sample Size
b Truly Representative Sample
c Association vs. Causation
d Blinding
c Association vs. Causation
Mr. Wonka's inventive ways are probably not the cause of increased chocolate candy
sales for his competitors at the end of October like Mr. Wonka has concluded.
There is a 90 percent chance that a package will arrive within three days of when
it was shipped. Also, there is a 75 percent chance that it will get wet. There is a 70
percent chance that it will get wet and will be delivered within three days. What is
the likelihood that at least one of these events occurs?
a 0.8
b 0.85
c 0.9
d 0.95
d 0.95
This is a union between P(on time) and P(wet). Therefore, P(on time∪wet)
=P(on time)+P(wet)−P(on time∩wet)
=0.90+0.75−0.70=0.95=95%
There is an 80 percent chance of snow. If it snows there is a 10 percent chance of
Todd walking to the store. If it doesn't snow there is a 60 percent chance of Todd
walking to the store. What is the likelihood that it will not snow and Todd will walk
to the store?
a 0.12
b 0.18
c 0.2
d 0.48
a 0.12
The probability of no snow is 20 percent ( P(no snow)=0.20 ), and the probability of
Todd walking to the store with no snow is 60 percent ( P(walk|no snow)=0.60 ).
Therefore to determine the likelihood of it not snowing and Todd walking to the store,
these probabilities are multiplied together. Therefore P(no snow∩walk)=P(no
snow)×P(walk|no snow)= 0.20×0.60=0.12=12%
Determine the Mode, Median, Mean (in that order) from the following data set. 4, 7,
11, 12, 14, 14, 15, 17, 17, 17, 18, 20, 23, 24, 26, 29, 35, 39, 40
a 17, 17, 20
b 14, 17.5, 17
, c 17, 17, 17
d 17, 17, 20.11
d 17, 17, 20.11
The mode is 17 as it shows up 3 times and no other data point shows up as many
times. The median is 17 as it is the 10th number out of 19 ordered numbers. The mean
is 20.11 as it is the result of the sum of the numbers, 382, divided by 19, the count of
numbers.
Standard deviation measures
a the average of the difference for the data set.
b the normal exceptions between the expected data points.
c the median of the data set distributed equally.
d the dispersion from the average for the data set.
d the dispersion from the average for the data set.
Elizabeth got a 75 on her performance review. The average was 80, but the
standard deviation was 3.5. Determine the z-score for her performance review.
a 1.43
b -1.43
c5
d 3.5
b -1.43
This is because she was 1.43 standard deviations below the average. Z-score is
determined by the distance a data point is from the mean divided by the standard
deviation. The data point was five below the mean ( therefore -5), and the standard
deviation was 3.5. Therefore, the z-score for Elizabeth's performance review was
−53.5=−1.43 .
Statistical analysis can be used to make __________ about entire populations
using samples.
a) conclusions
b) inferences
c) decisions
d) references
b) inferences
What percent of the data of a normal distribution is within two standard
deviations of the mean?
a) 68.3
b) 75
c) 95.4
d) 99.7
c) 95.4
Module 2 Exam (Answered)
Which of the following is NOT an application of statistics in business?
a Determine the internet advertiser that will reach the most people from extensive
page view data.
b Determine a target market based on information on household incomes
throughout the country.
c Predict trends in certain investments of big companies from previous results of
Fortune 100 companies.
d Forecast likelihood of board of director decisions from previous voting habits.
d Forecast likelihood of board of director decisions from previous voting habits.
Although this might be in a business context, this is the use of statistics in politics. The
other three examples are applications of statistics in business.
Bethany notices that her husband is wearing a blue sweater on Tuesday. She
cannot remember what he has worn previous Tuesdays. The next Tuesday she
notices he is wearing another blue sweater. She concludes that if it is Tuesday,
he will wear a blue sweater having data from her experiment to support this. What
is the flaw with this experiment?
a Small Sample Size
b Operationalization
c Missing Data
d Assumptions
a Small Sample Size
This is an experiment with too few data entries to form a statistically relevant
conclusion.
Doctor Andrews has been trying to measure the likelihood of heart attack risk.
Doctor Andrews decides to monitor hair length in people to determine those at
high risk of heart attack. What is the flaw in this experiment?
a Assumptions
b Association vs. Causation
c Response Bias
d Operationalization
d Operationalization
Monitoring hair growth does not measure the risk of heart attack. There is a flaw in the
experiment because the experiment is not measuring what the objective is trying to
determine.
Mr. Wonka notices that the last twenty times he invented a new chocolate candy,
his major competitors, Count Chocula, and the Easter Bunny, have big sales in
,late October. Mr. Wonka feels directly responsible for the profit of his
competitors. What is the flaw in this experiment?
a Small Sample Size
b Truly Representative Sample
c Association vs. Causation
d Blinding
c Association vs. Causation
Mr. Wonka's inventive ways are probably not the cause of increased chocolate candy
sales for his competitors at the end of October like Mr. Wonka has concluded.
There is a 90 percent chance that a package will arrive within three days of when
it was shipped. Also, there is a 75 percent chance that it will get wet. There is a 70
percent chance that it will get wet and will be delivered within three days. What is
the likelihood that at least one of these events occurs?
a 0.8
b 0.85
c 0.9
d 0.95
d 0.95
This is a union between P(on time) and P(wet). Therefore, P(on time∪wet)
=P(on time)+P(wet)−P(on time∩wet)
=0.90+0.75−0.70=0.95=95%
There is an 80 percent chance of snow. If it snows there is a 10 percent chance of
Todd walking to the store. If it doesn't snow there is a 60 percent chance of Todd
walking to the store. What is the likelihood that it will not snow and Todd will walk
to the store?
a 0.12
b 0.18
c 0.2
d 0.48
a 0.12
The probability of no snow is 20 percent ( P(no snow)=0.20 ), and the probability of
Todd walking to the store with no snow is 60 percent ( P(walk|no snow)=0.60 ).
Therefore to determine the likelihood of it not snowing and Todd walking to the store,
these probabilities are multiplied together. Therefore P(no snow∩walk)=P(no
snow)×P(walk|no snow)= 0.20×0.60=0.12=12%
Determine the Mode, Median, Mean (in that order) from the following data set. 4, 7,
11, 12, 14, 14, 15, 17, 17, 17, 18, 20, 23, 24, 26, 29, 35, 39, 40
a 17, 17, 20
b 14, 17.5, 17
, c 17, 17, 17
d 17, 17, 20.11
d 17, 17, 20.11
The mode is 17 as it shows up 3 times and no other data point shows up as many
times. The median is 17 as it is the 10th number out of 19 ordered numbers. The mean
is 20.11 as it is the result of the sum of the numbers, 382, divided by 19, the count of
numbers.
Standard deviation measures
a the average of the difference for the data set.
b the normal exceptions between the expected data points.
c the median of the data set distributed equally.
d the dispersion from the average for the data set.
d the dispersion from the average for the data set.
Elizabeth got a 75 on her performance review. The average was 80, but the
standard deviation was 3.5. Determine the z-score for her performance review.
a 1.43
b -1.43
c5
d 3.5
b -1.43
This is because she was 1.43 standard deviations below the average. Z-score is
determined by the distance a data point is from the mean divided by the standard
deviation. The data point was five below the mean ( therefore -5), and the standard
deviation was 3.5. Therefore, the z-score for Elizabeth's performance review was
−53.5=−1.43 .
Statistical analysis can be used to make __________ about entire populations
using samples.
a) conclusions
b) inferences
c) decisions
d) references
b) inferences
What percent of the data of a normal distribution is within two standard
deviations of the mean?
a) 68.3
b) 75
c) 95.4
d) 99.7
c) 95.4
