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WGU C207 DATA DRIVEN DECISION MAKING MODULE 2 EXAM QUESTIONS AND ANSWERS 100% CORRECT!!

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The interquartile range for these numbers is 29. The 75th percentile is 43, and the 25th percentile is 14. Subtracting 14 from 43, we arrive at 29. What is the mean of the following data set? 9, 7, 23, 20, 4, 6, 26, 5, 21, 14 a) 13.5 b) 12 c) 26 d) 14 - ANSWER a) 13.5 The mean is the same as the average, which we get by adding up the values of each data point and dividing by the number of data points. There are 10 data points that add up to 135. 135/10=13.5 . Which of the following numbers is closest to the 70th percentile? 3, 4, 5, 6, 7, 9, 10, 14, 20, 21, 23, 26, 65 a) 65 b) 20 c) 26 d) 14 - ANSWER b) 20 The number 20 in this series of numbers is the closest value to the 70th percentile. There are 13 numbers, and therefore the 70th percentile would be at the 9.1th number in the series (from small to large). The 9th number in that series is 20. If a man owns two sports cars, two luxury cars, an

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WGU C207 DATA DRIVEN DECISION MAKING
MODULE 2 EXAM QUESTIONS AND ANSWERS
100% CORRECT!!

,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. -
ANSWER 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 - ANSWER 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 - ANSWER 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 - ANSWER 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 - ANSWER 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%

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 - ANSWER b -1.43

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