WGU D207 DATA DRIVEN DECISION MAKING OA EXAM
AND STUDY GUIDE LATEST 2024-2025 ACTUAL EXAM 250
QUESTIONS AND CORRECT DETAILED ANSWERS WITH
RATIONALES (VERIFIED ANSWERS) |ALREADY GRADED A+
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%
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 - ANSWER-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 - ANSWER-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. - ANSWER-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 c 5 d
3.5 - ANSWER-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 .
If a man owns two sports cars, two luxury cars, and one SUV, what is the probability that the man will not
take a luxury car out this evening?
a) 1/3
b) 5/3
c) 3/5
d) 3/1 - ANSWER-c) 3/5
There is an 80 percent chance Georgina will wear her jean jacket. There is a 60 percent chance it won't
rain. What are the chances that Georgina's jacket gets wet?
a) 48%
b) 140%
c) 32%
d) 80% - ANSWER-c) 32%
This is an intersection between it raining and Georgina wearing her jean jacket. This means multiplying
the likelihoods of both events. P(rain∩jacket) =P(rain)×P(jacket)=0.40×0.80=0.32
In the next hour, there is a 25 percent chance that Bruce will put on a suit. Also in the next hour, there is
a 60 percent chance that Clark will take off his glasses. There is a 9 percent chance that both of these
events might happen. What is the likelihood that either Clark will take off his glasses or Bruce will put on
a suit?
a) 76%
b) 24%
c) 15%
d) 94% - ANSWER-a) 76%
AND STUDY GUIDE LATEST 2024-2025 ACTUAL EXAM 250
QUESTIONS AND CORRECT DETAILED ANSWERS WITH
RATIONALES (VERIFIED ANSWERS) |ALREADY GRADED A+
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%
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 - ANSWER-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 - ANSWER-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. - ANSWER-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 c 5 d
3.5 - ANSWER-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 .
If a man owns two sports cars, two luxury cars, and one SUV, what is the probability that the man will not
take a luxury car out this evening?
a) 1/3
b) 5/3
c) 3/5
d) 3/1 - ANSWER-c) 3/5
There is an 80 percent chance Georgina will wear her jean jacket. There is a 60 percent chance it won't
rain. What are the chances that Georgina's jacket gets wet?
a) 48%
b) 140%
c) 32%
d) 80% - ANSWER-c) 32%
This is an intersection between it raining and Georgina wearing her jean jacket. This means multiplying
the likelihoods of both events. P(rain∩jacket) =P(rain)×P(jacket)=0.40×0.80=0.32
In the next hour, there is a 25 percent chance that Bruce will put on a suit. Also in the next hour, there is
a 60 percent chance that Clark will take off his glasses. There is a 9 percent chance that both of these
events might happen. What is the likelihood that either Clark will take off his glasses or Bruce will put on
a suit?
a) 76%
b) 24%
c) 15%
d) 94% - ANSWER-a) 76%
