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WGU C784 Applied Healthcare Statistics OA and Pre-Assessment Practice Exam | Latest Version — Comprehensive Mastery Assessment

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This document contains study material and comprehensive practice questions for the WGU C784 Applied Healthcare Statistics Objective Assessment and Pre-Assessment. Topics include descriptive statistics, probability, sampling, data distributions, confidence intervals, hypothesis testing, correlation, regression, statistical interpretation, healthcare data analysis, and application of statistical methods to clinical and healthcare scenarios. It is designed to help WGU students review core statistical concepts and prepare for C784 assessments.

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WGU C784 Applied Healthcare Statistics OA
and Pre-Assessment Practice Exam | Latest
Version — Comprehensive Mastery Assessment

EXAM: WGU C784 Applied Healthcare Statistics OA and Pre-Assessment
TOTAL QUESTIONS: 150
PASSING SCORE: 100% (150 out of 150 must be correct)
TIME ALLOWED: Untimed (comprehensive practice)

OVERVIEW:
This exam covers descriptive statistics, probability, inferential statistics, hypothesis testing,
confidence intervals, correlation and regression, and healthcare applications of statistics.
Mastery of all concepts is required for success on the official WGU C784 Objective Assessment.

INSTRUCTIONS:
Read each question carefully. Select the best answer for each question. All questions must be
answered correctly to achieve a passing score. For calculation questions, show your work and
round as specified. For Select All That Apply (SATA) questions, select all correct options. For
scenario-based questions, use the provided data to answer all associated questions.



SECTION 1: DESCRIPTIVE STATISTICS

Q1: Which of the following best describes the purpose of descriptive statistics in healthcare
research?
A. To make inferences about a population based on sample data
B. To organize, summarize, and present data in a meaningful way
C. To test hypotheses about relationships between variables
D. To determine the statistical significance of research findings
Correct Answer: B
Rationale: Descriptive statistics are used to organize, summarize, and present data in a
meaningful way, such as calculating measures of central tendency and dispersion. Inferential
statistics (options A, C, and D) involve making predictions or testing hypotheses about
populations.

Q2: A nurse researcher collects the following patient satisfaction scores from 8 participants: 78,
82, 85, 85, 88, 90, 92, 95. What is the mean score?

,A. 85.5
B. 86.5
C. 87.5
D. 88.5
Correct Answer: C
Rationale: The mean is calculated by summing all values and dividing by the number of values.
(78 + 82 + 85 + 85 + 88 + 90 + 92 + 95) = 695. 695 ÷ 8 = 86.875, which rounds to 87.5 if rounded
to one decimal place. However, the exact mean is 86.875. Wait — let me recalculate:
78+82=160, +85=245, +85=330, +88=418, +90=508, +92=600, +95=695. 695/8 = 86.875. None of
the options match exactly. Let me adjust the question.

Q2: A nurse researcher collects the following patient satisfaction scores from 8 participants: 78,
82, 85, 85, 88, 90, 92, 95. What is the mean score? (Round to one decimal place)
A. 85.5
B. 86.5
C. 86.9
D. 88.5
Correct Answer: C
Rationale: The mean is calculated by summing all values and dividing by the number of values.
(78 + 82 + 85 + 85 + 88 + 90 + 92 + 95) = 695. 695 ÷ 8 = 86.875, which rounds to 86.9 when
rounded to one decimal place.

Q3: Using the same data set from Q2 (78, 82, 85, 85, 88, 90, 92, 95), what is the median?
A. 85
B. 86.5
C. 87
D. 88
Correct Answer: B
Rationale: The median is the middle value when data is ordered. With 8 values (even number),
the median is the average of the 4th and 5th values: (85 + 88) ÷ 2 = 86.5.

Q4: Using the same data set from Q2 (78, 82, 85, 85, 88, 90, 92, 95), what is the mode?
A. 82
B. 85
C. 88
D. There is no mode
Correct Answer: B
Rationale: The mode is the value that appears most frequently. In this data set, 85 appears
twice while all other values appear only once, making 85 the mode.

,Q5: [CALCULATION] A hospital tracks the length of stay (in days) for 10 patients: 3, 5, 2, 7, 4, 6,
5, 8, 4, 5. Calculate the range.
Formula: Range = Maximum value − Minimum value
Answer: 6 days
Rationale: The range is calculated by subtracting the minimum value from the maximum value.
Maximum = 8 days, Minimum = 2 days. Range = 8 − 2 = 6 days.

Q6: [CALCULATION] Using the same length of stay data from Q5 (3, 5, 2, 7, 4, 6, 5, 8, 4, 5),
calculate the variance. (Round to two decimal places)
Formula: s² = Σ(x − x̄)² / (n − 1)
Answer: 3.33
Rationale: First, calculate the mean: (3+5+2+7+4+6+5+8+4+5) = 49/10 = 4.9. Then calculate
squared deviations: (3−4.9)²=3.61, (5−4.9)²=0.01, (2−4.9)²=8.41, (7−4.9)²=4.41, (4−4.9)²=0.81,
(6−4.9)²=1.21, (5−4.9)²=0.01, (8−4.9)²=9.61, (4−4.9)²=0.81, (5−4.9)²=0.01. Sum = 28.9. Variance
= 28. = 3.211... which rounds to 3.21. Let me recalculate: 3.61+0.01=3.62, +8.41=12.03,
+4.41=16.44, +0.81=17.25, +1.21=18.46, +0.01=18.47, +9.61=28.08, +0.81=28.89, +0.01=28.90.
28.90/9 = 3.211... rounds to 3.21.

Q6: [CALCULATION] Using the same length of stay data from Q5 (3, 5, 2, 7, 4, 6, 5, 8, 4, 5),
calculate the variance. (Round to two decimal places)
Formula: s² = Σ(x − x̄)² / (n − 1)
Answer: 3.21
Rationale: First, calculate the mean: 49/10 = 4.9. Then calculate squared deviations from the
mean: (3−4.9)²=3.61, (5−4.9)²=0.01, (2−4.9)²=8.41, (7−4.9)²=4.41, (4−4.9)²=0.81, (6−4.9)²=1.21,
(5−4.9)²=0.01, (8−4.9)²=9.61, (4−4.9)²=0.81, (5−4.9)²=0.01. Sum of squared deviations = 28.90.
Variance = 28.90 / (10−1) = 28. = 3.21.

Q7: [CALCULATION] Using the variance from Q6 (3.21), calculate the standard deviation. (Round
to two decimal places)
Formula: s = √s²
Answer: 1.79
Rationale: The standard deviation is the square root of the variance. s = √3.21 = 1.7916..., which
rounds to 1.79 when rounded to two decimal places.

Q8: In a normal distribution, approximately what percentage of data falls within one standard
deviation of the mean?
A. 50%
B. 68%
C. 95%
D. 99.7%

, Correct Answer: B
Rationale: According to the empirical rule (68-95-99.7 rule), approximately 68% of data in a
normal distribution falls within one standard deviation of the mean, 95% within two standard
deviations, and 99.7% within three standard deviations.

Q9: A healthcare administrator is analyzing patient wait times. The data set has a mean of 45
minutes and a standard deviation of 12 minutes. A patient waited 33 minutes. What is the z-
score for this patient's wait time?
A. −0.5
B. −1.0
C. −1.5
D. −2.0
Correct Answer: B
Rationale: The z-score formula is z = (x − μ) / σ. Substituting the values: z = (33 − 45) / 12 = −12 /
12 = −1.0. This indicates the patient's wait time is exactly one standard deviation below the
mean.

Q10: [CALCULATION] A laboratory reports the following cholesterol levels (mg/dL) for 12
patients: 185, 192, 198, 200, 205, 210, 215, 220, 225, 230, 235, 240. Calculate the interquartile
range (IQR).
Formula: IQR = Q3 − Q1
Answer: 37.5 mg/dL
Rationale: With 12 ordered values, Q1 is the median of the lower half (positions 1-6):
(192+198)/2 = 195. Q3 is the median of the upper half (positions 7-12): (225+230)/2 = 227.5.
IQR = 227.5 − 195 = 32.5. Let me recalculate: Lower half: 185, 192, 198, 200, 205, 210. Median
of lower half = (198+200)/2 = 199. Upper half: 215, 220, 225, 230, 235, 240. Median of upper
half = (225+230)/2 = 227.5. IQR = 227.5 − 199 = 28.5.

Q10: [CALCULATION] A laboratory reports the following cholesterol levels (mg/dL) for 12
patients: 185, 192, 198, 200, 205, 210, 215, 220, 225, 230, 235, 240. Calculate the interquartile
range (IQR).
Formula: IQR = Q3 − Q1
Answer: 28.5 mg/dL
Rationale: With 12 ordered values, Q1 is the median of the lower half (positions 1-6): (198 +
200) / 2 = 199. Q3 is the median of the upper half (positions 7-12): (225 + 230) / 2 = 227.5. IQR =
Q3 − Q1 = 227.5 − 199 = 28.5 mg/dL.

Q11: Which measure of central tendency is most appropriate when a data set contains extreme
outliers?
A. Mean

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