Biostatistics Quiz: Mean Median And Mode
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Mean Median And ModeQuestion 1 of 20

In a study of genetic variants, researchers found that 60% of subjects are homozygous dominant (AA), 35% are heterozygous (Aa), and 5% are homozygous recessive (aa). If they assign numerical codes of 2, 1, and 0 respectively for statistical analysis, what is the mode of this coded dataset?

2, representing the homozygous dominant genotype with highest frequency
1, representing the heterozygous genotype as the middle category
1.55, calculated as the weighted average of the coded values
AA, since mode refers to the original categorical data
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Biostatistics Quiz

Biostatistics Quiz: Mean Median And Mode

Practice Mean Median And Mode in Biostatistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Mean Median And Mode, giving you a quick way to practice the rules, question types, and explanations that matter most for Biostatistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

In a study of genetic variants, researchers found that 60% of subjects are homozygous dominant (AA), 35% are heterozygous (Aa), and 5% are homozygous recessive (aa). If they assign numerical codes of 2, 1, and 0 respectively for statistical analysis, what is the mode of this coded dataset?

  1. 2, representing the homozygous dominant genotype with highest frequency (correct answer)
  2. 1, representing the heterozygous genotype as the middle category
  3. 1.55, calculated as the weighted average of the coded values
  4. AA, since mode refers to the original categorical data
Explanation: The mode is the value that appears most frequently. With 60% coded as 2, 35% as 1, and 5% as 0, the value 2 has the highest frequency and is therefore the mode. Choice B incorrectly identifies the middle value rather than most frequent. Choice C confuses mode with mean (2×0.6 + 1×0.35 + 0×0.05 = 1.55). Choice D incorrectly refers to the original categories rather than the numerical codes being analyzed.

Question 2

A dataset contains the following values: 12, 15, 18, 18, 21, 24, 27. If each value is transformed by the function f(x) = 2x - 6, what is the relationship between the original median and the transformed median?

  1. The transformed median equals twice the original median
  2. The transformed median equals twice the original median minus 6 (correct answer)
  3. The transformed median equals the original median plus 6
  4. The transformed median equals the original median minus 3
  5. The transformed median equals half the original median plus 6
Explanation: When you encounter questions about data transformations in biostatistics, focus on how mathematical operations affect measures of central tendency. The key insight is that linear transformations applied to data follow predictable patterns for statistics like the median. Let's work through this systematically. First, find the original median from the dataset: 12, 15, 18, 18, 21, 24, 27. With 7 values, the median is the 4th value when ordered, which is 18. Now apply the transformation f(x) = 2x - 6 to each value: 18, 24, 30, 30, 36, 42, 48. The transformed median is the 4th value: 30. For any linear transformation f(x) = ax + b, the transformed median equals a(original median) + b. Here, with f(x) = 2x - 6, the transformed median = 2(18) - 6 = 36 - 6 = 30. This confirms our calculation. Choice A suggests the transformed median equals twice the original median (2 × 18 = 36), but this ignores the "-6" component of the transformation. Choice C proposes adding 6 to the original median (18 + 6 = 24), which incorrectly treats the transformation as if it were f(x) = x + 6. Choice D suggests subtracting 3 from the original median (18 - 3 = 15), which has no logical basis in the given transformation. Choice B correctly states that the transformed median equals twice the original median minus 6: 2(18) - 6 = 30. Remember: linear transformations preserve the relationship between the transformation function and all measures of central tendency, including the median.

Question 3

In a study of reaction times, 9 participants had the following times (in seconds): 0.8, 1.2, 1.4, 1.4, 1.6, 1.8, 2.0, 2.4, 3.2. If the participant with the reaction time of 3.2 seconds is replaced by a participant with a reaction time of 1.4 seconds, how do the measures of central tendency change?

  1. Mean decreases, median decreases, mode changes from 1.4 to no mode
  2. Mean decreases, median stays the same, mode frequency increases but value stays at 1.4
  3. Mean decreases, median decreases, mode frequency increases and value stays at 1.4 (correct answer)
  4. Mean stays the same, median decreases, mode changes from no mode to 1.4
  5. Mean decreases, median increases, mode changes from 1.4 to multiple modes
Explanation: When analyzing how changes to a dataset affect measures of central tendency, you need to calculate each measure before and after the change to see the impact. Let's work through this systematically. Originally: 0.8, 1.2, 1.4, 1.4, 1.6, 1.8, 2.0, 2.4, 3.2. After replacement: 0.8, 1.2, 1.4, 1.4, 1.4, 1.6, 1.8, 2.0, 2.4. For the mean: Original sum = 15.8, so mean = 15.8/9 = 1.76. New sum = 13.0, so mean = 13.0/9 = 1.44. The mean decreases because we replaced a large value (3.2) with a smaller one (1.4). For the median: Both datasets have 9 values, so the median is the 5th value when ordered. Originally, the 5th value is 1.6. After the change, the 5th value is 1.4. The median decreases. For the mode: Originally, 1.4 appears twice while all other values appear once, making 1.4 the mode. After replacement, 1.4 appears three times, so it's still the mode but with higher frequency. Answer C correctly identifies all three changes. Answer A wrongly claims the mode disappears. Answer B incorrectly states the median stays the same—it actually drops from 1.6 to 1.4. Answer D falsely claims the mean stays constant and that there was originally no mode. Study tip: When a dataset changes, always recalculate each measure separately rather than trying to predict the changes. The median is particularly tricky since it depends on position, not just the values involved in the change.

Question 4

A dataset contains the values 2, 4, 6, 8, 10, 12, 14. If three additional identical values are added to make the mean equal to the median, what value must be added three times?

  1. 6
  2. 7
  3. 8 (correct answer)
  4. 9
  5. 10
Explanation: When you encounter problems involving changes to mean and median, you need to understand how adding data points affects these measures of central tendency. The median is the middle value when data is ordered, while the mean is the average of all values. Starting with the dataset {2, 4, 6, 8, 10, 12, 14}, the current median is 8 (the middle value of 7 numbers), and the current mean is 567=8\frac{56}{7} = 8. Notice they're already equal! However, adding three identical values will change both measures. After adding three identical values xx, you'll have 10 total values. The new mean will be 56+3x10\frac{56 + 3x}{10}. For the median of 10 values, you need the average of the 5th and 6th values when ordered. If x=8x = 8, the ordered dataset becomes {2, 4, 6, 8, 8, 8, 8, 10, 12, 14}. The median is 8+82=8\frac{8 + 8}{2} = 8, and the mean is 56+2410=8\frac{56 + 24}{10} = 8. They're equal, confirming answer C. Answer A (6) would create a median of 6+82=7\frac{6 + 8}{2} = 7 but a mean of 7410=7.4\frac{74}{10} = 7.4. Answer B (7) gives a median of 7+82=7.5\frac{7 + 8}{2} = 7.5 and mean of 7710=7.7\frac{77}{10} = 7.7. Answer D (9) produces a median of 8+92=8.5\frac{8 + 9}{2} = 8.5 and mean of 8310=8.3\frac{83}{10} = 8.3. Remember: when adding multiple identical values to a dataset, systematically check how they affect the position of the median and recalculate both measures to verify equality.

Question 5

A researcher has two datasets: Dataset A (n=20) with mean=50 and median=48, and Dataset B (n=30) with mean=60 and median=62. When the datasets are combined, which statement about the combined measures of central tendency is most likely correct?

  1. The combined mean will be 56 and the combined median will be approximately 55
  2. The combined mean will be 55 and the combined median will be approximately 56
  3. The combined mean will be 56 and the combined median cannot be determined from given information (correct answer)
  4. Both the combined mean and median will equal 55 due to the averaging effect
  5. The combined median will be 55 and the combined mean cannot be determined from given information
Explanation: When combining datasets with different sample sizes, you need to understand how measures of central tendency behave differently. The mean can always be calculated using weighted averages, but the median requires knowledge of the actual data distribution. For the combined mean, you calculate a weighted average based on sample sizes: (20×50)+(30×60)20+30=1000+180050=56\frac{(20 \times 50) + (30 \times 60)}{20 + 30} = \frac{1000 + 1800}{50} = 56. This calculation is straightforward because means are additive when weighted by sample size. However, the combined median cannot be determined from the given information. To find a median, you need to arrange all 50 values in order and find the middle value(s). Knowing only the individual medians tells you nothing about how the datasets will interleave when combined. The combined median could range anywhere from 48 to 62 depending on the actual distribution of values. Answer C correctly identifies that the mean is 56 and acknowledges that the median cannot be determined. Answer A incorrectly assumes the median will be 55 (the average of individual medians), which ignores sample size weighting. Answer B makes the same median error while also miscalculating the weighted mean as 55. Answer D incorrectly suggests both measures will equal 55 due to some "averaging effect" that doesn't exist in statistics. Remember this key distinction: when combining datasets, means can always be calculated using weighted averages, but medians require knowledge of the complete data distribution. Never assume you can simply average the individual medians.

Question 6

A researcher calculates that for a set of positive values, the geometric mean is 12 and the harmonic mean is 8. If the dataset contains exactly 4 observations and one of them is 18, which measure of central tendency (arithmetic mean, median, or mode) can be definitively determined?

  1. Only the arithmetic mean can be determined; it equals 15.0
  2. Only the median can be determined; it equals 13.5
  3. Only the arithmetic mean can be determined; it equals 18.0
  4. Both the arithmetic mean and median can be determined
  5. None of the measures can be definitively determined from the given information (correct answer)
Explanation: When you encounter problems involving multiple measures of central tendency, remember that these measures are interconnected through mathematical relationships that can help you solve for unknown values. Given information: geometric mean = 12, harmonic mean = 8, n = 4 observations, one value is 18. You need to use the relationship between means to find what can be determined. For any set of positive numbers, there's a fundamental relationship: AM×HM=GM2AM \times HM = GM^2 (where AM = arithmetic mean, HM = harmonic mean, GM = geometric mean). Substituting the known values: AM×8=122=144AM \times 8 = 12^2 = 144, so AM=18AM = 18. Now you know the sum of all four observations: 4×18=724 \times 18 = 72. Since one observation is 18, the sum of the other three is 72−18=5472 - 18 = 54. You can also use the geometric mean formula: 18×a×b×c4=12\sqrt[4]{18 \times a \times b \times c} = 12, which gives 18×a×b×c=20,73618 \times a \times b \times c = 20,736, so a×b×c=1,152a \times b \times c = 1,152. With these constraints (three unknowns summing to 54 with product 1,152), you can solve the system to find the other values are 6, 12, and 36. Therefore, the complete dataset is {6, 12, 18, 36}, making the median = 15. Answer A is wrong because the arithmetic mean is 18, not 15. Answer B incorrectly calculates the median as 13.5. Answer C gives the wrong arithmetic mean value. Answer D is correct since both arithmetic mean (18) and median (15) can be determined. Study tip: Remember the AM-GM-HM relationship: AM×HM=GM2AM \times HM = GM^2 for positive values. This identity frequently appears in biostatistics problems involving multiple central tendency measures.

Question 7

A clinical trial measures recovery times (days) for 9 patients: 3, 5, 7, 8, 10, 12, 15, 18, 22. The research team wants to report a measure of central tendency that would be most affected by potential outliers in future similar studies. Which measure should they choose and what is its value for this dataset?

  1. Median; 10 days because it represents the middle value and is outlier-resistant
  2. Mode; there is no mode because all values appear exactly once
  3. Mean; 11.1 days because it incorporates all values and is sensitive to outliers (correct answer)
  4. Median; 11.1 days because it balances all data points equally
  5. Mean; 10 days because it provides the most representative central value
Explanation: When you encounter questions about measures of central tendency and outliers, focus on how each measure responds to extreme values. The question specifically asks for the measure "most affected by potential outliers," which is your key clue. The mean is calculated by summing all values and dividing by the number of observations. For this dataset: 3+5+7+8+10+12+15+18+229=1009=11.1\frac{3+5+7+8+10+12+15+18+22}{9} = \frac{100}{9} = 11.1 days. The mean incorporates every single data point in its calculation, making it highly sensitive to outliers. If one patient had taken 50 days to recover instead of 22, the mean would jump dramatically to 14.2 days, while other measures would be much less affected. Option A incorrectly identifies the median as outlier-sensitive when it's actually outlier-resistant. The median (10 days, the middle value when data is ordered) would barely change even with extreme outliers. Option B correctly notes there's no mode, but the question asks specifically for the measure most affected by outliers, not whether a mode exists. Option D gives the wrong value for the median—it confuses the median calculation with the mean's value of 11.1 days. Remember this pattern: the mean uses all data points in its calculation, making it the most outlier-sensitive measure. The median only depends on the middle position(s), and the mode only cares about frequency. When biostatistics questions mention outlier sensitivity, think "mean" first—it's the measure that gets pulled toward extreme values.

Question 8

In a dose-escalation study, the following doses (mg) were administered: 2, 4, 8, 8, 16, 32, 64. If the study protocol requires that the geometric mean dose be exactly twice the median dose, how should the dataset be modified by changing exactly one value?

  1. Change the value 2 to 4, making the geometric mean approximately 16.0 mg
  2. Change the value 64 to 32, making the geometric mean approximately 16.0 mg
  3. Change the value 4 to 8, making the geometric mean approximately 16.0 mg
  4. Change the value 32 to 16, making the geometric mean approximately 16.0 mg
  5. No single value change can satisfy the protocol requirement (correct answer)
Explanation: When you encounter dose-escalation studies in biostatistics, you're often working with data that spans multiple orders of magnitude, making geometric means particularly relevant since they're less sensitive to extreme values than arithmetic means. Let's first establish our baseline. The current dataset is: 2, 4, 8, 8, 16, 32, 64. The median of these 7 values is the 4th value when ordered: 8 mg. For the protocol requirement to be met, the geometric mean must equal twice the median, so 2 × 8 = 16 mg. The geometric mean equals ∏xin\sqrt[n]{\prod x_i}. For our data: 2×4×8×8×16×32×647=1,048,5767≈11.31\sqrt[7]{2 \times 4 \times 8 \times 8 \times 16 \times 32 \times 64} = \sqrt[7]{1,048,576} ≈ 11.31 mg. We need this to equal 16 mg. Since we need the geometric mean to increase from 11.31 to 16, we must replace a smaller value with a larger one (changing a larger value to smaller would decrease the geometric mean). Looking at the options systematically, none of them actually achieve a geometric mean of 16 mg when calculated properly. Option A changes 2→4, giving a geometric mean of about 12.6 mg, not 16.0. Option B changes 64→32, which would decrease the geometric mean. Option C changes 4→8, yielding approximately 13.2 mg. Option D changes 32→16, also decreasing the geometric mean. Since none of the provided modifications achieve the required geometric mean of 16 mg, the answer is E. Remember: always verify geometric mean calculations rather than accepting stated values, especially when dealing with exponential relationships in dose studies.

Question 9

A research team collected data on protein concentrations (mg/mL): 2.1, 2.3, 2.3, 2.5, 2.7, 2.9, 3.1. Due to a calibration error, each measurement was 0.4 mg/mL higher than the true value. After correcting for this systematic error, what is the corrected median?

  1. 2.1 mg/mL (correct answer)
  2. 2.3 mg/mL
  3. 2.5 mg/mL
  4. 2.9 mg/mL
  5. The median cannot be determined without knowing the distribution shape
Explanation: When you encounter systematic measurement errors in biostatistics, remember that these errors affect all data points equally, so measures of central tendency shift by the same amount as the individual values. First, let's find the median of the original (incorrect) data: 2.1, 2.3, 2.3, 2.5, 2.7, 2.9, 3.1. With 7 values, the median is the 4th value when arranged in order, which is 2.5 mg/mL. Since each measurement was 0.4 mg/mL too high, we subtract 0.4 from every value to get the corrected data: 1.7, 1.9, 1.9, 2.1, 2.3, 2.5, 2.7. The corrected median is the 4th value: 2.1 mg/mL. Here's why the other answers are wrong: Answer B (2.3 mg/mL) represents the 5th value in the corrected dataset, not the median. Answer C (2.5 mg/mL) is a common trap—this is the original uncorrected median, showing the student forgot to apply the systematic correction. Answer D (2.9 mg/mL) doesn't correspond to any meaningful position in either the original or corrected dataset. Notice that we could have solved this more efficiently: since systematic errors shift all values equally, the median simply shifts by the same amount. Original median (2.5) minus systematic error (0.4) equals corrected median (2.1). Study tip: For systematic measurement errors, all measures of central tendency (mean, median, mode) shift by exactly the error amount. You can either correct the entire dataset first or just subtract the error from the original statistic.

Question 10

A frequency distribution shows: Value 5 appears 3 times, Value 7 appears 5 times, Value 9 appears 5 times, Value 11 appears 2 times. Which statement about the measures of central tendency is correct?

  1. The dataset is bimodal with modes at 7 and 9, and the mean is greater than the median (correct answer)
  2. The dataset is unimodal with mode at 7, and the median equals the mean
  3. The dataset is bimodal with modes at 7 and 9, and the median is greater than the mean
  4. The dataset has no mode because multiple values have the same highest frequency
  5. The dataset is unimodal with mode at 9, and the mean is less than the median
Explanation: When analyzing frequency distributions, you need to calculate three measures of central tendency: mode (most frequent value), median (middle value), and mean (average). Understanding how these relate to each other reveals important patterns about your data's shape. Let's work through this systematically. First, identify the mode by finding which value(s) appear most frequently. Both 7 and 9 appear 5 times each (the highest frequency), making this dataset bimodal with modes at 7 and 9. Next, find the median by arranging all 15 values in order: 5,5,5,7,7,7,7,7,9,9,9,9,9,11,11. The median is the 8th value (middle of 15), which is 7. For the mean, calculate: (5×3)+(7×5)+(9×5)+(11×2)15=15+35+45+2215=11715=7.8\frac{(5×3) + (7×5) + (9×5) + (11×2)}{15} = \frac{15 + 35 + 45 + 22}{15} = \frac{117}{15} = 7.8 Since the mean (7.8) is greater than the median (7), answer A is correct. Answer B is wrong because the dataset has two modes, not one. Answer C incorrectly states the median is greater than the mean, when actually 7 < 7.8. Answer D shows a common misconception—when multiple values tie for the highest frequency, they're all modes, creating a multimodal distribution. Study tip: Always calculate all three measures systematically. In biostatistics, understanding whether data is skewed (mean ≠ median) helps you choose appropriate statistical tests and interpret results correctly.

Question 11

A research study collected systolic blood pressure measurements (mmHg) from 7 participants: 118, 142, 138, 142, 125, 160, 142. If one additional participant with a systolic blood pressure of 200 mmHg is added to the dataset, which statement best describes the effect on the measures of central tendency?

  1. The mean increases more than the median, and the mode remains unchanged (correct answer)
  2. The median increases more than the mean, and the mode increases by 8 mmHg
  3. Both the mean and median increase by the same amount, while the mode decreases
  4. The mean decreases due to the outlier effect, while the median and mode both increase
  5. The mode increases more than the median, and the mean remains approximately constant
Explanation: When analyzing how adding a data point affects measures of central tendency, you need to consider how sensitive each measure is to extreme values. This question tests your understanding of how outliers differently impact the mean, median, and mode. Let's calculate the original values. The dataset {118, 142, 138, 142, 125, 160, 142} has a mean of 9677=138.1\frac{967}{7} = 138.1 mmHg. When ordered {118, 125, 138, 142, 142, 160, 142}, the median is 142 mmHg (middle value), and the mode is 142 mmHg (appears three times). Adding 200 mmHg creates {118, 142, 138, 142, 125, 160, 142, 200}. The new mean becomes 11678=145.9\frac{1167}{8} = 145.9 mmHg (increase of 7.8). The ordered dataset {118, 125, 138, 142, 142, 160, 142, 200} now has 8 values, so the median is 142+1422=142\frac{142 + 142}{2} = 142 mmHg (no change). The mode remains 142 mmHg since it still appears most frequently. Option A correctly identifies that the mean increases more than the median (7.8 vs. 0), and the mode stays unchanged. Option B incorrectly claims the median increases more and that the mode increases. Option C wrongly states both measures increase equally and the mode decreases. Option D falsely suggests the mean decreases and other measures increase. Remember that the mean is most sensitive to outliers because every data point affects its calculation equally, while the median only depends on middle values and the mode reflects frequency patterns. When you see extreme values added to datasets, expect the mean to show the largest change.

Question 12

A researcher reports that for a dataset of 11 observations, the median is 45 and the mode is 38. If the sum of all observations is 495, what can be concluded about the relationship between the mean and median?

  1. The mean equals the median, indicating a symmetric distribution (correct answer)
  2. The mean is greater than the median, suggesting a right-skewed distribution
  3. The mean is less than the median, suggesting a left-skewed distribution
  4. The relationship cannot be determined without knowing the individual data values
  5. The mean must equal the mode since the distribution is unimodal
Explanation: When analyzing the relationship between measures of central tendency, you need to calculate what's given and compare the results systematically. First, let's find the mean. With 11 observations and a sum of 495, the mean is 49511=45\frac{495}{11} = 45. Since the median is also 45, the mean equals the median exactly. This equality tells us something important about the distribution's shape. When the mean and median are equal, it indicates the data is symmetrically distributed around the center value. The mode being 38 (different from both mean and median) doesn't contradict this - a distribution can still be symmetric even when the mode differs, especially with small sample sizes where multiple values might appear with the same frequency. Let's examine why the other options fail: Option B incorrectly assumes the mean is greater than the median, which would indicate right skewness, but our calculation shows they're equal. Option C makes the opposite error, suggesting left skewness when the mean is less than the median - again, not supported by the data. Option D claims we can't determine the relationship, but we absolutely can since we have all the information needed to calculate the mean and compare it to the given median. Study tip: Always calculate what you can from the given information before making assumptions. The relationship between mean and median is a reliable indicator of distribution shape: mean = median suggests symmetry, mean > median suggests right skew, and mean < median suggests left skew.

Question 13

A dataset has the following properties: (1) It contains 8 values, (2) The median is 12, (3) The mode is 8, and (4) Three values are below the median and three values are above the median. Which statement about the mean is most likely correct?

  1. The mean is less than 12 because the mode pulls it down
  2. The mean equals 12 because the distribution is symmetric around the median
  3. The mean is greater than 12 because there are more values above the median
  4. The mean cannot be determined from the given information (correct answer)
  5. The mean is between 8 and 12 due to the influence of the mode
Explanation: When analyzing relationships between measures of central tendency, you need to consider what information is actually provided versus what appears to be given. This question tests whether you can distinguish between sufficient and insufficient data for determining the mean. Let's examine what we know: 8 values total, median = 12, mode = 8, with three values below and three above the median. Since there are 8 values with 3 below and 3 above the median, the median of 12 must be the average of the 4th and 5th values when arranged in order. However, we don't know the actual values - only their positions relative to the median and that 8 appears most frequently. The mean depends on the sum of all values, and we simply don't have enough specific information about the individual data points. The values above the median could range from just slightly above 12 to extremely large numbers, dramatically affecting the mean. Similarly, the values below could cluster near 12 or be much smaller. Answer choice A incorrectly assumes the mode directly "pulls" the mean downward. The mode's influence on the mean depends on how many times it appears and what the other values are. Choice B wrongly concludes the distribution is symmetric just because equal numbers of values fall above and below the median - the actual spacing of these values matters for symmetry. Choice C makes an unfounded assumption that having equal counts above and below the median somehow favors higher values. Remember: the mean requires knowing actual values or their sum, not just their relative positions. Don't confuse positional information with the quantitative data needed for calculating means.

Question 14

A pharmaceutical company is analyzing the time to peak drug concentration in plasma samples. The data shows a right-skewed distribution with the following summary statistics from 50 patients:

Given that the mean time to peak concentration is 4.2 hours and the median is 3.1 hours, if the company wants to report a measure that best represents the 'typical' patient experience while minimizing the influence of a few patients who had unusually delayed absorption, which measure should they choose and why?

  1. The mean, because it incorporates information from all patients equally
  2. The median, because it resists the influence of extreme values (correct answer)
  3. The mode, because it represents the most frequently occurring response
  4. The geometric mean, because it accounts for the multiplicative nature
Explanation: In a right-skewed distribution, the mean is pulled toward the tail by extreme values, making it higher than the median (4.2 vs 3.1 hours confirms right skew). The median is robust to outliers and better represents the center of the distribution when skewness is present. Choice A is wrong because equal weighting of all values is problematic when outliers exist. Choice C is incorrect as the mode isn't given and may not be representative. Choice D misapplies geometric mean concepts inappropriately.

Question 15

A dataset contains the values 3, 5, 7, 7, 9, 11, 13. If each value is transformed by the function f(x) = 2x - 1, what happens to the relationship between the mean and median?

  1. The mean increases by more than the median due to the linear transformation
  2. The transformation affects the mean more because it's sensitive to extreme values
  3. The median changes by exactly 13 while the mean changes by approximately 14.8
  4. Both the mean and median undergo identical transformations, preserving their relationship (correct answer)
Explanation: When you encounter questions about data transformations in biostatistics, focus on how linear transformations affect measures of central tendency. Linear transformations of the form f(x) = ax + b apply the same mathematical operation to every data point. Let's work through this systematically. For the original dataset {3, 5, 7, 7, 9, 11, 13}, the mean is 3+5+7+7+9+11+137=7.86\frac{3+5+7+7+9+11+13}{7} = 7.86 and the median is 7 (the middle value). After applying f(x) = 2x - 1, the transformed dataset becomes {5, 9, 13, 13, 17, 21, 25}. The new mean is 2(7.86)−1=14.712(7.86) - 1 = 14.71 and the new median is 2(7)−1=132(7) - 1 = 13. Both measures changed by exactly the same transformation rule: they were doubled and then decreased by 1. This preserves their original relationship, confirming answer D. Answer A incorrectly suggests the mean increases more than the median, but both follow the identical transformation formula. Answer B misapplies the concept that means are sensitive to outliers—while true in general, it's irrelevant here because linear transformations affect all values equally, including any outliers. Answer C provides incorrect calculations: the median changes by 13−7=613 - 7 = 6, not 13, and the mean changes by 14.71−7.86=6.8514.71 - 7.86 = 6.85, not 14.8. Remember this key principle: linear transformations preserve the mathematical relationships between statistics. If you transform data by f(x) = ax + b, then f(mean) = a(mean) + b and f(median) = a(median) + b.

Question 16

In a clinical trial, the number of adverse events per patient follows this distribution: 40% of patients have 0 events, 30% have 1 event, 20% have 2 events, and 10% have 3 events. If the trial includes 200 patients, what is the relationship between the mean and median number of adverse events?

  1. The mean is 0.5 events higher than the median
  2. The mean equals the median at 1.0 event
  3. The mean is 1.0 event higher than the median (correct answer)
  4. The median is 0.5 events higher than the mean
Explanation: Mean = (0×0.4) + (1×0.3) + (2×0.2) + (3×0.1) = 0 + 0.3 + 0.4 + 0.3 = 1.0. For the median with 200 patients: 40% (80 patients) have 0 events, 30% (60 patients) have 1 event. The 100th and 101st patients (median positions) both fall in the group with 0 events, so median = 0. Mean - Median = 1.0 - 0 = 1.0. Choice A gives an incorrect calculation. Choice B incorrectly assumes they're equal. Choice D reverses the relationship.

Question 17

A researcher studying pain scores (0-10 scale) collected data from 15 patients: 2, 3, 3, 4, 4, 4, 5, 6, 6, 7, 7, 7, 7, 8, 9. Due to a data entry error, one of the 7s was mistakenly recorded as 17. What is the difference between the median of the corrupted dataset and the median of the correct dataset?

  1. The median increases by 1.0
  2. The median remains unchanged at 6.0 (correct answer)
  3. The median decreases by 0.5
  4. The median increases by 0.67
Explanation: In both datasets, we have 15 values, so the median is the 8th value when arranged in order. In the correct dataset: 2, 3, 3, 4, 4, 4, 5, 6, 6, 7, 7, 7, 7, 8, 9 - the 8th value is 6. In the corrupted dataset: 2, 3, 3, 4, 4, 4, 5, 6, 6, 7, 7, 7, 8, 9, 17 - the 8th value is still 6. The outlier (17) doesn't affect the median position. Choice A assumes the median shifts upward. Choice C incorrectly calculates a decrease. Choice D represents the mean change divided by something, showing confusion between mean and median.

Question 18

A bimodal dataset of enzyme activity levels has modes at 15 U/L and 25 U/L, with equal frequencies of 12 observations each. The remaining 16 observations are symmetrically distributed around a mean of 20 U/L. What is the mean of the entire dataset?

  1. 20.0 U/L due to perfect symmetry around this value (correct answer)
  2. 19.5 U/L accounting for the weighted modal influences
  3. 20.5 U/L reflecting the upward skew from higher mode
  4. 18.75 U/L calculated from the harmonic mean approach
Explanation: Total observations = 12 + 12 + 16 = 40. Sum = (15×12) + (25×12) + (20×16) = 180 + 300 + 320 = 800. Mean = 800/40 = 20.0 U/L. The symmetric distribution of the middle group around 20 U/L, combined with the two modes being equidistant from 20 (15 and 25), creates perfect balance. Choice B incorrectly applies a weighting scheme. Choice C assumes asymmetric influence from the higher mode. Choice D misapplies harmonic mean concepts.

Question 19

A pharmaceutical company conducted a dose-response study with the following results:

Based on the table shown, if the company wants to report a single 'typical' response value that is most resistant to the outlying response at the highest dose, which measure should they use and what is its value?

  1. Mean; the value is 24.3 because it incorporates all data points equally
  2. Median; the value is 22 because it is unaffected by extreme values (correct answer)
  3. Mode; the value is 20 because it represents the most common response
  4. Mean; the value is 22 because outliers have minimal impact on this measure
  5. Median; the value is 24 because it represents the middle value when outliers are removed
Explanation: From the data: 18, 20, 20, 22, 24, 26, 60. The value 60 is clearly an outlier. Mean = (18+20+20+22+24+26+60)/7 = 190/7 ≈ 27.1 (heavily influenced by outlier). Median = 22 (4th value in ordered list, unaffected by the outlier). Mode = 20 (appears twice). The median is the most resistant to outliers, and its value is 22.

Question 20

A researcher calculates that removing a single data point from a dataset of 21 values changes the mean from 45.2 to 44.8. What was the value of the removed data point?

  1. 50.0, estimated from the approximate change magnitude
  2. 8.4, representing twice the mean difference
  3. 37.6, reflecting the below-average contribution
  4. 53.2, calculated from the difference in totals (correct answer)
Explanation: When you encounter questions about data points affecting summary statistics, you need to think about the relationship between individual values and the calculated measures like means. To find the removed data point, you must work backwards from the change in means. The original dataset of 21 values had a mean of 45.2, so the total of all values was 21×45.2=949.221 \times 45.2 = 949.2. After removing one point, 20 values remained with a mean of 44.8, giving a new total of 20×44.8=89620 \times 44.8 = 896. The difference between these totals (949.2−896=53.2949.2 - 896 = 53.2) is exactly the value of the removed data point. Answer D correctly identifies 53.2 as the removed value, calculated precisely from the difference in totals. Answer A (50.0) represents a rough estimation that ignores the exact mathematical relationship—while it's in the right ballpark, biostatistics requires precision. Answer B (8.4) incorrectly doubles the mean difference (2×4.4=8.82 \times 4.4 = 8.8, approximately), which has no mathematical basis for this problem. Answer C (37.6) might tempt you because it's below the mean, but this backward reasoning assumes the removed point must be below average when it's actually above average. Remember this formula: when removing a data point changes the mean, the removed value equals the difference in the dataset totals before and after removal. This direct calculation method eliminates guesswork and ensures accuracy in biostatistics problems involving changing sample compositions.