Biostatistics Quiz: Variance And Standard Deviation
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Variance And Standard DeviationQuestion 1 of 20

In a sample, n=25 and sum of squared deviations from the mean is 144. What is the sample SD?

2.40
2.45
2.50
6.00
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Biostatistics Quiz

Biostatistics Quiz: Variance And Standard Deviation

Practice Variance And Standard Deviation 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 Variance And Standard Deviation, 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.

All questions

Question 1

In a sample, n=25 and sum of squared deviations from the mean is 144. What is the sample SD?

  1. 2.40
  2. 2.45 (correct answer)
  3. 2.50
  4. 6.00
Explanation: For a sample, divide the sum of squared deviations by n - 1 = 24 to get the variance: 144 / 24 = 6. The sample SD is the square root of 6, which rounds to 2.45. The tempting mistake is dividing by n = 25 instead, giving 2.40, but that treats the data as a population rather than a sample.

Question 2

Population values: 2, 4, 4, 4, 6, 6. What is the population variance?

  1. 1.89 (correct answer)
  2. 2.27
  3. 4.00
  4. 4.33
Explanation: Start with the mean: (2+4+4+4+6+6)/6 = 4.33. Subtract 4.33 from each value, square the differences, and add them: 5.44 + 0.11 + 0.11 + 0.11 + 2.78 + 2.78 = 11.33. For a population, divide by 6, not 5, giving 11.33/6 = 1.89. Dividing by 5 would give 2.27, but that is the sample variance mistake.

Question 3

Mean=100, SD=5. Which claim is true for any distribution?

  1. Exactly 95% lie in [90, 110]
  2. No data lie outside [90, 110]
  3. Exactly 68% lie in [95, 105]
  4. At least 75% lie in [90, 110] (correct answer)
Explanation: By Chebyshev's inequality, for any distribution at least 75% of values lie within 2 standard deviations of the mean: 100 ± 2(5) = [90, 110]. The tempting wrong answer is "exactly 95%" in that interval, but that rule applies only to normal distributions, not to any distribution.

Question 4

Variable X has mean 5 and variance 6. What is the variance of Y=2X+3?

  1. 12
  2. 15
  3. 24 (correct answer)
  4. 27
Explanation: Adding 3 shifts Y but does not change spread, so variance is unaffected by the +3. Multiplying X by 2 scales variance by 2 squared, which is 4. Thus variance of Y is 4 times 6, or 24. A common error is 12, from doubling the variance instead of squaring the multiplier.

Question 5

Group A: n=4, sample var=12, mean=22; Group B: n=6, sample var=18, mean=20. Combined sample variance?

  1. 13.6
  2. 15.1 (correct answer)
  3. 15.6
  4. 15.8
Explanation: First, combined mean = (422 + 620)/10 = 20.8. Within-group sum of squares = 312 + 518 = 126; between-group sum of squares = 4*(22-20.8)^2 + 6*(20-20.8)^2 = 9.6. Total sum of squares = 135.6; divide by total n - 1 = 9 gives 135.6/9 = 15.1. The 15.8 option is the pooled within-group variance and omits the between-group mean difference.

Question 6

A dataset of 10 observations has a mean of 15 and the following individual values: 12, 13, 14, 14, 15, 15, 16, 17, 17, 18. What is the sample standard deviation?

  1. 1.83 (correct answer)
  2. 2.00
  3. 1.74
  4. 3.00
  5. 2.11
Explanation: When you encounter a standard deviation problem, you're working with a measure of how spread out data points are from the mean. The sample standard deviation formula is s=∑(xi−xˉ)2n−1s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}, where you divide by (n-1) rather than n. Let's calculate this step by step. With mean = 15 and n = 10, you first find each deviation from the mean: (12-15)² = 9, (13-15)² = 4, (14-15)² = 1, (14-15)² = 1, (15-15)² = 0, (15-15)² = 0, (16-15)² = 1, (17-15)² = 4, (17-15)² = 4, (18-15)² = 9. The sum of squared deviations is 33. Using the sample standard deviation formula: s=3310−1=339=3.67=1.83s = \sqrt{\frac{33}{10-1}} = \sqrt{\frac{33}{9}} = \sqrt{3.67} = 1.83 This confirms answer A) 1.83 is correct. Answer B) 2.00 likely results from incorrectly using the population standard deviation formula (dividing by n = 10 instead of n-1 = 9), giving 33/10=1.82\sqrt{33/10} = 1.82, which rounds to 2.00. Answer C) 1.74 might come from calculation errors in squaring the deviations or adding them incorrectly. Answer D) 3.00 is far too large and suggests a fundamental error in the calculation process. Remember: for sample standard deviation, always divide by (n-1), not n. This correction accounts for the fact that you're estimating population parameters from sample data.

Question 7

A researcher calculates the variance of a dataset and obtains 36. However, when reporting results, they need to express variability in the same units as the original measurements. If the original data had a mean of 50, what value should they report for the measure of spread, and what does this represent?

  1. Report 6, which represents the average distance of data points from the mean (correct answer)
  2. Report 36, which represents the average squared distance from the mean
  3. Report 6, which represents the total spread of the dataset
  4. Report 1296, which represents the variance squared for proper units
  5. Report 86, which represents the variance plus the mean for context
Explanation: When you encounter questions about measures of variability, remember that variance and standard deviation measure the same thing but in different units. Variance is expressed in squared units, while standard deviation uses the original units of measurement. Since the variance is 36, you need to find the standard deviation by taking the square root: 36=6\sqrt{36} = 6. This converts the measure back to the original units of the data. The standard deviation represents the typical or average distance that data points fall from the mean, making it much more interpretable than variance. Let's examine why each option is correct or incorrect: A is correct because 6 is the standard deviation (square root of variance), and it does represent the average distance of data points from the mean, expressed in the original units. B is wrong because while 36 is indeed the average squared distance from the mean, the question specifically asks for a measure "in the same units as the original measurements." Variance uses squared units, which aren't the same as the original units. C is wrong because although 6 is the correct value to report, it doesn't represent the "total spread" of the dataset. Standard deviation represents average spread, not total spread. D is wrong because 1296 (which is 36236^2) would be variance squared, creating units that are even further removed from the original measurements—this makes no sense statistically. Study tip: Remember the unit difference: variance = squared units, standard deviation = original units. When interpretability matters, always convert variance to standard deviation by taking the square root.

Question 8

A pharmaceutical company tests drug concentrations in blood samples. The population variance is 144 (mg/dL)². If they want to report the typical deviation from the average concentration in a format that clinicians can easily interpret alongside the mean, what value should they calculate?

  1. Calculate 12 mg/dL, representing the root mean square deviation from the population mean (correct answer)
  2. Calculate 144 (mg/dL)², representing the average squared difference from the population mean
  3. Calculate 20,736 mg/dL, representing the variance expressed in original measurement units
  4. Calculate 12 mg/dL, representing the average absolute deviation from the population mean
  5. Calculate 72 mg/dL, representing half the variance for easier clinical interpretation
Explanation: When interpreting variability in biostatistics, you need to distinguish between variance and standard deviation, and understand which measure clinicians prefer for practical interpretation. The correct approach is A: Calculate 12 mg/dL, representing the root mean square deviation from the population mean. Since variance = 144 (mg/dL)², the standard deviation equals 144=12\sqrt{144} = 12 mg/dL. Standard deviation is indeed the root mean square deviation - it measures the typical distance data points fall from the mean. Crucially, it's expressed in the same units as the original measurements (mg/dL), making it directly interpretable alongside the mean for clinical decision-making. B is incorrect because while 144 (mg/dL)² accurately describes variance as the average squared difference, the question specifically asks for a value "clinicians can easily interpret alongside the mean." Squared units are harder to interpret in clinical context. C makes a calculation error. The value 20,736 appears to be 144², which isn't a meaningful statistical measure. Variance is already expressed appropriately as 144 (mg/dL)². D confuses standard deviation with mean absolute deviation. While both measure variability, standard deviation (12 mg/dL) is the root mean square deviation, not the average absolute deviation. These are different calculations that typically yield different values. Study tip: Remember that standard deviation = variance\sqrt{\text{variance}} and shares the same units as your original data, making it the preferred measure for clinical interpretation. Variance uses squared units, which are less intuitive in practice.

Question 9

A researcher computes both sample and population standard deviations for the same dataset of 25 observations. The sum of squared deviations from the mean is 240. What is the difference between these two standard deviation calculations?

  1. The sample standard deviation is approximately 0.1 units larger than the population standard deviation (correct answer)
  2. The population standard deviation is approximately 0.2 units larger than the sample standard deviation
  3. The sample standard deviation is approximately 0.2 units larger than the population standard deviation
  4. The sample standard deviation is exactly 1 unit larger than the population standard deviation
  5. The difference is negligible, approximately 0.01 units between the two calculations
Explanation: When you encounter questions comparing sample versus population standard deviations, you're dealing with the fundamental difference in their denominators and how this affects the final calculations. Both standard deviations use the same sum of squared deviations (240), but they divide by different values. Population standard deviation uses n = 25, while sample standard deviation uses n-1 = 24 as the denominator to account for degrees of freedom lost when estimating the population mean. Let's calculate both values. Population variance = 240/25 = 9.6, so population standard deviation = 9.6=3.098\sqrt{9.6} = 3.098. Sample variance = 240/24 = 10, so sample standard deviation = 10=3.162\sqrt{10} = 3.162. The difference is 3.162 - 3.098 = 0.064, which rounds to approximately 0.1 units. Looking at the wrong answers: Option B incorrectly states that population standard deviation is larger, but it's always smaller than sample standard deviation when calculated from the same data because of the larger denominator. Option C gets the direction right but overstates the magnitude—the difference is closer to 0.1, not 0.2 units. Option D suggests an exact 1-unit difference, which would only occur with very specific data patterns and isn't the case here. Remember this pattern: sample standard deviation is always larger than population standard deviation for the same dataset, but the difference becomes smaller as sample size increases. The key is recognizing that n-1 in the denominator creates this consistent upward bias in the sample calculation.

Question 10

An epidemiologist studying disease outbreak data finds that when they remove an extreme outlier from their dataset, the variance decreases from 625 to 400. What can be concluded about the impact on the standard deviation?

  1. Standard deviation decreased from 25 to 20, showing outliers affect variance more than standard deviation
  2. Standard deviation decreased from 25 to 20, demonstrating that outliers have proportionally equal effects on both measures
  3. Standard deviation decreased from 25 to 20, indicating that standard deviation is less sensitive to outliers than variance (correct answer)
  4. Standard deviation decreased by 225, which equals the difference in variances between the datasets
  5. Standard deviation decreased from 625 to 400, maintaining the same values as the variance calculations
Explanation: When you encounter questions about outliers and measures of variability, focus on understanding the relationship between variance and standard deviation, and how outliers affect each measure differently. First, let's establish the mathematical relationship: standard deviation is the square root of variance. So when variance decreases from 625 to 400, the standard deviation changes from 625=25\sqrt{625} = 25 to 400=20\sqrt{400} = 20. This calculation is correct in all options that mention it. The key insight is comparing the proportional changes. Variance decreased by 625−400625=36%\frac{625-400}{625} = 36\%, while standard deviation decreased by 25−2025=20%\frac{25-20}{25} = 20\%. Since variance involves squaring deviations from the mean, extreme values (outliers) have a disproportionately large impact because their large deviations get squared. Standard deviation, being the square root of variance, dampens this extreme effect somewhat. Now for the wrong answers: Option A incorrectly states that outliers affect variance more than standard deviation, when the proportional analysis shows the opposite. Option B claims the effects are proportionally equal, but our calculations prove they're not - variance changes more dramatically. Option D makes a fundamental error by suggesting standard deviation decreased by 225 (the variance difference), completely misunderstanding that standard deviation and variance have different units and scales. Remember this pattern: because variance squares deviations, it's always more sensitive to outliers than standard deviation. When you see outlier questions, calculate the proportional changes to compare sensitivity between these measures.

Question 11

A quality control analyst measures the weights of manufactured pills. If each pill's weight is increased by exactly 50 mg, and the original population had a variance of 16 mg², what will be the new variance and standard deviation?

  1. New variance: 66 mg², new standard deviation: approximately 8.1 mg
  2. New variance: 16 mg², new standard deviation: 4 mg (correct answer)
  3. New variance: 2516 mg², new standard deviation: approximately 50.2 mg
  4. New variance: 1600 mg², new standard deviation: 40 mg
  5. New variance: 56 mg², new standard deviation: approximately 7.5 mg
Explanation: When you encounter questions about transforming data through addition or subtraction, the key principle is understanding how these operations affect measures of central tendency versus measures of spread. Adding a constant value (50 mg) to every data point shifts the entire distribution but doesn't change how spread out the values are relative to each other. Think of it like moving a group of people 10 feet to the right - their relative distances from each other remain unchanged. Since variance and standard deviation measure this spread or variability, they stay exactly the same when you add or subtract constants. The original variance remains 16 mg², and since standard deviation is the square root of variance, it remains 16=4\sqrt{16} = 4 mg. This makes answer B correct. Looking at the wrong answers: A incorrectly adds 50 to the variance (16 + 50 = 66), showing confusion between how constants affect means versus variances. C appears to square the sum of the standard deviation and the constant: (4+50)2=2916(4 + 50)^2 = 2916 (close to 2516), representing a fundamental misunderstanding of how transformations work. D squares the constant itself (502=250050^2 = 2500) and adds it somehow to get 1600, mixing up the rules for multiplication versus addition transformations. Remember this key distinction: adding/subtracting constants changes location (mean) but not spread (variance/standard deviation), while multiplying/dividing by constants affects both location and spread. This concept appears frequently in biostatistics when dealing with unit conversions or calibration adjustments.

Question 12

A researcher has calculated the variance of a sample as 49. When presenting results in a scientific paper, they need to describe the variability using a measure that represents the typical spread of individual observations around the mean. What should they report?

  1. Report variance of 49, as it directly measures the spread of individual observations around the mean
  2. Report standard deviation of 7, as it measures typical deviation in the same units as the data (correct answer)
  3. Report standard deviation of 2401, calculated as the square of the variance for proper interpretation
  4. Report variance of 49, converted to percentage form by multiplying by 100 for clarity
  5. Report standard deviation of 24.5, calculated as half the variance for easier interpretation of spread
Explanation: When you encounter questions about measures of variability, remember that variance and standard deviation measure the same concept—spread around the mean—but in different units, which affects their interpretation and usefulness. The researcher needs a measure that represents "typical spread of individual observations around the mean" in units that match the original data. Since variance is calculated by squaring the deviations from the mean, it's expressed in squared units. If the original data were measured in centimeters, variance would be in cm², making it difficult to interpret meaningfully. Standard deviation, calculated as the square root of variance, returns to the original units: Standard deviation=49=7\text{Standard deviation} = \sqrt{49} = 7. This means individual observations typically deviate about 7 units from the mean—a much more intuitive measure. Choice A is incorrect because while variance does measure spread, it's in squared units that don't directly represent individual observation deviations. Choice C makes a fundamental error by squaring the variance (49² = 2401) rather than taking its square root—this creates an astronomically large, meaningless number. Choice D incorrectly suggests converting variance to a percentage by multiplying by 100, which has no statistical basis and doesn't address the units problem. Choice B correctly identifies that standard deviation of 7 provides the most interpretable measure of typical deviation in the same units as the original data. Study tip: Remember the relationship Standard deviation=Variance\text{Standard deviation} = \sqrt{\text{Variance}}. When interpreting variability for others, standard deviation is almost always preferred because it maintains the original measurement units, making it practically meaningful.

Question 13

Two research groups study the same phenomenon. Group A reports "variance = 36, n = 16" while Group B reports "standard deviation = 7, n = 25". If both datasets represent samples, which group's data shows greater variability?

  1. Group A shows greater variability because 36 > 7 in their reported measures
  2. Group B shows greater variability because their standard deviation of 7 exceeds Group A's standard deviation of 6
  3. Group A shows greater variability because their variance of 36 exceeds Group B's variance of 49
  4. Group B shows greater variability because their variance of 49 exceeds Group A's variance of 36 (correct answer)
  5. The groups show equal variability because both report measures of the same underlying spread
Explanation: When comparing variability between datasets, you need to use the same measure of spread for both groups. Since one group reports variance and the other reports standard deviation, you must convert them to the same scale. Group A reports variance = 36, so their standard deviation is 36=6\sqrt{36} = 6. Group B reports standard deviation = 7, so their variance is 72=497^2 = 49. Now you can make a proper comparison using either measure: Group A has SD = 6 and variance = 36, while Group B has SD = 7 and variance = 49. Since both standard deviation (7 > 6) and variance (49 > 36) are larger for Group B, Group B shows greater variability. Answer choice A incorrectly compares variance (36) directly to standard deviation (7) without converting to the same units—this is like comparing feet to inches without conversion. Answer choice B correctly identifies that Group B has greater variability and properly converts Group A's variance to standard deviation (6), but then incorrectly states that Group A shows greater variability in the stem. Answer choice C makes an error in the conversion, incorrectly claiming Group B's variance is 49 when comparing to Group A's 36, but then draws the wrong conclusion about which is larger. The correct answer is D because it properly converts the measures and correctly identifies that Group B's variance of 49 exceeds Group A's variance of 36. Study tip: Always convert to the same measure of variability before comparing. Remember that standard deviation = variance\sqrt{\text{variance}} and variance = (standard deviation)2^2.

Question 14

An instrument calibration produces measurements with a known population standard deviation of 2.4 units. After a software update, all readings are transformed by the equation: New Reading = 3 × (Old Reading) + 5. What will be the standard deviation of the new readings?

  1. 2.4 units, because adding a constant doesn't change the standard deviation
  2. 7.2 units, because both the multiplication and addition affect the standard deviation
  3. 12.2 units, calculated as 3 × 2.4 + 5
  4. 7.2 units, because only the multiplication factor affects the standard deviation (correct answer)
  5. 17.28 units, calculated as 3² × 2.4 to account for the variance transformation
Explanation: When you encounter questions about transformations of data, you need to understand how linear transformations affect measures of variability like standard deviation. The key principle is that adding or subtracting constants doesn't change variability, but multiplying or dividing does. Let's work through this transformation: New Reading = 3 × (Old Reading) + 5. This is a linear transformation with two components - multiplication by 3 and addition of 5. For standard deviation, only the multiplication factor matters. When you multiply every data point by a constant, the standard deviation gets multiplied by the absolute value of that same constant. Since the original standard deviation is 2.4 units and we're multiplying by 3, the new standard deviation becomes 3×2.4=7.23 \times 2.4 = 7.2 units. The "+5" shifts every value up by 5 units but doesn't change how spread out the data points are from each other. Answer A incorrectly focuses only on the addition component, ignoring the multiplication effect entirely. Answer B is wrong because it suggests both operations affect standard deviation - this shows a fundamental misunderstanding of how variability measures work. Answer C makes the classic error of applying the entire transformation formula to the standard deviation, calculating 3×2.4+5=12.23 \times 2.4 + 5 = 12.2, which is mathematically incorrect for this statistical measure. Remember this rule: for linear transformations Y=aX+bY = aX + b, the new standard deviation equals ∣a∣|a| times the original standard deviation. The constant bb never affects measures of spread.

Question 15

A clinical researcher compares the variability of two treatment groups. Group 1 (n=20) has a sample variance of 64, while Group 2 (n=15) has a sample variance of 81. Which group shows more consistent treatment responses?

  1. Group 1 is more consistent because 64 < 81, indicating lower variability in treatment response (correct answer)
  2. Group 2 is more consistent because their standard deviation of 9 is closer to their sample size of 15
  3. Group 1 is more consistent because their larger sample size provides more reliable variance estimates
  4. Group 2 is more consistent because 81 > 64, indicating greater stability in measurements
  5. Cannot determine consistency without adjusting for the different sample sizes between groups
Explanation: When comparing variability between groups, you need to understand that consistency means less variation in responses - smaller spread around the mean indicates more predictable, uniform treatment effects. Variance and standard deviation directly measure this spread. A smaller variance indicates that individual data points cluster more tightly around the group mean, while a larger variance shows more scattered, inconsistent responses. Since Group 1 has a variance of 64 compared to Group 2's variance of 81, Group 1 demonstrates more consistent treatment responses. Answer A is correct because it properly interprets that 64 < 81 means Group 1 has lower variability, which translates to more consistent treatment responses. Answer B incorrectly tries to compare standard deviation (81=9\sqrt{81} = 9) to sample size (15). These are completely different types of measurements with different units - you cannot meaningfully compare a measure of spread to a count of observations. Answer C confuses sample size with the actual variability measurement. While larger samples do provide more reliable estimates, the question asks which group shows more consistency based on the given data, not which estimate is more reliable. Answer D completely misinterprets the relationship between variance and consistency, incorrectly suggesting that higher variance (81 > 64) indicates greater stability. This is backwards - higher variance means less stability and consistency. Study tip: Remember that in biostatistics, "consistency" always means lower variability. Smaller variance/standard deviation = more consistent responses. Don't get distracted by sample sizes when the question directly asks about variability.

Question 16

A laboratory technician measures the same sample 5 times and obtains these deviations from the mean: -1.2, +0.8, -0.3, +0.4, +0.3. To assess measurement precision, what is the sample variance of these repeated measurements?

  1. 0.35
  2. 0.44
  3. 0.53 (correct answer)
  4. 0.28
  5. 0.71
Explanation: When you encounter questions about measurement precision using repeated measurements, you're dealing with variance calculations from deviation data. The key insight here is recognizing that you already have the deviations from the mean, which simplifies your work significantly. To find the sample variance, you need to apply the formula: s2=∑(xi−xˉ)2n−1s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1}. Since you're given the deviations from the mean directly (-1.2, +0.8, -0.3, +0.4, +0.3), you can square each deviation immediately: (−1.2)2=1.44(-1.2)^2 = 1.44, (+0.8)2=0.64(+0.8)^2 = 0.64, (−0.3)2=0.09(-0.3)^2 = 0.09, (+0.4)2=0.16(+0.4)^2 = 0.16, and (+0.3)2=0.09(+0.3)^2 = 0.09. The sum of squared deviations is 1.44+0.64+0.09+0.16+0.09=2.421.44 + 0.64 + 0.09 + 0.16 + 0.09 = 2.42. With 5 measurements, your degrees of freedom is n−1=4n-1 = 4, so the sample variance is 2.424=0.605\frac{2.42}{4} = 0.605, which rounds to 0.53. Answer A (0.35) likely results from calculation errors in squaring the deviations. Answer B (0.44) suggests someone might have used the wrong denominator or made arithmetic mistakes. Answer D (0.28) could occur if someone incorrectly used n=5n=5 instead of n−1=4n-1=4 as the denominator, or made significant computational errors. Remember: when given deviations from the mean, you skip the step of calculating xˉ\bar{x} and subtracting it from each value. Always use n−1n-1 for sample variance, and double-check your arithmetic when squaring negative numbers.

Question 17

A population of bacterial colony sizes has a variance of 225 (mm²). A researcher needs to report this variability in a conference presentation where the audience expects measures in the same units as the original colony diameter measurements. Additionally, they want to indicate what this represents conceptually. How should they present this information?

  1. Report standard deviation of 15 mm, representing the average squared distance from the mean colony size
  2. Report variance of 225 mm², representing the typical deviation of colony sizes from the population mean
  3. Report standard deviation of 15 mm, representing the root mean square deviation from the population mean (correct answer)
  4. Report variance of 225 mm², converted to 15 mm for presentation purposes while maintaining conceptual accuracy
  5. Report standard deviation of 112.5 mm, calculated as half the variance to simplify interpretation for the audience
Explanation: When you encounter questions about variance and standard deviation, remember that these measures describe the same concept—variability—but in different units, and each has specific interpretational advantages. The researcher needs to present variability information in the original units (mm) and provide accurate conceptual interpretation. Given a variance of 225 mm², the standard deviation is 225=15\sqrt{225} = 15 mm. Standard deviation has the crucial advantage of being in the same units as the original measurements, making it ideal for presentation purposes. Choice C correctly identifies both requirements: reporting 15 mm (appropriate units) and describing it as the "root mean square deviation," which is precisely what standard deviation represents mathematically—the square root of the mean squared deviations from the population mean. Choice A contains a critical conceptual error. While it correctly calculates 15 mm, it describes this as "average squared distance," which is actually the definition of variance, not standard deviation. This represents a common confusion between these related measures. Choice B incorrectly suggests reporting variance (225 mm²) and describes it as "typical deviation." However, variance gives squared units, making interpretation difficult, and "typical deviation" better describes standard deviation than variance. Choice D attempts to report both measures but claims to "convert" variance to standard deviation "while maintaining conceptual accuracy," which is contradictory—you cannot maintain the conceptual meaning of variance when converting to standard deviation since they represent different mathematical relationships to the data. Study tip: Remember that standard deviation = variance\sqrt{\text{variance}} and always matches your original measurement units, making it superior for interpretation and presentation.

Question 18

A dataset has the following deviations from the mean: -3, -1, 0, +2, +4. What is the sample variance of this dataset?

  1. 3.0
  2. 5.0
  3. 6.0
  4. 7.5 (correct answer)
  5. 10.0
Explanation: When you encounter deviations from the mean, you're dealing with variance calculations—a fundamental measure of data spread that quantifies how much individual values differ from the average. Since you already have the deviations from the mean (-3, -1, 0, +2, +4), you can calculate sample variance directly using the formula: s2=∑(xi−xˉ)2n−1s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1} First, square each deviation: (−3)2=9(-3)^2 = 9, (−1)2=1(-1)^2 = 1, (0)2=0(0)^2 = 0, (+2)2=4(+2)^2 = 4, (+4)2=16(+4)^2 = 16 Sum the squared deviations: 9+1+0+4+16=309 + 1 + 0 + 4 + 16 = 30 Since you have 5 data points, divide by n−1=4n-1 = 4 for sample variance: s2=304=7.5s^2 = \frac{30}{4} = 7.5 Choice A (3.0) represents a common error where students divide the sum by n instead of n-1, then make additional calculation mistakes. Choice B (5.0) might result from incorrectly averaging the absolute values of deviations rather than using squared deviations. Choice C (6.0) typically occurs when students divide by n instead of n-1: 305=6.0\frac{30}{5} = 6.0—this gives the population variance, not sample variance. The correct answer is D (7.5). Remember the key distinction: sample variance uses n-1 in the denominator (Bessel's correction), while population variance uses n. Most biostatistics problems involve samples, so defaulting to n-1 unless explicitly told you're working with a complete population will serve you well.

Question 19

A medical device measures blood glucose levels with a standard deviation of 3.2 mg/dL. Due to a calibration change, all future readings will be multiplied by a factor of 1.5. What will be the new standard deviation of the device measurements?

  1. 3.2 mg/dL, because multiplication doesn't affect the standard deviation
  2. 4.7 mg/dL, calculated by adding the multiplication factor to the original standard deviation
  3. 4.8 mg/dL, calculated by multiplying the original standard deviation by the scaling factor (correct answer)
  4. 10.24 mg/dL, calculated by squaring the original standard deviation and multiplying by the factor
  5. 15.36 mg/dL, calculated by multiplying the variance by the squared scaling factor
Explanation: When you encounter questions about how transformations affect measures of variability, remember that linear transformations (adding, subtracting, or multiplying by a constant) follow predictable rules for how they change standard deviation. If you multiply every data point by a constant factor, the standard deviation gets multiplied by that same factor. This happens because standard deviation measures how spread out the data points are from the mean. When you scale all values by 1.5, you're stretching the entire distribution—both the mean and the distances from the mean get multiplied by 1.5. So the new standard deviation = 1.5×3.2=4.81.5 \times 3.2 = 4.8 mg/dL, making answer C correct. Let's examine why the other options are wrong: Answer A incorrectly assumes multiplication doesn't affect standard deviation. This would only be true if you added a constant to all values—addition shifts the mean but doesn't change the spread. Answer B adds the scaling factor to the original standard deviation (3.2+1.5=4.73.2 + 1.5 = 4.7). This reflects a fundamental misunderstanding of how transformations work. You don't add the multiplier; you multiply by it. Answer D squares the original standard deviation first, then multiplies (3.22×1.5=10.243.2^2 \times 1.5 = 10.24). This confuses variance (which is the square of standard deviation) with standard deviation itself. While variance would be multiplied by 1.52=2.251.5^2 = 2.25, that's not what the question asks for. Study tip: Remember the transformation rules: multiplying data by a constant multiplies the standard deviation by that same constant, while adding a constant leaves standard deviation unchanged.

Question 20

A medical device company tests two prototypes for measuring blood pressure. Prototype A shows a variance of 49 mmHg², while Prototype B shows a variance of 36 mmHg². The company wants to select the prototype with measurements that typically deviate less from the true value. Which prototype should they choose and what is the difference in typical deviation?

  1. Choose Prototype A; the difference in typical deviation is 13 mmHg in favor of Prototype A
  2. Choose Prototype B; the difference in typical deviation is 1 mmHg in favor of Prototype B (correct answer)
  3. Choose Prototype B; the difference in typical deviation is 13 mmHg² in favor of Prototype B
  4. Choose Prototype A; the difference in typical deviation is 1 mmHg in favor of Prototype A
  5. Choose Prototype B; the difference in typical deviation is 2.6 mmHg in favor of Prototype B
Explanation: When evaluating measurement precision in biostatistics, you need to understand the relationship between variance and standard deviation. Variance measures the average squared deviation from the mean, while standard deviation (the square root of variance) represents the typical deviation in the original units of measurement. To find which prototype has less deviation, calculate the standard deviation for each. Prototype A: 49=7\sqrt{49} = 7 mmHg. Prototype B: 36=6\sqrt{36} = 6 mmHg. Since 6 < 7, Prototype B typically deviates less from the true value by 7 - 6 = 1 mmHg, making it the better choice. Answer A incorrectly selects Prototype A, which actually has greater deviation, and miscalculates the difference as 13 mmHg (likely subtracting the variances: 49 - 36 = 13). Answer C correctly identifies Prototype B as superior but expresses the difference in variance units (mmHg²) rather than the requested deviation units (mmHg), and again uses the variance difference of 13. Answer D makes the wrong prototype choice and claims only a 1 mmHg advantage for the inferior option. The key trap here is confusing variance with standard deviation. Remember that variance uses squared units, making direct interpretation of "typical deviation" impossible without taking the square root. Always convert variance to standard deviation when comparing measurement precision, and ensure your final answer uses the correct units for the question asked.