GMAT Quiz: Interpret Standard Deviation

What this quiz covers

This quiz focuses on Interpret Standard Deviation, giving you a quick way to practice the rules, question types, and explanations that matter most for GMAT.

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.

Question 1

1. Modified Dataset P has higher relative variability since its coefficient of variation is approximately 0.23 compared to Q's 0.10 (correct answer)
2. Modified Dataset P has lower relative variability since its coefficient of variation is approximately 0.19 compared to Q's 0.25
3. The datasets have similar relative variability since both coefficients of variation are approximately 0.15
4. Relative variability cannot be determined without knowing the exact new mean of modified Dataset P

Explanation: To compare relative variability, we use the coefficient of variation (CV = standard deviation/mean). For Dataset Q: CV = 6/60 = 0.10. For modified Dataset P, the new mean = (20×50 + 120)/21 = 1120/21 ≈ 53.3. The CV = 12/53.3 ≈ 0.23. Therefore, modified Dataset P has higher relative variability. Choice B uses incorrect CV calculations. Choice C incorrectly suggests they're similar. Choice D is wrong because we can calculate the new mean from the given information.

Question 2

1. The new standard deviation is approximately 0.6 mm, representing a 25% improvement in precision over the original process (correct answer)
2. The new standard deviation is approximately 0.6 mm, representing a 33% improvement in precision over the original process
3. The new standard deviation is approximately 0.3 mm, representing a 63% improvement in precision over the original process
4. Insufficient information is provided to determine the new standard deviation since the sample size affects the distribution

Explanation: In a normal distribution, approximately 68% of values fall within one standard deviation of the mean. Since 68% of bolts fall within 0.6 mm of the mean, the new standard deviation is approximately 0.6 mm. The improvement is (0.8 - 0.6)/0.8 = 0.2/0.8 = 25%. Choice B uses the correct standard deviation but incorrectly calculates improvement as (0.8 - 0.6)/0.6 ≈ 33%. Choice C misinterprets the ±0.6 mm range as being two standard deviations. Choice D is incorrect because the 68% rule directly gives us the standard deviation regardless of sample size.

Question 3

1. Class A's new standard deviation will be approximately 9.6 points and Class B's will be approximately 6.4 points
2. Both classes will have approximately the same standard deviation after trimming the extreme values
3. Class A's new standard deviation will decrease proportionally more than Class B's standard deviation (correct answer)
4. The ratio of Class A's standard deviation to Class B's will remain approximately 1.5

Explanation: Removing extreme values (outliers) reduces standard deviation, but the effect is more pronounced for datasets that initially had higher variability. Class A, starting with higher standard deviation (12 vs 8), likely had more extreme values contributing to that variability. Removing the same percentage of extreme values will cause a larger proportional decrease in Class A's standard deviation. Choice A assumes a fixed proportional decrease (20%) which isn't necessarily accurate. Choice B is incorrect because trimming doesn't equalize standard deviations. Choice D incorrectly assumes the ratio remains constant after trimming.

Question 4

1. The conclusion is definitely correct since City B has a higher standard deviation, which directly measures inequality
2. The conclusion is likely incorrect because standard deviation can be influenced by extreme values and may not reflect typical income differences (correct answer)
3. The conclusion cannot be evaluated without knowing the mean incomes of both cities to calculate relative variability
4. The conclusion is correct only if both cities have approximately normal income distributions with similar sample sizes

Explanation: Standard deviation measures variability but can be heavily influenced by outliers. Income distributions are typically right-skewed with high earners creating large standard deviations without necessarily indicating greater inequality among typical households. Since both cities have the same median, City B's higher standard deviation might reflect a few very high earners rather than broader inequality. Choice A oversimplifies the relationship between standard deviation and inequality. Choice C is partially relevant but doesn't address the main issue of outlier sensitivity. Choice D incorrectly suggests the conclusion is conditionally correct when the fundamental issue is outlier sensitivity.

Question 5

1. Machine 2 is clearly superior because lower standard deviation indicates better precision, and the 0.03 mm bias can be easily corrected
2. Machine 1 is superior because accuracy (hitting the target mean) is more important than precision (low standard deviation) in manufacturing
3. Both machines are equivalent in quality since Machine 1's better accuracy compensates for Machine 2's better precision in manufacturing contexts
4. The evaluation depends on whether the tolerance specifications prioritize accuracy or precision, as Machine 1 is more accurate while Machine 2 is more precise (correct answer)

Explanation: When you encounter questions about manufacturing quality control, you need to distinguish between accuracy (how close measurements are to the target) and precision (how consistent measurements are). These are independent concepts that both matter in manufacturing. Machine 1 hits the target diameter of 10.00 mm perfectly but has higher variability (0.05 mm standard deviation). Machine 2 is more consistent (0.03 mm standard deviation) but systematically misses the target by 0.03 mm. Neither machine is objectively "better" without knowing the specific requirements. The correct answer is D because the relative importance of accuracy versus precision depends entirely on the application's tolerance specifications. Some manufacturing processes can tolerate systematic bias but require tight consistency, while others need perfect centering but can handle more variation. Without knowing these specifications, you cannot definitively rank the machines. A is wrong because it assumes precision automatically trumps accuracy and that bias correction is always feasible or cost-effective. B makes the opposite error, assuming accuracy is inherently more important than precision. C incorrectly suggests these qualities can simply "compensate" for each other when they serve different manufacturing needs. The key trap here is thinking one aspect of quality (accuracy or precision) is universally more important than the other. On quantitative reasoning questions involving manufacturing or quality control, look for answer choices that acknowledge context-dependent trade-offs rather than making absolute judgments about which metric matters most.

Question 6

1. The claim about effectiveness, because the difference in means might not be statistically significant given the sample sizes
2. The claim about variability, because higher standard deviation in the treatment group could indicate better responsiveness in some patients rather than just inconsistency (correct answer)
3. Both claims equally, because the standard deviations overlap significantly and the means are too close to draw conclusions
4. Neither claim, because the sample sizes are equal and the differences in both mean and standard deviation are substantial

Explanation: The effectiveness claim (higher mean) is straightforward to evaluate statistically. However, interpreting higher variability requires careful consideration: it could indicate inconsistent treatment effects (negative interpretation) or that the treatment helps some patients much more than others (potentially positive interpretation). The increase from 8 to 12 in standard deviation could reflect a beneficial increase in the upper tail of responses. Choice A focuses on statistical significance but misses the interpretive complexity of variability. Choice C incorrectly suggests the differences aren't meaningful. Choice D dismisses the need for careful evaluation.

Question 7

1. The combined standard deviation will be exactly 2.1 grams, the average of the two individual standard deviations
2. The combined standard deviation will be approximately 1.5 grams, calculated using the formula for combined variances
3. The combined standard deviation could be anywhere from 0.6 grams to 4.2 grams, depending on the relationship between the datasets (correct answer)
4. The combined standard deviation will be exactly 3.0 grams, calculated by taking the square root of the sum of variances

Explanation: The standard deviation of combined datasets depends on the correlation between them and the difference in their means. The minimum possible value occurs when the datasets are perfectly negatively correlated and have the same mean (|2.4 - 1.8| = 0.6). The maximum occurs when they are perfectly positively correlated or have very different means (up to 2.4 + 1.8 = 4.2). Choice A incorrectly assumes you can average standard deviations. Choice B miscalculates the combined variance formula. Choice D incorrectly applies the formula √(σ₁² + σ₂²) which only works for independent variables with the same mean.

Question 8

1. The combined portfolio's standard deviation will be exactly 15%, the average of the two individual standard deviations
2. The combined portfolio's standard deviation will be approximately 15.3%, assuming the portfolios are uncorrelated
3. The combined portfolio's standard deviation could range from 3% to 15%, depending on the correlation between the portfolios (correct answer)
4. The combined portfolio's standard deviation will be approximately 21.6%, calculated using the formula for independent risks

Explanation: For a portfolio with equal weights (0.5 each), the standard deviation formula is σₚ = √[w₁²σ₁² + w₂²σ₂² + 2w₁w₂σ₁σ₂ρ]. With perfect negative correlation (ρ = -1): σₚ = |0.5×12% - 0.5×18%| = 3%. With perfect positive correlation (ρ = +1): σₚ = 0.5×12% + 0.5×18% = 15%. The actual value depends on correlation. Choice A incorrectly averages standard deviations. Choice B assumes zero correlation: √[0.25×144 + 0.25×324] = 15.3%. Choice D incorrectly uses √[12² + 18²] = 21.6%.

Question 9

1. Company A's standard deviation will be $28,500 and Company B's will be $9,500
2. Company A's standard deviation will be $27,500 and Company B's will be $8,500
3. Company A's standard deviation will be $16,500 and Company B's will be $5,500 (correct answer)
4. Both companies will have the same standard deviation after the adjustments

Explanation: When a constant is added to all values in a dataset, the standard deviation remains unchanged. When all values are multiplied by a constant, the standard deviation is multiplied by that same constant. Here: (1) Add $10,000 - no change to standard deviation, (2) Multiply by 1.10 - standard deviation multiplies by 1.10. Company A: $15,000 × 1.10 = $16,500. Company B: $5,000 × 1.10 = $5,500. Choice A incorrectly adds the $10,000 to the standard deviation and then multiplies. Choice B incorrectly adds $10,000 then multiplies by 1.10. Choice D ignores that different initial standard deviations remain proportionally different.

Question 10

1. It will be between 6 and 9, closer to 7 since Class 2 has more students and lower variability
2. It will be between 6 and 9, but could be closer to either value depending on the distribution of individual scores
3. It will be approximately 7.4, calculated as the weighted average of the two standard deviations based on class sizes
4. It will likely be greater than 9 due to the difference in means between the classes creating additional variability (correct answer)

Explanation: When combining data from multiple groups, you need to consider not just the variability within each group, but also how the groups differ from each other. This question tests whether you understand that pooled standard deviation can exceed the individual group standard deviations when group means are sufficiently different. The key insight is that when you combine classes with different means (78 vs. 82), you're introducing additional variability beyond what exists within each class. The combined dataset will include scores clustered around 78 AND scores clustered around 82, creating a more spread-out distribution than either class alone. This extra variability from the mean differences can push the pooled standard deviation above the higher individual standard deviation of 9. Choice A incorrectly assumes the pooled standard deviation must fall between the individual values, ignoring the impact of different group means. Choice B makes the same fundamental error, suggesting the result depends only on score distributions rather than recognizing that different means automatically create additional spread. Choice C represents a common misconception—you cannot calculate pooled standard deviation by simply taking a weighted average of individual standard deviations. This approach completely ignores the between-group variability. Choice D correctly recognizes that the 4-point difference in means (78 vs. 82) will likely contribute enough additional variability to push the combined standard deviation above 9, despite Class 2's lower individual variability. Study tip: Remember that pooled standard deviation isn't bounded by the individual group standard deviations when group means differ significantly. Always consider both within-group and between-group sources of variability.

Question 11

1. Classroom 1, because its scores cluster around the middle values. (correct answer)
2. Classroom 2, because it has more scores at the extremes.
3. Both classrooms, since they have equal sample sizes of 25 students.
4. Cannot be determined from the bar chart information alone.

Explanation: Classroom 1's bars show a pronounced peak at the center (scores 6-7) with few extreme values, indicating low variability. Classroom 2 shows a flatter distribution spread across scores 2-10, including many low and high scores, indicating higher variability. Therefore Classroom 1 has the smaller standard deviation. B – More extreme scores increase, not decrease, standard deviation. C – Equal sample sizes do not determine equal standard deviations. D – Distribution shape in bar charts does indicate relative variability.

Question 12

1. Warehouse X, because its weights are distributed over a wider interval. (correct answer)
2. Warehouse Y, because its tallest bar is at the center of the distribution.
3. Both warehouses must have exactly the same standard deviation.
4. Neither; without computing numerical values, no comparison is possible.

Explanation: Warehouse X's histogram covers a spread from roughly 10 kg to 50 kg, while Warehouse Y's data are tightly clustered between about 20 kg and 30 kg. The broader range and more even spread of X imply a larger standard deviation. B – A central peak alone does not imply greater variability. C – The differing spreads contradict identical standard deviations. D – Visual evidence about dispersion is usually sufficient to compare variability qualitatively.

Question 13

1. Company A, because its revenue line fluctuates less from the horizontal trend. (correct answer)
2. Company B, because it earned more total revenue overall.
3. The two companies have equal standard deviations, since each has 12 data points.
4. There is insufficient information; the mean revenues are not provided.

Explanation: Visually, Company A's revenue line is nearly flat, staying within about $3 million of its mean, whereas Company B swings almost $10 million above and below its average. Less fluctuation translates to a smaller standard deviation for Company A. B – Total revenue does not determine variability. C – Having the same number of observations says nothing about their spread. D – Even without the exact mean, relative variability is evident from the plot.

Question 14

1. Data Set R has the larger standard deviation, because its interquartile range is wider. (correct answer)
2. Data Set S has the larger standard deviation, because its median exceeds that of R.
3. The two standard deviations are equal, since the means coincide.
4. No relationship between the standard deviations can be inferred from the plots.

Explanation: With identical means, variability decides the standard deviation. Data Set R's box spans a wider interval and its whiskers extend farther, indicating more dispersed data and therefore a larger standard deviation. B – A higher median does not guarantee greater spread. C – Equal means do not entail equal variability. D – The visual evidence of spread is sufficient for comparison.

Question 15

1. Department M, because its monthly shares appear more evenly distributed. (correct answer)
2. Department N, because its largest slice exceeds 35 percent of the pie.
3. Both departments necessarily have the same standard deviation, as each pie totals 100 percent.
4. Neither; a pie chart never offers clues about variability over time.

Explanation: Department M's pie shows 12 wedges of nearly equal size (around 8–10 percent each), implying monthly expenses are consistent. Department N has one dominant slice and several very small ones, signaling larger month-to-month fluctuations; hence its standard deviation is higher and Department M's is lower. B – A large single slice indicates greater, not smaller, variation. C – Percent totals do not dictate variability. D – The distribution of slice sizes reveals consistency.