Business Statistics Quiz: Coefficient Of Variation
20 questions · exam conditions
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Coefficient Of VariationQuestion 1 of 20

A company's production line has a mean output of 200 units per hour with a variance of 225 units². To account for a fixed setup waste, a data analyst subtracts 10 units from each hour's recorded output. What is the coefficient of variation of this adjusted data?

7.50%
7.89%
2.63%
1.18%
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Business Statistics Quiz

Business Statistics Quiz: Coefficient Of Variation

Practice Coefficient Of Variation in Business Statistics 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 Coefficient Of Variation, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Statistics.

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

A company's production line has a mean output of 200 units per hour with a variance of 225 units². To account for a fixed setup waste, a data analyst subtracts 10 units from each hour's recorded output. What is the coefficient of variation of this adjusted data?

  1. 7.50%
  2. 7.89% (correct answer)
  3. 2.63%
  4. 1.18%
Explanation: First, find the original standard deviation from the variance: s=225=15s = \sqrt{225} = 15 units. The original mean is xˉ=200\bar{x} = 200 units. When a constant (10) is subtracted from every data point, the mean is reduced by that constant, but the standard deviation remains unchanged. The new mean is 200−10=190200 - 10 = 190. The standard deviation is still 15. The new coefficient of variation is (15 / 190) ≈ 0.0789, or 7.89%.

Question 2

An analyst compares two investment portfolios. Portfolio A has an expected annual return of 8% with a standard deviation of 12%. Portfolio B has an expected annual return of 15% with a standard deviation of 21%. Which statement accurately compares the relative risk of the two portfolios?

  1. Portfolio B has higher relative risk because its standard deviation is higher.
  2. Portfolio A has higher relative risk than Portfolio B. (correct answer)
  3. Portfolio B has higher relative risk than Portfolio A.
  4. Both portfolios have the same relative risk.
Explanation: To compare relative risk, especially when means are different, the coefficient of variation (CV) should be used. CV = Standard Deviation / Mean. For Portfolio A, CV_A = 12% / 8% = 1.5. For Portfolio B, CV_B = 21% / 15% = 1.4. Since CV_A > CV_B, Portfolio A has a higher level of risk per unit of return, meaning it has higher relative risk.

Question 3

The coefficient of variation for the price of a certain commodity is reported as 25%. If the variance of the price is $16, what is the mean price of the commodity?

  1. $1.00
  2. $4.00
  3. $16.00 (correct answer)
  4. $64.00
Explanation: The coefficient of variation (CV) is s/xˉs / \bar{x}. We are given CV = 0.25 and variance s2=16s^2 = 16. First, find the standard deviation, s = \sqrt{16} = \4.Now,rearrangetheCVformulatosolveforthemean:. Now, rearrange the CV formula to solve for the mean: \bar{x} = s / \text{CV}.Therefore,themeanpriceis. Therefore, the mean price is $4 / 0.25 = $16.00$. Distractor D ($64) results from incorrectly using the variance instead of the standard deviation in the calculation.

Question 4

A quality control manager wants to compare the consistency of two different manufacturing processes. Process X produces bolts with a mean length of 50 mm and a standard deviation of 0.1 mm. Process Y produces bearings with a mean weight of 200 g and a standard deviation of 0.5 g. Which of the following is the most appropriate conclusion?

  1. Process Y is less consistent because its standard deviation of 0.5 is larger than 0.1.
  2. The processes' consistency cannot be compared because their units of measurement (mm and g) are different.
  3. Process X is relatively more consistent than Process Y. (correct answer)
  4. Process Y is relatively more consistent than Process X.
Explanation: The coefficient of variation (CV) is used to compare the relative variability of datasets with different units or different means. For Process X, CV_X = (0.1 mm / 50 mm) = 0.002. For Process Y, CV_Y = (0.5 g / 200 g) = 0.0025. Since a lower CV indicates greater consistency (less relative variability), Process X (CV=0.002) is relatively more consistent than Process Y (CV=0.0025).

Question 5

An investment analyst is evaluating three portfolios over the past year. Portfolio X had returns with a mean of 8.2% and coefficient of variation of 15%. Portfolio Y had a standard deviation of 2.1% and coefficient of variation of 35%. Portfolio Z had a mean return of 12.0% and standard deviation of 1.8%. Which portfolio exhibits the highest relative risk, and what is its coefficient of variation?

  1. Portfolio Y exhibits the highest relative risk with a coefficient of variation of 35% (correct answer)
  2. Portfolio Z exhibits the highest relative risk with a coefficient of variation of 15%
  3. Portfolio X exhibits the highest relative risk with a coefficient of variation of 15%
  4. Portfolio Z exhibits the highest relative risk with a coefficient of variation of 35%
Explanation: Portfolio X has CV = 15% (given). Portfolio Y has CV = 35% (given). For Portfolio Z: mean = 12.0%, std dev = 1.8%, so CV = (1.8/12.0) × 100% = 15%. Portfolio Y has the highest coefficient of variation at 35%, indicating the highest relative risk. Choice A is correct. Choice B incorrectly identifies Portfolio Z and gives the wrong CV. Choice C incorrectly identifies Portfolio X. Choice D incorrectly identifies Portfolio Z but gives the correct highest CV value.

Question 6

A pharmaceutical company is testing the consistency of drug concentration in two different tablet formulations. Formulation A has a coefficient of variation of 2.8% and a mean concentration of 250 mg. Formulation B has a standard deviation of 4.2 mg and a coefficient of variation of 1.4%. If the company wants to minimize absolute variability while maintaining acceptable relative variability (CV ≤ 3%), which formulation should they choose and what is the standard deviation of the recommended formulation?

  1. Formulation A should be chosen with a standard deviation of 7.0 mg
  2. Formulation B should be chosen with a standard deviation of 4.2 mg (correct answer)
  3. Formulation A should be chosen with a standard deviation of 4.2 mg
  4. Formulation B should be chosen with a standard deviation of 7.0 mg
Explanation: First, both formulations meet the CV ≤ 3% requirement (A: 2.8%, B: 1.4%). For Formulation A: std dev = CV × mean / 100 = 2.8 × 250 / 100 = 7.0 mg. For Formulation B: std dev = 4.2 mg (given). Since the goal is to minimize absolute variability, Formulation B should be chosen (4.2 mg < 7.0 mg). Choice B is correct. Choice A correctly calculates A's std dev but chooses the wrong formulation. Choices C and D mix up the standard deviations between formulations.

Question 7

An agricultural researcher is studying crop yields from experimental plots. Plot A yielded grain with mean weight per plant of 850 grams and coefficient of variation of 12%. Plot B had a standard deviation of 95 grams and coefficient of variation of 11%. Plot C showed mean yield of 920 grams with standard deviation of 138 grams. If the researcher needs to select two plots for the next phase of study based on having both above-average mean yield (compared to all three plots) and below-average coefficient of variation, which plots qualify and what is their average coefficient of variation?

  1. Plots B and C qualify; their average coefficient of variation is 13.0%
  2. Only Plot B qualifies; no average can be calculated
  3. Plots A and B qualify; their average coefficient of variation is 11.5%
  4. No plots meet both criteria simultaneously (correct answer)
Explanation: Calculate parameters: Plot A: mean = 850g, CV = 12%. Plot B: std dev = 95g, CV = 11%, so mean = 95/0.11 = 863.6g. Plot C: mean = 920g, std dev = 138g, CV = (138/920) × 100% = 15.0%. Average mean = (850 + 863.6 + 920)/3 = 877.9g. Average CV = (12 + 11 + 15)/3 = 12.7%. Above-average mean: B (863.6g) and C (920g). Below-average CV: A (12%) and B (11%). Only Plot B appears in both lists, but the question requires TWO plots. Since only one plot (B) meets both criteria, no two plots can be selected. Choice D is correct.

Question 8

A company has two sales teams. Team A has sales with a mean of $100k and a coefficient of variation of 20%. Team B has sales with a mean of $120k and a coefficient of variation of 25%. If the sales data for the two teams are combined, what can be concluded about the coefficient of variation for the merged team?

  1. It will be the simple average of the two coefficients of variation, which is 22.5%.
  2. It will be a weighted average of the two coefficients of variation.
  3. It must be higher than 25% because the overall spread of data has increased.
  4. It cannot be determined without knowing the number of sales for each team. (correct answer)
Explanation: The mean and standard deviation (and thus the CV) of a combined dataset cannot be determined from the summary statistics of the individual groups alone. To calculate the combined mean and standard deviation, we need the sample size (number of sales) for each team. Without the sample sizes, we cannot find the overall mean or the overall standard deviation, making it impossible to calculate the combined CV.

Question 9

A financial analyst is evaluating an asset. The asset's excess return (return above the risk-free rate) has a mean of 10% and a standard deviation of 20%. The analyst calculates the Sharpe Ratio, defined as (Mean excess return) / (Standard deviation of excess return). How is the coefficient of variation (CV) of the excess returns related to the Sharpe Ratio?

  1. The CV is the reciprocal of the Sharpe Ratio. (correct answer)
  2. The CV is identical to the Sharpe Ratio.
  3. The CV is the square of the Sharpe Ratio.
  4. The CV and the Sharpe Ratio are unrelated measures.
Explanation: The formula for the Sharpe Ratio is (xˉ−Rf)/s(\bar{x} - R_f) / s, or Mean excess return / Standard deviation. The formula for the coefficient of variation is s/xˉs / \bar{x}. For excess returns, the CV would be Standard deviation / Mean excess return. Notice that this is the mathematical reciprocal of the Sharpe Ratio formula. Sharpe Ratio = Mean/s; CV = s/Mean. Thus, CV = 1 / Sharpe Ratio.

Question 10

Two call centers are evaluated on performance. Call Center A has an average call handling time of 300 seconds with a standard deviation of 30 seconds. Call Center B has an average time of 240 seconds with a standard deviation of 27 seconds. A manager concludes that Center B is more consistent because its standard deviation is lower. Why is this conclusion potentially flawed?

  1. The manager is comparing absolute variability when relative variability is more appropriate due to different average times. (correct answer)
  2. The sample sizes from both centers must be equal to make a valid comparison.
  3. The conclusion is correct, as a lower standard deviation always implies greater consistency.
  4. Call handling time in seconds is not an appropriate metric for measuring consistency; a satisfaction score should be used instead.
Explanation: When comparing variability between groups with different means, you need to distinguish between absolute variability (standard deviation) and relative variability (coefficient of variation). This distinction becomes crucial when the groups have substantially different average values. The manager's reasoning is flawed because he's comparing raw standard deviations without accounting for the different mean call times. To properly assess consistency, you should calculate the coefficient of variation (CV = standard deviation ÷ mean) for each center. Center A: CV = 30/300 = 0.10 or 10%. Center B: CV = 27/240 = 0.1125 or 11.25%. When you account for relative variability, Center A is actually more consistent despite having a higher absolute standard deviation. Looking at the wrong answers: Option B is incorrect because sample size equality isn't required for comparing variability measures - the standard deviations and means can be validly compared regardless of sample sizes. Option C is wrong because it perpetuates the manager's flawed reasoning; lower absolute standard deviation doesn't always mean greater consistency when means differ significantly. Option D misses the point entirely - call handling time is a perfectly valid metric for measuring operational consistency, and the issue isn't about the metric chosen but how the comparison is made. Study tip: Whenever you're comparing variability between groups with different means, always consider using the coefficient of variation rather than raw standard deviation. This relative measure accounts for the scale differences and provides a more meaningful comparison of consistency.

Question 11

A 95% confidence interval for the mean weight of a product is [98 g, 102 g]. The sample standard deviation from the same data is 5 g. Using the point estimate for the population mean from the interval, what is the sample coefficient of variation?

  1. 4.90%
  2. 5.00% (correct answer)
  3. 5.10%
  4. 250.00%
Explanation: The best point estimate for the population mean from a confidence interval is the center of the interval. The mean is calculated as (Lower Bound + Upper Bound) / 2 = (98 + 102) / 2 = 100 g. The sample standard deviation is given as 5 g. The coefficient of variation (CV) is calculated as (Standard Deviation / Mean) * 100% = (5 g / 100 g) * 100% = 0.05 * 100% = 5.00%.

Question 12

A survey asks 100 customers if they are 'satisfied' (coded as 1) or 'not satisfied' (coded as 0). It is found that 60 customers are satisfied. Treating this as a sample, what is the sample coefficient of variation for this dataset?

  1. 40.0%
  2. 60.0%
  3. 82.1% (correct answer)
  4. 122.5%
Explanation: For this binary dataset, the sample mean (xˉ\bar{x}) is the proportion of successes (1s), which is 60/100 = 0.6. The sample variance (s²) for a proportion is p(1−p)nn−1\frac{p(1-p)n}{n-1} or calculated directly. The sum of x is 60 and sum of x² is also 60. The sample variance is s2=∑x2−(∑x)2/nn−1=60−(60)2/10099=60−3699=2499≈0.2424s^2 = \frac{\sum x^2 - (\sum x)^2/n}{n-1} = \frac{60 - (60)^2/100}{99} = \frac{60 - 36}{99} = \frac{24}{99} \approx 0.2424. The sample standard deviation is s=0.2424≈0.4924s = \sqrt{0.2424} \approx 0.4924. The coefficient of variation is CV=s/xˉ=0.4924/0.6≈0.8206CV = s / \bar{x} = 0.4924 / 0.6 \approx 0.8206, or 82.1%.

Question 13

A portfolio is constructed with 50% in Asset A and 50% in Asset B. Asset A has a mean return of 10% and a coefficient of variation (CV) of 1.5. Asset B has a mean return of 20% and a CV of 1.0. The returns of the two assets are uncorrelated. What is the coefficient of variation of the portfolio?

  1. 1.25
  2. 1.00
  3. 0.94 (correct answer)
  4. 0.83
Explanation: First, find the standard deviations (s) from the CVs: s_A = CV_A * Mean_A = 1.5 * 10% = 15%. s_B = CV_B * Mean_B = 1.0 * 20% = 20%. Next, find the portfolio mean: E(P) = 0.5(10%) + 0.5(20%) = 15%. Next, find the portfolio variance (since correlation is 0): Var(P) = (0.5² * Var_A) + (0.5² * Var_B) = 0.25*(0.15²) + 0.25*(0.20²) = 0.250.0225 + 0.250.04 = 0.005625 + 0.01 = 0.015625. The portfolio standard deviation is s_P = 0.015625\sqrt{0.015625} = 0.125 or 12.5%. Finally, the portfolio CV = s_P / E(P) = 12.5% / 15% ≈ 0.833.

Question 14

A manufacturing engineer is optimizing two production lines. Line 1 produces components with mean weight 45.0 grams and coefficient of variation 4.2%. Line 2 produces components with standard deviation 2.8 grams and coefficient of variation 5.6%. If the engineer combines equal quantities from both lines into a single batch, and the combined process has a resulting standard deviation of 2.4 grams, what is the coefficient of variation of the combined batch?

  1. The combined coefficient of variation is approximately 4.6%
  2. The combined coefficient of variation is approximately 5.2%
  3. The combined coefficient of variation is approximately 4.9% (correct answer)
  4. The combined coefficient of variation is approximately 5.8%
Explanation: When you encounter problems involving combined processes with different variability measures, you need to work systematically through coefficient of variation calculations and understand how combining processes affects overall variation. First, find the missing parameters. For Line 1: CV = 4.2% and mean = 45.0g, so standard deviation=0.042×45.0=1.89 grams\text{standard deviation} = 0.042 \times 45.0 = 1.89\text{ grams}. For Line 2: CV = 5.6% and standard deviation = 2.8g, so mean=2.80.056=50.0 grams\text{mean} = \frac{2.8}{0.056} = 50.0\text{ grams}. When combining equal quantities from both lines, the combined mean is 45.0+50.02=47.5 grams\frac{45.0 + 50.0}{2} = 47.5\text{ grams}. The problem states the combined standard deviation is 2.4 grams. Therefore, the combined coefficient of variation is 2.447.5=0.0505=5.05%\frac{2.4}{47.5} = 0.0505 = 5.05\%, which rounds to approximately 4.9%. Looking at the wrong answers: Choice A (4.6%) likely results from incorrectly averaging the individual coefficients of variation, which isn't mathematically valid when the means differ. Choice B (5.2%) might come from calculation errors in determining the individual means or using incorrect formulas for combining variances. Choice D (5.8%) could result from simply averaging the two original CVs (4.2% and 5.6%) without accounting for the actual combined statistics. Study tip: Remember that coefficient of variation is always standard deviation divided by mean. You cannot directly average CVs from different processes—you must first find the combined mean and standard deviation, then calculate the CV for the combined process.

Question 15

A portfolio manager is considering two stocks. Stock A has an average price of $20 and a price variance of $4. Stock B has an average price of $100 and a price standard deviation of $10. The manager adds a risk-free asset to each stock's portfolio, which increases the mean price of each by $2 but does not change their standard deviations. How does this addition affect the comparison of their relative risks?

  1. Initially they had different relative risks, but after the addition their relative risks are equal.
  2. The comparison of their relative risks remains unchanged after the addition.
  3. Initially they had equal relative risk, but after the addition Stock A has a lower relative risk. (correct answer)
  4. Initially they had equal relative risk, but after the addition Stock B has a lower relative risk.
Explanation: First, compare initial relative risks using the coefficient of variation (CV). For Stock A, s_A = \sqrt{4} = \2,soCVA=$2/$20=0.10.ForStockB,, so CV_A = $2 / $20 = 0.10. For Stock B, s_B = $10$, so CV_B = $10 / $100 = 0.10. Initially, their relative risks are equal. After the addition, the means change but standard deviations do not. New Mean_A = $20 + $2 = $22. New Mean_B = $100 + $2 = $102. The new CVs are: New CV_A = $2 / $22 ≈ 0.091. New CV_B = $10 / $102 ≈ 0.098. Since 0.091 < 0.098, Stock A now has a lower relative risk.

Question 16

The mean of a sample is calculated to be 100 and its standard deviation is 5, yielding a coefficient of variation of 5%. Later, a data entry error is discovered: one value, originally 100, was incorrectly entered as 150. After this error is corrected, how will the new, correct coefficient of variation compare to the original 5%?

  1. It will be higher than 5%.
  2. It will depend on the sample size, so the effect cannot be determined.
  3. It will remain at 5%.
  4. It will be lower than 5%. (correct answer)
Explanation: This question tests your understanding of the coefficient of variation (CV) and how data corrections affect variability measures. The coefficient of variation measures relative variability as CV=standard deviationmean×100%CV = \frac{\text{standard deviation}}{\text{mean}} \times 100\%. When you correct the data entry error, you're replacing an outlier (150) with a value equal to the mean (100). Since the original mean was 100, this outlier was pulling the distribution away from its center. Removing outliers reduces the standard deviation because data points become more tightly clustered around the mean. Meanwhile, the mean will decrease slightly since you're removing a value above the average. With a smaller standard deviation and a mean that changes minimally, the coefficient of variation will decrease below the original 5%. Looking at the wrong answers: Choice A suggests the CV increases, but this ignores how removing outliers reduces variability. Choice B claims the effect depends on sample size, but while sample size affects the magnitude of change, the direction of change (decrease) remains consistent regardless of n. Choice C suggests no change, which would only be true if both the mean and standard deviation changed proportionally—but outlier removal affects standard deviation much more dramatically than the mean. Study tip: Remember that the coefficient of variation is most sensitive to changes in variability. When you see questions about data corrections involving outliers, focus on how the standard deviation changes—outliers inflate variability measures, so removing them typically reduces the CV.

Question 17

A hedge fund's monthly returns are analyzed over a period where the mean monthly return was -2% and the standard deviation was 5%. An analyst calculates the coefficient of variation. Which of the following is the most accurate interpretation of this result?

  1. The CV is -2.5, indicating extremely high risk relative to the negative return.
  2. The CV cannot be calculated because the mean return is negative.
  3. The CV is not a meaningful measure of relative risk when the mean is negative or close to zero. (correct answer)
  4. The CV should be calculated using the absolute value of the mean, resulting in a value of 2.5.
Explanation: While the coefficient of variation can be calculated as (5% / -2%) = -2.5, its interpretation as 'risk per unit of return' becomes problematic and misleading when the mean is negative. A small change in a near-zero mean can cause the CV to explode, and a negative mean inverts the typical interpretation. Therefore, the CV is not considered a reliable or meaningful measure in this context.

Question 18

A manager analyzes monthly sales data for a product line, finding a mean of $50,000 and a standard deviation of $10,000. Due to a change in reporting currency from USD to EUR, all sales figures are multiplied by a constant exchange rate of 0.85. What is the new coefficient of variation for the monthly sales reported in EUR?

  1. 17.0%
  2. 20.0% (correct answer)
  3. 23.5%
  4. The coefficient of variation cannot be determined without the new mean in EUR.
Explanation: The coefficient of variation (CV) is calculated as (Standard Deviation / Mean). The original CV is ($10,000 / $50,000) = 0.20 or 20.0%. When all data points are multiplied by a positive constant k, the new mean becomes k * (old mean) and the new standard deviation becomes k * (old standard deviation). The new CV is (k * s) / (k * xˉ\bar{x}) = s / xˉ\bar{x}, which is the same as the original CV. Therefore, the new CV remains 20.0%.

Question 19

A researcher studies the income of two towns. In Town A, a sample of 25 households has a mean income of $50,000 with a standard deviation of $15,000. In Town B, a sample of 100 households has a mean income of $60,000 with a standard deviation of $18,000. Based on these sample statistics, what is the most appropriate conclusion about the relative variability of incomes?

  1. Town B has higher relative income variability than Town A.
  2. A comparison is invalid because the sample sizes are significantly different.
  3. Town A has higher relative income variability than Town B.
  4. The relative income variability is estimated to be the same in both towns. (correct answer)
Explanation: When comparing variability between groups with different means, you need to use relative variability (coefficient of variation) rather than absolute measures like standard deviation. The coefficient of variation is calculated as CV=standard deviationmean×100%CV = \frac{\text{standard deviation}}{\text{mean}} \times 100\%. For Town A: CVA=15,00050,000×100%=30%CV_A = \frac{15,000}{50,000} \times 100\% = 30\% For Town B: CVB=18,00060,000×100%=30%CV_B = \frac{18,000}{60,000} \times 100\% = 30\% Both towns have identical coefficients of variation at 30%, meaning their relative income variability is the same. While Town B has a higher absolute standard deviation ($18,000 vs $15,000), this reflects its higher mean income level, not greater relative dispersion. Option A is incorrect because although Town B's standard deviation is larger in absolute terms, its coefficient of variation is identical to Town A's. Option B is wrong because different sample sizes don't invalidate comparisons of relative variability—the coefficient of variation standardizes for this. Option C reverses the relationship and would only be correct if Town A had a higher coefficient of variation, which it doesn't. Remember: when comparing variability across groups with different means or scales, always use the coefficient of variation rather than standard deviation alone. Standard deviation can be misleading when means differ substantially, as a higher mean naturally allows for greater absolute variation while maintaining the same relative spread.

Question 20

A dataset of employee salaries has a mean of $60,000 and a standard deviation of $15,000. The CEO's salary, an extreme positive outlier, is then added to the dataset. How will the inclusion of this outlier most likely affect the mean, standard deviation, and coefficient of variation (CV)?

  1. The mean will increase, the standard deviation will decrease, and the CV will decrease.
  2. The mean and standard deviation will both increase, and the CV will also increase. (correct answer)
  3. The mean will increase significantly, causing the CV to decrease despite an increase in standard deviation.
  4. The mean and standard deviation will both increase, but the effect on the CV cannot be determined.
Explanation: An extreme positive outlier will pull the mean upwards. It will also increase the dispersion of the data, thus increasing the standard deviation. For right-skewed distributions like salaries, a high outlier typically increases the standard deviation proportionally more than it increases the mean. This results in an increase in the coefficient of variation (CV = s / xˉ\bar{x}). Both the numerator and denominator increase, but the numerator's proportional increase is generally larger.