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Coefficient of Variation - Business Statistics

  1. 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?
    Answer: The relative income variability is estimated to be the same in both towns. 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.
  2. A business analyst notes that the coefficient of variation for annual sales in Division A is 35% and for Division B is 75%. Which of the following can be validly inferred from this information alone?
    Answer: The annual sales of Division B are more variable relative to their average than the sales of Division A. When you encounter coefficient of variation (CV) questions, remember that CV measures relative variability - how spread out data points are compared to the mean. It's calculated as CV=standard deviationmean×100%CV = \frac{\text{standard deviation}}{\text{mean}} \times 100\%, which allows you to compare variability across datasets with different scales or means. Division B's CV of 75% versus Division A's 35% tells you that Division B's sales vary more relative to its average than Division A's sales do relative to its average. This is exactly what answer choice A states - it's the direct interpretation of what coefficient of variation measures. Let's examine why the other options are incorrect. Choice B assumes you can determine skewness from CV alone, but CV only measures relative spread, not the shape or asymmetry of the distribution. Choice C incorrectly suggests that CV tells you about absolute mean values. However, a division could have a low CV with either a high or low mean - CV standardizes for the mean, so you can't compare actual mean values between groups. Choice D makes a similar error by assuming CV reveals information about the range (maximum minus minimum). CV uses standard deviation, not range, and again standardizes relative to the mean, so absolute measures of spread cannot be compared. Study tip: Remember that coefficient of variation is specifically designed for comparing relative variability between datasets. When you see CV comparisons, focus only on which group has more variability relative to its own mean - avoid drawing conclusions about absolute values, distribution shapes, or other descriptive statistics.
  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?
    Answer: $16.00 The coefficient of variation (CV) is s/xˉ s / \bar{x} . We are given CV = 0.25 and variance s2=16 s^2 = 16 . First, find the standard deviation, s=16=$4 s = \sqrt{16} = \$4 . Now, rearrange the CV formula to solve for the mean: xˉ=s/CV \bar{x} = s / \text{CV} . Therefore, the mean price is $4/0.25=$16.00 \$4 / 0.25 = \$16.00 . Distractor D ($64) results from incorrectly using the variance instead of the standard deviation in the calculation.
  4. 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?
    Answer: The manager is comparing absolute variability when relative variability is more appropriate due to different average times. 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.
  5. 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?
    Answer: 0.94 First, find the standard deviations (s) from the CVs: sAs_A = CV_A Mean_A = 1.5 10% = 15%. sBs_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) = 0.25*0.0225 + 0.25*0.04 = 0.005625 + 0.01 = 0.015625. The portfolio standard deviation is sPs_P = 0.015625 \sqrt{0.015625} = 0.125 or 12.5%. Finally, the portfolio CV = sPs_P / E(P) = 12.5% / 15% ≈ 0.833.
  6. 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?
    Answer: 7.89% First, find the original standard deviation from the variance: s=225=15 s = \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=190 200 - 10 = 190 . The standard deviation is still 15. The new coefficient of variation is (15 / 190) ≈ 0.0789, or 7.89%.
  7. 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?
    Answer: Process X is relatively more consistent than Process Y. 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).
  8. A quality assurance team is monitoring the thickness of silicon wafers from three suppliers. Supplier X reports wafers with mean thickness 0.725 mm and CV of 1.2%. Supplier Y provides wafers with standard deviation 0.008 mm and mean thickness 0.680 mm. Supplier Z has CV of 2.1% and standard deviation 0.015 mm. If the team needs to rank suppliers from most consistent to least consistent in relative terms, and then determine which supplier has the smallest absolute variation, what is the correct ranking and identification?
    Answer: Ranking: Y, X, Z; smallest absolute variation is Supplier Y with 0.008 mm Calculate CV and std dev for each: Supplier X: CV = 1.2%, std dev = 0.012 × 0.725 = 0.0087 mm. Supplier Y: std dev = 0.008 mm, CV = (0.008/0.680) × 100% = 1.18%. Supplier Z: CV = 2.1%, std dev = 0.015 mm. Ranking by CV (most to least consistent): Y (1.18%), X (1.2%), Z (2.1%). Smallest absolute variation (std dev): Y (0.008 mm) < X (0.0087 mm) < Z (0.015 mm). Choice B is correct. Choice A has the wrong ranking. Choices C and D incorrectly identify the smallest absolute variation.
  9. 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?
    Answer: 20.0% 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%.
  10. 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)?
    Answer: The mean and standard deviation will both increase, and the CV will also increase. 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.
  11. 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?
    Answer: Portfolio Y exhibits the highest relative risk with a coefficient of variation of 35% 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.
  12. 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?
    Answer: 5.00% 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%.
  13. 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?
    Answer: No plots meet both criteria simultaneously 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.
  14. A dataset consists of the values {10, 20, 30, 40, 100}. An analyst calculates the coefficient of variation (CV). If the outlier (100) is removed from the dataset, what will be the effect on the CV?
    Answer: The CV will decrease. First, calculate the original CV. Mean = (10+20+30+40+100)/5 = 40. The sample standard deviation is ((10−40)2+...+(100−40)2)/4=1250≈35.36\sqrt{((10-40)^2+...+(100-40)^2)/4} = \sqrt{1250} \approx 35.36. Original CV = 35.36 / 40 ≈ 88.4%. After removing 100, the new dataset is {10, 20, 30, 40}. The new mean is 25. The new sample standard deviation is ((10−25)2+...+(40−25)2)/3=500/3≈12.91\sqrt{((10-25)^2+...+ (40-25)^2)/3} = \sqrt{500/3} \approx 12.91. The new CV = 12.91 / 25 ≈ 51.6%. The CV decreases significantly because removing the outlier reduces the standard deviation more than it reduces the mean in proportional terms.
  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?
    Answer: Initially they had equal relative risk, but after the addition Stock A has a lower relative risk. First, compare initial relative risks using the coefficient of variation (CV). For Stock A, sA=4=$2 s_A = \sqrt{4} = \$2 , so CV_A = $2 / $20 = 0.10. For Stock B, sB=$10 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.
  16. 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?
    Answer: The CV is not a meaningful measure of relative risk when the mean is negative or close to zero. 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.
  17. 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?
    Answer: Formulation B should be chosen with a standard deviation of 4.2 mg 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.
  18. 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?
    Answer: 82.1% 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≈0.2424 s^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≈0.4924 s = \sqrt{0.2424} \approx 0.4924 . The coefficient of variation is CV=s/xˉ=0.4924/0.6≈0.8206 CV = s / \bar{x} = 0.4924 / 0.6 \approx 0.8206 , or 82.1%.
  19. 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?
    Answer: The combined coefficient of variation is approximately 4.9% 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.
  20. A retail chain is analyzing customer satisfaction scores across three regions. Region North has scores with mean 7.5 and standard deviation 1.2. Region South has mean 8.4 and coefficient of variation 10%. Region West has coefficient of variation 18% and standard deviation 1.44. If management wants to identify the region with the most consistent customer satisfaction relative to its average performance, which region should they focus on improving, and what is the range of coefficients of variation across all regions?
    Answer: Focus on improving Region West; CV range is 8.0 percentage points When analyzing variability across different groups with different means, you need the coefficient of variation (CV), which measures relative variability as a percentage. CV = (standard deviation ÷ mean) × 100%. Let's calculate each region's CV. Region North: CV = (1.2 ÷ 7.5) × 100% = 16.0%. Region South already gives us CV = 10%. For Region West, we have CV = 18% and standard deviation = 1.44, so we can find the mean: 18% = (1.44 ÷ mean) × 100%, which gives us mean = 8.0. The most consistent region has the lowest CV: South at 10%, followed by North at 16.0%, then West at 18.0%. Since management wants to improve the region with the most consistent performance, they should focus on Region South (the best performer) to maintain its advantage. The CV range is 18.0% - 10% = 8.0 percentage points. Answer A incorrectly suggests focusing on West (the least consistent region) and miscalculates the range as 12.0 points. Answer B correctly identifies South but uses the wrong range of 12.0 points. Answer D incorrectly suggests focusing on South but gets the range right at 8.0 points - however, the question asks for the most consistent region to focus on improving, which is South. Study tip: Always distinguish between absolute variability (standard deviation) and relative variability (coefficient of variation). When comparing groups with different means, CV is the appropriate measure for consistency.

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