Measures of Dispersion: Range, Quartiles, Mean Deviation, Standard Deviation

TAKEAWAYS:

• Dispersion tells us how much data values vary from each other—high dispersion means data is spread out, low means it is clustered.
• Range is the simplest measure (max – min), but it only uses two values and ignores the rest.
• Quartiles divide data into four equal parts (Q1, Q2/median, Q3) and help find the interquartile range (IQR), which is better than range.
• Mean deviation and standard deviation use all data points—standard deviation is the most powerful but requires more calculation.
• Coefficient of variation (CV) compares dispersion relative to the mean, useful for comparing datasets of different units or sizes.
• NEB exams test calculations, interpretations, and real-world applications (e.g., comparing variability in sales data or exam scores).

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What is Dispersion?

Dispersion (or variability) measures how spread out the values in a dataset are. If all values are close to the mean, dispersion is low. If values are far apart, dispersion is high.

Why is it important?

• Helps compare two datasets with the same mean but different spreads.
• Used in business to analyze risk (e.g., stock prices), quality control (e.g., product weights), and finance (e.g., investment returns).

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1. Range

Definition: The range is the difference between the highest (H) and lowest (L) values in a dataset. $$\text{Range}=H-L$$

Advantages:

• Easy to calculate.
• Quick to understand.

Disadvantages:

• Only uses two values (ignores the rest).
• Sensitive to extreme values (outliers).

Worked Example 1: Calculating Range

Data: Exam scores of 10 students: 45, 56, 67, 78, 89, 90, 92, 95, 98, 100 Steps:

1. Identify the highest (H) and lowest (L) values:
   • $H=100$
   • $L=45$
2. Calculate range: $$\text{Range}=100-45=55$$

Answer: The range is 55.

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2. Quartiles and Interquartile Range (IQR)

Quartiles divide data into four equal parts:

• Q1 (First Quartile): 25% of data below this value.
• Q2 (Median): 50% of data below this value.
• Q3 (Third Quartile): 75% of data below this value.

Interquartile Range (IQR): $$\text{IQR}=Q3-Q1$$ IQR measures the spread of the middle 50% of data, ignoring extreme values.

Worked Example 2: Finding Quartiles and IQR

Data: Same exam scores (ordered): 45, 56, 67, 78, 89, 90, 92, 95, 98, 100 Steps:

1. Find Q2 (median):
   • For 10 data points, median is the average of 5th and 6th values: $$Q2=\frac{89+90}{2}=89.5$$
2. Find Q1 (median of first 5 values: 45, 56, 67, 78, 89):
   • Q1 is the 3rd value: 67.
3. Find Q3 (median of last 5 values: 90, 92, 95, 98, 100):
   • Q3 is the 3rd value: 95.
4. Calculate IQR: $$\text{IQR}=95-67=28$$

Answer: Q1 = 67, Q2 = 89.5, Q3 = 95, IQR = 28.

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3. Mean Deviation

Definition: Mean deviation measures the average distance of each data point from the mean. $$\text{Mean Deviation (MD)}=\frac{\sum |x_i-\bar{x}|}{n}$$ where:

• $x_i$ = each data point,
• $\bar{x}$ = mean,
• $n$ = number of data points.

Advantages:

• Uses all data points.
• Easy to understand.

Disadvantages:

• Not as widely used as standard deviation in advanced statistics.

Worked Example 3: Calculating Mean Deviation

Data: 5, 7, 8, 9, 12 Steps:

1. Calculate the mean ($\bar{x}$): $$\bar{x}=\frac{5+7+8+9+12}{5}=\frac{41}{5}=8.2$$
2. Find deviations from the mean and take absolute values: $$|5-8.2|=3.2,|7-8.2|=1.2,|8-8.2|=0.2,|9-8.2|=0.8,|12-8.2|=3.8$$
3. Sum the absolute deviations: $$3.2+1.2+0.2+0.8+3.8=9.2$$
4. Calculate MD: $$\text{MD}=\frac{9.2}{5}=1.84$$

Answer: Mean deviation = 1.84.

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4. Standard Deviation

Definition: Standard deviation ($\sigma$) measures the average squared distance from the mean. It is the square root of variance. $$\sigma =\sqrt{\frac{\sum (x_i-\bar{x}{)}^2}{n}}$$ For a sample, divide by $n-1$ (not covered in Class 11).

Advantages:

• Uses all data points.
• More reliable than range or IQR for large datasets.
• Used in probability and advanced statistics.

Disadvantages:

• More complex to calculate.
• Sensitive to extreme values (though less than range).

Worked Example 4: Calculating Standard Deviation

Data: Same as above (5, 7, 8, 9, 12), mean = 8.2 Steps:

1. Calculate squared deviations: $$(5-8.2{)}^2=10.24,(7-8.2{)}^2=1.44,(8-8.2{)}^2=0.04,(9-8.2{)}^2=0.64,(12-8.2{)}^2=14.44$$
2. Sum the squared deviations: $$10.24+1.44+0.04+0.64+14.44=26.8$$
3. Calculate variance: $$\text{Variance}=\frac{26.8}{5}=5.36$$
4. Take the square root for standard deviation: $$\sigma =\sqrt{5.36}\approx 2.31$$

Answer: Standard deviation ≈ 2.31.

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Comparison of Measures of Dispersion

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5. Coefficient of Variation (CV)

Definition: CV compares dispersion relative to the mean. It is useful for comparing datasets with different units or different means. $$\text{CV}=\left(\frac{\sigma }{\bar{x}}\right)\times 100\%$$

Worked Example 5: Calculating CV

Data Set 1: Exam scores (mean = 8.2, $\sigma \approx 2.31$) Data Set 2: Salary data (mean = 50,000, $\sigma =5,000$) Steps:

1. Calculate CV for exam scores: $$\text{CV}=\left(\frac{2.31}{8.2}\right)\times 100\approx 28.17\%$$
2. Calculate CV for salaries: $$\text{CV}=\left(\frac{5,000}{50,000}\right)\times 100=10\%$$ Interpretation:

• Exam scores have higher relative variability (28.17%) than salaries (10%), even though the absolute standard deviation of salaries is larger.

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NEB-Style Questions and Solutions

Question 1 (Short Answer)

Calculate the range and IQR for the following dataset: Data: 12, 15, 14, 16, 18, 20, 22, 25 Solution:

1. Range:
   • $H=25$, $L=12$
   • Range = $25-12=13$
2. Quartiles:
   • Ordered data: 12, 14, 15, 16, 18, 20, 22, 25
   • $n=8$ (even), so:
     • Q1 = median of first 4: $\frac{14+15}{2}=14.5$
     • Q3 = median of last 4: $\frac{20+22}{2}=21$
   • IQR = $21-14.5=6.5$

Answer:

• Range = 13
• IQR = 6.5

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Question 2 (Long Answer)

The following are the monthly salaries (in thousands) of 10 employees in a company: Data: 25, 30, 28, 32, 35, 40, 45, 50, 55, 60 Calculate:

1. Mean deviation about the mean.
2. Standard deviation.
3. Which measure of dispersion (range, IQR, or standard deviation) would you use to describe the variability in salaries? Justify your answer.

Solution:

1. Mean ($\bar{x}$): $$\bar{x}=\frac{25+30+28+32+35+40+45+50+55+60}{10}=\frac{390}{10}=39$$

2. Mean Deviation:

   • Calculate $|x_i-39|$ for each value: $$14,9,11,7,4,1,6,11,16,21$$
   • Sum = $14+9+11+7+4+1+6+11+16+21=100$
   • MD = $\frac{100}{10}=10$

3. Standard Deviation:

   • Calculate $(x_i-39{)}^2$: $$196,81,121,49,16,1,36,121,256,441$$
   • Sum = $196+81+121+49+16+1+36+121+256+441=1,318$
   • Variance = $\frac{1,318}{10}=131.8$
   • $\sigma =\sqrt{131.8}\approx 11.48$

4. Choice of Measure:

   • Range is sensitive to outliers (e.g., 25 and 60).
   • IQR ignores extreme values but only uses the middle 50%.
   • Standard deviation uses all data points and is more reliable for large datasets.
   • Best choice: Standard deviation, as it provides a comprehensive measure of variability.

Answer:

1. Mean deviation = 10
2. Standard deviation ≈ 11.48
3. Standard deviation is the best measure because it considers all data points and gives a complete picture of variability.

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Question 3 (Application)

A shopkeeper records the daily sales (in hundreds of rupees) for 8 days: Data: 5, 7, 6, 8, 9, 10, 12, 15

1. Calculate the coefficient of variation.
2. If another shopkeeper has a mean daily sale of 1200 and a standard deviation of 150, which shop has more relative variability in sales?

Solution:

1. For the first shopkeeper:

   • Mean ($\bar{x}$): $$\bar{x}=\frac{5+7+6+8+9+10+12+15}{8}=\frac{72}{8}=9$$
   • Standard deviation ($\sigma$):
     • Calculate $(x_i-9{)}^2$: $$16,4,9,1,0,1,9,36$$
     • Sum = $16+4+9+1+0+1+9+36=76$
     • Variance = $\frac{76}{8}=9.5$
     • $\sigma =\sqrt{9.5}\approx 3.08$
   • CV: $$\text{CV}=\left(\frac{3.08}{9}\right)\times 100\approx 34.22\%$$

2. For the second shopkeeper:

   • Given: $\bar{x}=1200$, $\sigma =150$
   • CV: $$\text{CV}=\left(\frac{150}{1200}\right)\times 100=12.5\%$$

Interpretation:

• The first shopkeeper has a higher CV (34.22%) compared to the second (12.5%), meaning their sales are more relatively variable.

Answer:

1. CV for first shopkeeper ≈ 34.22%
2. The first shopkeeper has more relative variability in sales.

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Exam Tip

1. Understand the Concepts:

   • Know when to use range (quick estimate), IQR (robust to outliers), or standard deviation (comprehensive).
   • Remember that standard deviation is the most widely used measure in advanced statistics.

2. Practice Calculations:

   • NEB often tests step-by-step calculations for mean deviation and standard deviation. Show all steps clearly.
   • For quartiles, remember:
     • If $n$ is odd, median is the middle value.
     • If $n$ is even, median is the average of the two middle values.
     • Q1 and Q3 are medians of the lower and upper halves, respectively.

3. Interpret Results:

   • A higher standard deviation means more spread out data.
   • A higher CV means more relative variability (useful for comparing datasets with different units).

4. Real-World Applications:

   • Business: Compare variability in sales, profits, or customer satisfaction scores.
   • Finance: Assess risk in investments (higher standard deviation = higher risk).
   • Quality Control: Check consistency in product weights or measurements.

5. Common Mistakes to Avoid:

   • Forgetting to take absolute values in mean deviation.
   • Misplacing the decimal in standard deviation calculations.
   • Confusing population standard deviation (divide by $n$) with sample standard deviation (divide by $n-1$)—Class 11 only covers population.

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A box plot showing quartiles, median, and outliers for a dataset (Image: SGDWN, CC BY-SA 4.0, via Wikimedia Commons)