What is the difference between variance and standard deviation?

Variance (σ²) is the average of squared deviations from the mean. Standard deviation (σ) is the square root of variance. SD has the same units as the data; variance has squared units.

What is coefficient of variation?

CV = (σ/x̄) × 100. It expresses standard deviation as a percentage of the mean, allowing comparison of variability between datasets with different units or scales.

How do you find the median for grouped data?

Median = L + [(N/2 − cf)/f] × h, where L is the lower limit of the median class, N is total frequency, cf is the cumulative frequency before the median class, f is the frequency of the median class, and h is the class width.

Statistics Exercise Solutions | KalviGavani

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

View Solutions →

How to Use This Page

This page provides complete, step-by-step solutions for the Statistics chapter in the Samacheer Kalvi Class 11 Business Maths textbook. Work through each section in order:

1. Key Concepts — quick formula recap before attempting problems 2. MCQs through Match the Following — test your recall and understanding 3. Brief and Detailed answers — build your problem-solving fluency 4. Common Mistakes — avoid the errors students make most often 5. Exam Checklist — verify you are exam-ready

Every formula is displayed in code blocks so you can copy them easily. Worked numerical examples use realistic data so you can verify intermediate steps on your own calculator.

Key Concepts (Quick Recap)

Measures of Central Tendency

Measure	Ungrouped Data	Grouped Data
Mean	x̄ = Σxᵢ / n	x̄ = Σfᵢxᵢ / Σfᵢ
Median	Middle value after sorting	L + [(N/2 − cf) / f] × h
Mode	Most frequent value	L + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h

Shortcut (Assumed Mean) Method:

x̄ = A + Σfᵢdᵢ / Σfᵢ
where dᵢ = xᵢ − A
A = chosen assumed mean (usually the middle value)

Step Deviation Method:

x̄ = A + h × (Σfᵢuᵢ / Σfᵢ)
where uᵢ = (xᵢ − A) / h
h = class width

Measures of Dispersion

Measure	Formula	Notes
Range	H − L	Simplest; ignores distribution shape
Mean Deviation about mean	MD(x̄) = Σfᵢ\	xᵢ − x̄\	/ Σfᵢ	Uses absolute deviations
Variance	σ² = Σfᵢ(xᵢ − x̄)² / N (population) or s² = Σfᵢ(xᵢ − x̄)² / (N − 1) (sample)	Squared units
Standard Deviation	σ = √σ²	Same units as data
Coefficient of Variation	CV = (σ / x̄) × 100	Unit-free; for comparing datasets

Empirical Relationship

Mode ≈ 3 × Median − 2 × Mean

This is an approximate relationship useful for estimating one measure when the other two are known.

Quartiles and Percentiles

Q₁ class → cumulative frequency just ≥ N/4
Q₃ class → cumulative frequency just ≥ 3N/4

Q₁ = L + [(N/4 − cf) / f] × h
Q₃ = L + [(3N/4 − cf) / f] × h

Quartile Deviation = (Q₃ − Q₁) / 2

Percentile:
Pₖ class → cumulative frequency just ≥ kN/100
Pₖ = L + [(kN/100 − cf) / f] × h

I. Choose the Best Answer (MCQs)

1. The arithmetic mean of 5, 8, 12, 15, 20 is

Answer

(a) 10 (b) 12 (c) 14 (d) 15

Answer(b) 12

x̄ = (5 + 8 + 12 + 15 + 20) / 5
   = 60 / 5
   = 12

2. The median of the data 6, 8, 10, 12, 13, 15, 19 is

Answer

(a) 10 (b) 12 (c) 13 (d) 15

Answer(b) 12

Data is already sorted.
n = 7 (odd)
Position = (7 + 1) / 2 = 4th value = 12

3. The mode of 3, 5, 7, 5, 9, 5, 11 is

Answer

(a) 3 (b) 5 (c) 7 (d) 9

Answer(b) 5

The value 5 appears three times, more than any other value.

4. If the mean of a distribution is 25 and the median is 23, then the mode (using empirical relation) is approximately

Answer

(a) 19 (b) 21 (c) 23 (d) 27

Answer(a) 19

Mode ≈ 3 × Median − 2 × Mean
     ≈ 3 × 23 − 2 × 25
     ≈ 69 − 50
     ≈ 19

5. The standard deviation of 5, 5, 5, 5, 5 is

Answer

(a) 0 (b) 5 (c) 25 (d) 1

Answer(a) 0

All values are identical. Deviations from the mean are all zero, so σ = 0.

6. The coefficient of variation for a distribution with mean 50 and standard deviation 5 is

Answer

(a) 5% (b) 10% (c) 50% (d) 0.1%

Answer(b) 10%

CV = (σ / x̄) × 100
   = (5 / 50) × 100
   = 10%

7. If the range of a dataset is 30 and the largest value is 50, then the smallest value is

Answer

(a) 10 (b) 20 (c) 30 (d) 80

Answer(b) 20

Range = Largest − Smallest
30 = 50 − Smallest
Smallest = 50 − 30 = 20

II. Fill in the Blanks

1. The sum of deviations of all observations from their mean is always zero.

Σ(xᵢ − x̄) = 0  (by definition of mean)

2. The median of an even number of observations is the average of the two middle values.

If n is even, Median = [(n/2)th value + (n/2 + 1)th value] / 2

3. The most frequently occurring value in a dataset is called the mode.

4. Variance is the square of standard deviation.

Variance = σ² = (Standard Deviation)²

5. The coefficient of variation is expressed in percentage.

CV = (σ / x̄) × 100 — always a percentage

6. For a symmetrical distribution, mean = median = mode.

7. The difference between the largest and smallest values in a dataset is called the range.

8. The formula for the median class for grouped data is the class where cumulative frequency is just greater than or equal to N/2.

9. Quartile deviation = (Q₃ − Q₁) / 2.

10. If every observation is increased by a constant k, the standard deviation remains unchanged (standard deviation is not affected by change of origin).

11. The mean deviation about the median is minimum compared to mean deviation about any other value.

12. If CV of distribution A is 15% and CV of distribution B is 20%, then distribution B is more variable.

III. True or False

1. Mean is always affected by extreme values.

Answer

True. The arithmetic mean considers every data point, so outliers pull it toward them. Median and mode are more resistant to extreme values.

2. Standard deviation can be negative.

Answer

False. Standard deviation is the square root of variance (which is a sum of squares) and is therefore always non-negative. It equals zero only when all observations are identical.

3. The median of 2, 4, 6, 8 is 5.

Answer

True.

n = 4 (even)
Median = [(4/2)th + (4/2 + 1)th] / 2
       = [2nd + 3rd] / 2
       = (4 + 6) / 2
       = 5

4. Coefficient of variation has the same unit as the original data.

Answer

False. CV is a dimensionless (unit-free) percentage. It divides standard deviation by the mean, canceling out the original units.

5. If each observation is multiplied by 2, the standard deviation also gets multiplied by 2.

Answer

True. Standard deviation is affected by change of scale. Multiplying every observation by k multiplies the standard deviation by k.

If yᵢ = k × xᵢ, then σ_y = k × σ_x

IV. Match the Following

Column A	Column B
1. Mean	(d) Σxᵢ / n
2. Median	(e) Middle value
3. Mode	(a) Most frequent value
4. Range	(f) H − L
5. Standard Deviation	(b) √(σ²)
6. Coefficient of Variation	(c) (σ/x̄) × 100

Answer

1 → d (Mean = Σxᵢ / n)

2 → e (Median = middle value of sorted data)

3 → a (Mode = most frequent value)

4 → f (Range = Highest − Lowest)

5 → b (Standard Deviation = √Variance)

6 → c (CV = (σ/x̄) × 100)

V. Answer Briefly

1. Define mean deviation. How does it differ from standard deviation?

Answer

Mean deviation is the arithmetic mean of the absolute deviations of observations from a central value (mean or median).

MD(about mean) = Σ|xᵢ − x̄| / n  (ungrouped)
MD(about mean) = Σfᵢ|xᵢ − x̄| / Σfᵢ  (grouped)

Difference: Mean deviation uses absolute values of deviations, while standard deviation uses squared deviations. Standard deviation gives more weight to large deviations because of the squaring, making it more sensitive to outliers.

2. What is the empirical relationship between mean, median, and mode?

Mode ≈ 3 × Median − 2 × Mean

Answer

This relationship holds approximately for moderately skewed (positively skewed) distributions. For a perfectly symmetrical distribution, all three are equal. It is useful for estimating one measure when the other two are known.

3. Write the formula for standard deviation using the shortcut method.

σ = h × √(Σfᵢuᵢ²/Σfᵢ − (Σfᵢuᵢ/Σfᵢ)²)

where:
  uᵢ = (xᵢ − A) / h
  A = assumed mean
  h = class width

Answer

This method simplifies computation by reducing the magnitude of numbers involved, especially when class marks are large.

4. What is a coefficient of variation? Why is it used?

Answer

The coefficient of variation (CV) expresses standard deviation as a percentage of the mean.

CV = (σ / x̄) × 100

It is used to compare the variability of two or more datasets that have different units or widely different means. A lower CV indicates greater consistency, while a higher CV indicates greater variability. Since it is unit-free, it allows meaningful comparison across unlike datasets (e.g., comparing variability in heights measured in cm versus weights measured in kg).

VI. Answer in Detail

1. Compute Mean, Median, and Mode for a Frequency Distribution

Consider the following frequency distribution of marks scored by 50 students:

Marks	0–10	10–20	20–30	30–40	40–50
No. of Students	6	10	14	12	8

Mean (Direct Method):

Class Marks (xᵢ): 5, 15, 25, 35, 45

| xᵢ | fᵢ | fᵢxᵢ |
|-----|-----|-------|
|  5  |  6  |  30   |
| 15  | 10  | 150   |
| 25  | 14  | 350   |
| 35  | 12  | 420   |
| 45  |  8  | 360   |
|-----|-----|-------|
|     | 50  | 1310  |

x̄ = Σfᵢxᵢ / Σfᵢ = 1310 / 50 = 26.2

Median:

Cumulative Frequency Table:

| Marks  | f  | cf  |
|--------|-----|-----|
| 0–10   |  6  |  6  |
| 10–20  | 10  | 16  |
| 20–30  | 14  | 30  |
| 30–40  | 12  | 42  |
| 40–50  |  8  | 50  |

N/2 = 50/2 = 25
The cumulative frequency just ≥ 25 is 30 (class 20–30).

Median = L + [(N/2 − cf) / f] × h
       = 20 + [(25 − 16) / 14] × 10
       = 20 + (9/14) × 10
       = 20 + 6.43
       = 26.43

Mode:

Maximum frequency = 14 (class 20–30)
L = 20, f₁ = 14, f₀ = 10, f₂ = 12, h = 10

Mode = L + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h
     = 20 + [(14 − 10) / (2×14 − 10 − 12)] × 10
     = 20 + [4 / (28 − 10 − 12)] × 10
     = 20 + [4 / 6] × 10
     = 20 + 6.67
     = 26.67

Verification using empirical relation:

Mode ≈ 3 × Median − 2 × Mean
     ≈ 3(26.43) − 2(26.2)
     ≈ 79.29 − 52.40
     ≈ 26.89

This is close to the calculated mode of 26.67. ✓

2. Calculating Standard Deviation and Coefficient of Variation

Using the same frequency distribution from above (50 students, marks 0–50):

x̄ = 26.2

| xᵢ | fᵢ | dᵢ = xᵢ − 26.2 | fᵢdᵢ  | fᵢdᵢ²  |
|-----|-----|------------------|--------|---------|
|  5  |  6  |   −21.2          | −127.2 | 2696.64 |
| 15  | 10  |   −11.2          | −112.0 | 1254.40 |
| 25  | 14  |    −1.2          |  −16.8 |   20.16 |
| 35  | 12  |     8.8          |  105.6 |  929.28 |
| 45  |  8  |    18.8          |  150.4 | 2827.52 |
|-----|-----|------------------|--------|---------|
|     | 50  |                  |    0   | 7728.00 |

σ² = Σfᵢdᵢ² / N = 7728 / 50 = 154.56

σ = √154.56 = 12.43

CV = (σ / x̄) × 100 = (12.43 / 26.2) × 100 = 47.44%

Interpretation: A CV of 47.44% indicates moderate variability in the marks distribution. The marks are spread quite widely around the mean.

3. Comparing Two Distributions Using Coefficient of Variation

Two batsmen scored runs in 8 innings each:

Batsman A	30	40	50	60	70	80	90	100
Batsman B	55	58	60	62	64	66	68	71

Batsman A:

x̄_A = (30+40+50+60+70+80+90+100) / 8 = 520/8 = 65

Σxᵢ² = 900+1600+2500+3600+4900+6400+8100+10000 = 38000

σ²_A = (Σxᵢ²/n) − x̄²
      = (38000/8) − (65)²
      = 4750 − 4225
      = 525

σ_A = √525 = 22.91

CV_A = (22.91 / 65) × 100 = 35.25%

Batsman B:

x̄_B = (55+58+60+62+64+66+68+71) / 8 = 504/8 = 63

Σxᵢ² = 3025+3364+3600+3844+4096+4356+4624+5041 = 31950

σ²_B = (31950/8) − (63)²
      = 3993.75 − 3969
      = 24.75

σ_B = √24.75 = 4.97

CV_B = (4.97 / 63) × 100 = 7.89%

Comparison:

CV_A = 35.25%   →  High variability, inconsistent scorer
CV_B = 7.89%    →  Low variability, consistent scorer

Although Batsman A has a slightly higher mean (65 vs 63),
Batsman B is far more consistent (CV 7.89% vs 35.25%).

For team selection where consistency matters, Batsman B is preferred.
For selecting a risk-reward player, Batsman A could be chosen.

Key principle: When comparing variability between two distributions, always compare CV values — not standard deviations alone — because the means may differ significantly.

Common Mistakes to Avoid

1. Forgetting to sort data before finding the median. The median is the middle value of the sorted data, not the original order. Always arrange observations in ascending or descending order first.

2. Using the wrong formula for grouped vs. ungrouped data. Ungrouped formulas do not involve class intervals or cumulative frequencies. Grouped data requires identifying the correct class (median class, modal class, quartile class) before applying L + [...] formulas.

3. Confusing population variance and sample variance. Population variance divides by N; sample variance divides by (N − 1). For TN Board exams, unless stated otherwise, use the population formula (divide by N).

4. Incorrect cumulative frequency in median/quartile formulas. The "cf" in the formula refers to the cumulative frequency before the median/quartile class — not the cumulative frequency of that class itself. Double-check this every time.

5. Using N instead of Σfᵢ in grouped data formulas. For grouped data, N = Σfᵢ (total frequency). Do not use the number of class intervals in place of total frequency.

6. Forgetting the class width (h) in step deviation method. The step deviation method multiplies by h at the end for the mean, and by h² for variance. Skipping these multipliers is a common computational error.

7. Comparing standard deviations instead of CVs across different datasets. A dataset with mean 1000 and SD 50 is more consistent than a dataset with mean 10 and SD 5, even though the absolute SD is larger. Always use CV for cross-dataset comparison.

8. Wrong sign in the empirical relationship. The formula is Mode ≈ 3Median − 2Mean. Students often write Mode ≈ 3Mean − 2Median, which is incorrect. Remember: the mode is approximately three times the median minus twice the mean.

Exam Checklist

Before walking into the exam, confirm every item below:

[ ] I can compute mean for both ungrouped and grouped data (direct, shortcut, and step deviation methods)
[ ] I can find the median class and apply the grouped median formula correctly
[ ] I can identify the modal class and apply the grouped mode formula
[ ] I know the empirical relationship: Mode ≈ 3Median − 2Mean
[ ] I can calculate variance and standard deviation using direct and shortcut methods
[ ] I can compute coefficient of variation and use it to compare two distributions
[ ] I can find Q₁, Q₃, and quartile deviation for grouped data
[ ] I know that mean deviation about the median is minimum
[ ] I can distinguish when to use population variance (N) vs. sample variance (N−1)
[ ] I can set up a full calculation table (xᵢ, fᵢ, dᵢ, fᵢdᵢ, fᵢdᵢ²) without skipping columns

Frequently Asked Questions

Mean = Σfᵢxᵢ / Σfᵢ using the direct method. For large values, use the shortcut method: Mean = A + Σfᵢdᵢ/Σfᵢ, where A is the assumed mean and dᵢ = xᵢ − A.