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Class 11 Maths Chapter 13 Statistics – Exercise 13.2 NCERT Solutio

Class 11 Maths Chapter 13 Statistics – Exercise 13.2 NCERT Solutio

Introduction

Exercise 13.2 introduces variance and standard deviation as measures of dispersion. You will learn how to calculate variance and standard deviation for ungrouped and grouped data. This exercise builds the foundation for advanced statistical analysis and probability.

Key Concepts

  1. Mean (xˉ):

xˉ=∑xin

  1. Variance (Ungrouped Data):

σ2=∑(xi−xˉ)2n

  1. Standard Deviation (Ungrouped Data):

σ=∑(xi−xˉ)2n

  1. Variance (Grouped Data):

σ2=∑fi(xi−xˉ)2∑fi

  1. Shortcut Method (Grouped Data):

σ2=∑fixi2∑fi−xˉ2

NCERT Questions with Solutions (10)

Q1. Find variance and standard deviation of data: 2, 4, 6, 8

Mean = 5

$$ \sigma^2 = \frac{(2-5)^2 + (4-5)^2 + (6-5)^2 + (8-5)^2}{4} = \frac{9 + 1 + 1 + 9}{4} = 5 $$ $$ \sigma = \sqrt{5} $$

Q2. Find variance and standard deviation of data: 1, 2, 3, 4, 5

Mean = 3

$$ \sigma^2 = \frac{(1-3)^2 + (2-3)^2 + (3-3)^2 + (4-3)^2 + (5-3)^2}{5} = \frac{4 + 1 + 0 + 1 + 4}{5} = 2 $$ $$ \sigma = \sqrt{2} $$

Q3. Find variance and standard deviation of data: 10, 20, 30, 40

Mean = 25

$$ \sigma^2 = \frac{(10-25)^2 + (20-25)^2 + (30-25)^2 + (40-25)^2}{4} = \frac{225 + 25 + 25 + 225}{4} = 125 $$ $$ \sigma = \sqrt{125} = 5\sqrt{5} $$

Q4. Find variance and standard deviation of data: 7, 9, 12, 15, 18

Mean = 12

$$ \sigma^2 = \frac{(7-12)^2 + (9-12)^2 + (12-12)^2 + (15-12)^2 + (18-12)^2}{5} = \frac{25 + 9 + 0 + 9 + 36}{5} = 15.8 $$ $$ \sigma = \sqrt{15.8} $$

Q5. Find variance and standard deviation of data: 5, 10, 15, 20, 25

Mean = 15

$$ \sigma^2 = \frac{(5-15)^2 + (10-15)^2 + (15-15)^2 + (20-15)^2 + (25-15)^2}{5} = \frac{100 + 25 + 0 + 25 + 100}{5} = 50 $$ $$ \sigma = \sqrt{50} = 5\sqrt{2} $$

Q6. Find variance and standard deviation of data: 2, 4, 6, 8, 10, 12

Mean = 7

$$ \sigma^2 = \frac{(2-7)^2 + (4-7)^2 + (6-7)^2 + (8-7)^2 + (10-7)^2 + (12-7)^2}{6} = \frac{25 + 9 + 1 + 1 + 9 + 25}{6} \approx 11.67 $$ $$ \sigma = \sqrt{11.67} $$

Q7. Find variance and standard deviation of data: 3, 6, 9, 12

Mean = 7.5

$$ \sigma^2 = \frac{(3-7.5)^2 + (6-7.5)^2 + (9-7.5)^2 + (12-7.5)^2}{4} = \frac{20.25 + 2.25 + 2.25 + 20.25}{4} = 11.25 $$ $$ \sigma = \sqrt{11.25} $$

Q8. Find variance and standard deviation of data: 1, 2, 2, 3, 4

Mean = 2.4

$$ \sigma^2 = \frac{(1-2.4)^2 + (2-2.4)^2 + (2-2.4)^2 + (3-2.4)^2 + (4-2.4)^2}{5} = \frac{1.96 + 0.16 + 0.16 + 0.36 + 2.56}{5} = 1.04 $$ $$ \sigma = \sqrt{1.04} $$

Q9. Find variance and standard deviation of data: 4, 8, 12, 16, 20

Mean = 12

$$ \sigma^2 = \frac{(4-12)^2 + (8-12)^2 + (12-12)^2 + (16-12)^2 + (20-12)^2}{5} = \frac{64 + 16 + 0 + 16 + 64}{5} = 32 $$ $$ \sigma = \sqrt{32} = 4\sqrt{2} $$

Q10. Find variance and standard deviation of data: 5, 7, 9, 11, 13, 15

Mean = 10

$$ \sigma^2 = \frac{(5-10)^2 + (7-10)^2 + (9-10)^2 + (11-10)^2 + (13-10)^2 + (15-10)^2}{6} = \frac{25 + 9 + 1 + 1 + 9 + 25}{6} \approx 11.67 $$ $$ \sigma = \sqrt{11.67} $$

FAQs (10)

  1. What is variance? Average of squared deviations from mean.

  2. What is standard deviation? Square root of variance.

  3. Why use variance? To measure spread of data.

  4. Why use standard deviation? To express dispersion in same units as data.

  5. Can variance be negative? No, always non‑negative.

  6. Can standard deviation be zero? Yes, if all values are equal.

  7. What is shortcut method for variance? ∑fixi2∑fi−xˉ2.

  8. Which is better: variance or standard deviation? Standard deviation, as it is in same units.

  9. What is relation between variance and standard deviation? σ=σ2.

  10. Why is this exercise important? It builds foundation for probability and advanced statistics.

Conclusion

Exercise 13.2 has 10 solved questions and 10 FAQs that strengthen your understanding of variance and standard deviation in statistics. This builds the foundation for probability and advanced statistical analysis in Class 11 Maths.

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New Syllabud-Class 11 Maths Chapter 13 Statistics – Exercise 13.2 NCERT

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