Class 11 Maths Chapter 13 Statistics – Exercise 13.1 NCERT Solutions

Introduction

Exercise 13.1 introduces the concept of measures of dispersion, focusing on mean deviation. You will learn how to calculate mean deviation for ungrouped and grouped data, both about the mean and the median. This exercise builds the foundation for variance and standard deviation in statistics.

Key Concepts

1. Mean Deviation (MD): Average of absolute deviations of observations from a central value (mean or median).

2. Formula for Mean Deviation (Ungrouped Data):

$$MD=\frac{\sum \mathrm{\mid }{x}_i-A\mathrm{\mid }}{n}$$

where $A$ is mean or median.

3. Formula for Mean Deviation (Grouped Data):

$$MD=\frac{\sum f_i\mathrm{\mid }{x}_i-A\mathrm{\mid }}{\sum f_i}$$

4. Choice of Central Value:

   • Mean deviation about mean.

   • Mean deviation about median.

NCERT Questions with Solutions (10)

Q1. Find mean deviation about mean for data: 2, 4, 6, 8. Mean = $\frac{20}{4}=5$.

$$MD=\frac{\mathrm{\mid }2-5\mathrm{\mid }+\mathrm{\mid }4-5\mathrm{\mid }+\mathrm{\mid }6-5\mathrm{\mid }+\mathrm{\mid }8-5\mathrm{\mid }}{4}=\frac{3+1+1+3}{4}=2$$

Q2. Find mean deviation about median for data: 1, 2, 3, 4, 5. Median = 3.

$$MD=\frac{\mathrm{\mid }1-3\mathrm{\mid }+\mathrm{\mid }2-3\mathrm{\mid }+\mathrm{\mid }3-3\mathrm{\mid }+\mathrm{\mid }4-3\mathrm{\mid }+\mathrm{\mid }5-3\mathrm{\mid }}{5}=\frac{2+1+0+1+2}{5}=1.2$$

Q3. Find mean deviation about mean for data: 10, 20, 30, 40. Mean = 25.

$$MD=\frac{15+5+5+15}{4}=10$$

Q4. Find mean deviation about median for data: 7, 9, 12, 15, 18. Median = 12.

$$MD=\frac{5+3+0+3+6}{5}=3.4$$

Q5. Find mean deviation about mean for data: 5, 10, 15, 20, 25. Mean = 15.

$$MD=\frac{10+5+0+5+10}{5}=6$$

Q6. Find mean deviation about median for data: 2, 4, 6, 8, 10, 12. Median = $\frac{6+8}{2}=7$.

$$MD=\frac{5+3+1+1+3+5}{6}=\frac{18}{6}=3$$

Q7. Find mean deviation about mean for data: 3, 6, 9, 12. Mean = 7.5.

$$MD=\frac{4.5+1.5+1.5+4.5}{4}=3$$

Q8. Find mean deviation about median for data: 1, 2, 2, 3, 4. Median = 2.

$$MD=\frac{1+0+0+1+2}{5}=0.8$$

Q9. Find mean deviation about mean for data: 4, 8, 12, 16, 20. Mean = 12.

$$MD=\frac{8+4+0+4+8}{5}=4.8$$

Q10. Find mean deviation about median for data: 5, 7, 9, 11, 13, 15. Median = $\frac{9+11}{2}=10$.

$$MD=\frac{5+3+1+1+3+5}{6}=\frac{18}{6}=3$$

FAQs (10)

1. What is mean deviation? Average of absolute deviations from mean or median.

2. What is formula for mean deviation (ungrouped data)? $\frac{\sum \mathrm{\mid }{x}_i-A\mathrm{\mid }}{n}$.

3. What is formula for mean deviation (grouped data)? $\frac{\sum f_i\mathrm{\mid }{x}_i-A\mathrm{\mid }}{\sum f_i}$.

4. What is central value in mean deviation? Mean or median.

5. Why use absolute deviations? To avoid cancellation of positive and negative differences.

6. Which is better: mean or median? Median often gives smaller mean deviation.

7. Can mean deviation be negative? No, always non‑negative.

8. What is relation between mean deviation and dispersion? Mean deviation measures dispersion.

9. Is mean deviation same as variance? No, variance uses squares of deviations.

10. Why is this exercise important? It builds foundation for variance and standard deviation.

Conclusion

Exercise 13.1 has 10 solved questions and 10 FAQs that strengthen your understanding of mean deviation in statistics. This builds the foundation for variance, standard deviation, and advanced statistical analysis in Class 11 Maths.

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New Syllabus-Class 11 Maths Chapter 13 Statistics – Exercise 13.1 NCERT