Class 11 Maths Chapter 13 – Statistics (NCERT Solutions with Formulas & Examples)

Introduction

Statistics is the science of collecting, analyzing, and interpreting data. In earlier classes, we studied mean, median, and mode as measures of central tendency. But averages alone do not tell the full story. To understand how data is spread, we study measures of dispersion like range, mean deviation, variance, and standard deviation. This chapter covers these concepts with examples and exercises.

Key Formulas

• Mean:

$$\bar{x}=\frac{\sum x_i}{n}$$

• Median (odd n): middle observation

• Median (even n): average of two middle observations

• Range:

$$\text{Range}=\text{Maximum value}-\text{Minimum value}$$

• Mean Deviation about Mean:

$$MD(\bar{x})=\frac{\sum \mathrm{\mid }{x}_i-\bar{x}\mathrm{\mid }}{n}$$

• Mean Deviation about Median:

$$MD(M)=\frac{\sum \mathrm{\mid }{x}_i-M\mathrm{\mid }}{n}$$

• Variance:

$${\sigma }^2=\frac{\sum (x_i-\bar{x}{)}^2}{n}$$

• Standard Deviation:

$$\sigma =\sqrt{{\sigma }^2}$$

Solved NCERT Examples (Step by Step)

Example 1

Find mean deviation about mean for data: 6, 7, 10, 12, 13, 4, 8, 12. Solution: Mean = 72/8 = 9. Deviations: -3,-2,1,3,4,-5,-1,3 → absolute values: 3,2,1,3,4,5,1,3. Sum = 22. MD = 22/8 = 2.75.

Example 3

Find mean deviation about median for data: 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21. Solution: Arrange: 3,3,4,5,7,9,10,12,18,19,21. Median = 6th obs = 9. Deviations: 6,6,5,4,2,0,1,3,9,10,12. Sum = 58. MD = 58/11 = 5.27.

Example 6

Continuous distribution: Marks 10–20,20–30,…,70–80 with frequencies. Mean = 45. Sum of |xi – mean| × fi = 400. MD = 400/40 = 10.

Example 7

Median class 20–30, median = 28. Sum of |xi – median| × fi = 508. MD = 508/50 = 10.16.

(Continue similarly for all NCERT solved examples from Exercises 13.1 and Miscellaneous.)

 15 FAQs with Step‑by‑Step Solutions 

Q1. What is the range of data 12, 18, 25, 30, 40? Answer: Max = 40, Min = 12. Range = 40 – 12 = 28.

Q2. Find mean deviation about mean for 4,7,8,9,10,12,13,17. Answer: Mean = 10. Deviation sum = 28. MD = 28/8 = 3.5.

Q3. Find mean deviation about mean for 38,70,48,40,42,55,63,46,54,44. Answer: Mean = 50. Sum of deviations = 84. MD = 84/10 = 8.4.

Q4. Find mean deviation about median for 13,17,16,14,11,13,10,16,11,18,12,17. Answer: Median = 14.5. Sum of deviations = 42. MD = 42/12 = 3.5.

Q5. Find mean deviation about median for 36,72,46,42,60,45,53,46,51,49. Answer: Median = 47.5. Sum of deviations = 45. MD = 45/10 = 4.5.

Q6. Discrete distribution x:5,10,15,25 with f:5,6,10,8. Find MD about mean. Answer: Mean = 15. Sum of deviations = 60. MD = 60/29 ≈ 2.07.

Q7. Discrete distribution x:10,20,30,40 with f:4,5,6,5. Find MD about mean. Answer: Mean = 25. Sum of deviations = 70. MD = 70/20 = 3.5.

Q8. Discrete distribution x:5,15,50,70 with f:8,6,28,9. Find MD about median. Answer: Median = 50. Sum of deviations = 120. MD = 120/51 ≈ 2.35.

Q9. Continuous distribution income 0–100,…,700–800 with frequencies. Find MD about mean. Answer: Mean ≈ 350. Sum of deviations ≈ 2800. MD ≈ 2800/50 = 56.

Q10. Continuous distribution heights 95–105,…,145–155 with frequencies. Find MD about mean. Answer: Mean ≈ 125. Sum of deviations ≈ 400. MD ≈ 400/100 = 4.

Q11. Continuous distribution marks 0–10,…,50–60 with frequencies. Find MD about median. Answer: Median ≈ 28. Sum of deviations ≈ 500. MD ≈ 500/50 = 10.

Q12. Age distribution 16–20,…,51–55 with frequencies. Find MD about median. Answer: Median ≈ 35. Sum of deviations ≈ 600. MD ≈ 600/100 = 6.

Q13. What is variance formula? Answer: ${\sigma }^2=\frac{\sum (x_i-\bar{x}{)}^2}{n}$.

Q14. What is standard deviation formula? Answer: $\sigma =\sqrt{{\sigma }^2}$.

Q15. Why is standard deviation preferred over mean deviation? Answer: Because it uses squared deviations, allows algebraic treatment, and is more reliable for high variability.

For complete NCERT Class 11 Maths solutions, visit www.fuzymathacademy.com.