Statistics Class 11 Notes

• Measures of dispersion: range, mean deviation, variance and standard deviation.
• Range is useful for quick comparisons but is unreliable for data with outliers.
• Mean deviation and standard deviation use every observation and are more robust.

• Mean deviation about the median:
• $M.D.(\text{median})=\frac{\sum |x_i-\text{median}|}{n}$
• .
• For symmetric data, M.D. about the mean and M.D. about the median are approximately equal.
• Mean deviation is easier to compute but less used in further statistics than variance.

• Discrete frequency distribution:
• $M.D.(\bar{x})=\frac{\sum f_i|x_i-\bar{x}|}{\sum f_i}$
• .
• For continuous data, use the class mark (midpoint) of each interval as x_i.
• The procedure is the same — compute the mean, find absolute deviations, weight by frequencies, and average.

• For grouped data:
• ${\sigma }^2=\frac{\sum f_i(x_i-\bar{x}{)}^2}{\sum f_i}$
• .
• Variance amplifies large deviations more than mean deviation because it squares the differences.
• Standard deviation is the most widely used measure of spread in statistics.

• Short-cut method: choose an assumed mean A and compute deviations d_i = x_i − A. Then variance =
• $\frac{\sum f_i{d}_i^2}{\sum f_i}-{\left(\frac{\sum f_i{d}_i}{\sum f_i}\right)}^2$
• .
• Step-deviation method: set
• $u_i=\frac{x_i-A}{h}$
• where h is the class width. This further simplifies calculations for large datasets.
• Both methods give the same result as the direct method but with smaller numbers.

• Combined mean of two groups:
• $\bar{x}=\frac{n_1{\bar{x}}_1+n_2{\bar{x}}_2}{n_1+n_2}$
• .
• Combined variance requires means, variances and sizes of both groups — a formula-heavy topic often tested in JEE.
• The coefficient of variation (CV) compares consistency across distributions:
• $CV=\frac{\sigma }{\bar{x}}\times 100$
• .
• The distribution with the lower CV is more consistent (less variable relative to its mean).

• To compare two distributions, examine their means (central tendency) and their standard deviations or CVs (spread).
• A higher standard deviation does not always mean more variability — use the CV when means differ significantly.
• When comparing marks of two students or performance of two teams, CV is the standard tool.

Example: Find the variance of the data set {2, 4, 6, 8, 10}.

Solution: Mean $\bar{x}=(2+4+6+8+10)/5=6$. Deviations from mean: −4, −2, 0, 2, 4. Squared deviations: 16, 4, 0, 4, 16. Variance ${\sigma }^2=40/5=8$, so $\sigma =2\sqrt{2}\approx 2.83$.

Example: Group A has mean 60, standard deviation 5; Group B has mean 75, standard deviation 6. Which is more consistent?

Solution: CV of A = (5/60) × 100 = 8.33%. CV of B = (6/75) × 100 = 8.00%. Group B has the lower CV, so it is slightly more consistent despite having a higher standard deviation.

Exercises (NCERT Q1–Q22)

• 1) Find the mean deviation about the mean for the data 4, 7, 8, 9, 10, 12, 13, 17.
• 2) Find the mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.
• 3) Find the mean deviation about the median for the data 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.
• 4) Find the mean deviation about the median for the data 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.
• 5) Find the mean deviation about the mean for the data xᵢ: 5, 10, 15, 20, 25; fᵢ: 7, 4, 6, 3, 5.
• 6) Find the mean deviation about the mean for the data xᵢ: 10, 30, 50, 70, 90; fᵢ: 4, 24, 28, 16, 8.
• 7) Find the mean deviation about the median for the data xᵢ: 5, 7, 9, 10, 12, 15; fᵢ: 8, 6, 2, 2, 2, 6.
• 8) Find the mean deviation about the median for the data xᵢ: 15, 21, 27, 30, 35; fᵢ: 3, 5, 6, 7, 8.
• 9) Find the mean deviation about the mean for the data: Income per day (in Rs) 0-100, 100-200, 200-300, 300-400, 400-500, 500-600, 600-700, 700-800; Number of persons 4, 8, 9, 10, 7, 5, 4, 3.
• 10) Find the mean deviation about the mean for the data: Height in cms 95-105, 105-115, 115-125, 125-135, 135-145, 145-155; Number of men 9, 13, 26, 30, 12, 10.
• 11) Find the mean deviation about the median for the data: Marks 0-10, 10-20, 20-30, 30-40, 40-50, 50-60; Number of girls 6, 8, 14, 16, 4, 2.
• 12) Calculate the mean deviation about median age for the age distribution of 100 persons: Age (in years) 16-20, 21-25, 26-30, 31-35, 36-40, 41-45, 46-50, 51-55; Number 5, 6, 12, 14, 26, 12, 16, 9. [Hint: make the data continuous by subtracting 0.5 from lower and adding 0.5 to upper limits.]
• 13) Find the mean and variance for each of the data: 6, 7, 10, 12, 13, 4, 8, 12.
• 14) Find the mean and variance for each of the data: first n natural numbers.
• 15) Find the mean and variance for each of the data: first 10 multiples of 3.
• 16) Find the mean and variance for each of the data xᵢ: 6, 10, 14, 18, 24, 28, 30; fᵢ: 2, 4, 7, 12, 8, 4, 3.
• 17) Find the mean and variance for each of the data xᵢ: 92, 93, 97, 98, 102, 104, 109; fᵢ: 3, 2, 3, 2, 6, 3, 3.
• 18) Find the mean and standard deviation using the shortcut method: xᵢ: 60, 61, 62, 63, 64, 65, 66, 67, 68; fᵢ: 2, 1, 12, 29, 25, 12, 10, 4, 5.
• 19) Find the mean and variance for the frequency distribution: Classes 0-30, 30-60, 60-90, 90-120, 120-150, 150-180, 180-210; Frequencies 2, 3, 5, 10, 3, 5, 2.
• 20) Find the mean and variance for the frequency distribution: Classes 0-10, 10-20, 20-30, 30-40, 40-50; Frequencies 5, 8, 15, 16, 6.
• 21) Find the mean, variance and standard deviation using the short-cut method: Height in cms 70-75, 75-80, 80-85, 85-90, 90-95, 95-100, 100-105, 105-110, 110-115; Number of students 3, 4, 7, 7, 15, 9, 6, 6, 3.
• 22) The diameters of circles (in mm) drawn in a design are given below: Diameters 33-36, 37-40, 41-44, 45-48, 49-52; Number of circles 15, 17, 21, 22, 25. Calculate the standard deviation and mean diameter of the circles.

Answers at a glance

1. 1
   Q1 → MD = 3.
2. 2
   Q2 → MD = 8.4.
3. 3
   Q3 → MD = 7/3 ≈ 2.33.
4. 4
   Q4 → MD = 7.
5. 5
   Q5 → MD = 6.32.
6. 6
   Q6 → MD = 16.
7. 7
   Q7 → MD = 42/13 ≈ 3.23.
8. 8
   Q8 → MD = 148/29 ≈ 5.10.
9. 9
   Q9 → MD = 157.92.
10. 10
    Q10 → MD ≈ 11.29.
11. 11
    Q11 → MD = 362/35 ≈ 10.34.
12. 12
    Q12 → MD = 7.35.
13. 13
    Q13 → Mean = 9, variance = 9.25.
14. 14
    Q14 → Mean = (n + 1)/2, variance = (n² − 1)/12.
15. 15
    Q15 → Mean = 16.5, variance = 74.25.
16. 16
    Q16 → Mean = 19, variance = 43.4.
17. 17
    Q17 → Mean = 100, variance = 320/11 ≈ 29.09.
18. 18
    Q18 → Mean = 64, SD ≈ 1.69.
19. 19
    Q19 → Mean = 107, variance = 2276.
20. 20
    Q20 → Mean = 27, variance = 132.
21. 21
    Q21 → Mean = 93, variance ≈ 105.58, SD ≈ 10.28.
22. 22
    Q22 → Mean = 43.5 mm, SD ≈ 5.55 mm.

MCQ Questions

1 Mark Questions

1. 1
   Find the median of 1, 2, 3, 4, 5. (Ans: 3)
2. 2
   Write the formula for the variance of n observations x₁, x₂, …, xₙ.
3. 3
   State the relationship between standard deviation and variance. (Ans: σ = √variance)
4. 4
   Find the range of {8, 3, 10, 1, 6}. (Ans: 9)

2 Mark Questions

1. 1
   Find the mean deviation about the mean of {2, 4, 6, 8}. (Ans: mean 5, M.D. = (3+1+1+3)/4 = 2)
2. 2
   Find the standard deviation of {2, 4, 6, 8}. (Ans: √5)
3. 3
   Find the mean of {1, 2, 3, …, 10}. (Ans: 5.5)

3 Mark Questions

1. 1
   Compute the mean and variance of the distribution: x = 1, 2, 3, 4 with frequencies 2, 3, 4, 1. (Ans: mean 2.6, variance = Σf x²/N − x̄² = 0.84)
2. 2
   Find the mean deviation about the median of {3, 5, 7, 9, 11}. (Ans: median 7, M.D. = (4+2+0+2+4)/5 = 2.4)
3. 3
   Prove that Σ(xᵢ − x̄) = 0 for any data set.

4 Mark Questions

1. 1
   Prove that the variance of the observations x₁, x₂, …, xₙ satisfies Σ(xᵢ − x̄)² = Σxᵢ² − n x̄², and use it to compute the variance of {1, 3, 5, 7, 9}. (Ans: Σx = 25, Σx² = 165, σ² = 165/5 − 25 = 8)
2. 2
   Scores of two batsmen A and B are A: 45, 55, 65, 75, 85 and B: 40, 60, 70, 80, 90. Compute means, standard deviations and coefficients of variation for both, and decide who is more consistent. (Ans: same mean 65; A lower S.D. so A more consistent)