Home > NCERT Solutions > NCERT Solutions for Class 11 Maths – Chapter 13 Statistics Exercise 13.1
NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.1 – Statistics focuses on the concept of Mean Deviation. This exercise covers calculating mean deviation about the mean and mean deviation about the median for both ungrouped data (individual observations) and grouped data (discrete and continuous frequency distributions).
Prepared according to the latest CBSE Class 11 Maths syllabus 2025-26, Exercise 13.1 helps students understand how to measure the spread or dispersion of data around a central value. These concepts are fundamental to statistics and are important for both board exams and competitive exams.
The solutions are explained in a clear, step-by-step format to help students understand each calculation and apply the correct method systematically.
Q1. Calculate the mean deviation about mean for the numbers 4, 7, 8, 9, 10, 12, 13, 17.
Ans:
Mean = (4+7+8+9+10+12+13+17)/8 = 80/8 = 10
Deviations (xᵢ − x̄): −6, −3, −2, −1, 0, 2, 3, 7
|xᵢ − x̄|: 6, 3, 2, 1, 0, 2, 3, 7
M.D.(x̄) = Σ|xᵢ − x̄| / n = (6+3+2+1+0+2+3+7)/8 = 24/8 = 3
Q2. Calculate the mean deviation about the mean for the numbers 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.
Mean = (38+70+48+40+42+55+63+46+54+44)/10 = 500/10 = 50
|xᵢ − x̄|: 12, 20, 2, 10, 8, 5, 13, 4, 4, 6
Σ|xᵢ − x̄| = 12+20+2+10+8+5+13+4+4+6 = 84
M.D.(x̄) = 84/10 = 8.4
Q3. Calculate the mean deviation about the median for the numbers 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.
Arranging in ascending order: 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
n = 12 (even)
Median = (6th term + 7th term)/2 = (13+14)/2 = 13.5
|xᵢ − M|: 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
Σ|xᵢ − M| = 3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5 = 28
M.D.(M) = 28/12 = 7/3 ≈ 2.33
Q4. Calculate the mean deviation about the median for the numbers 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.
Arranging in ascending order: 36, 42, 45, 46, 46, 49, 51, 53, 60, 72
n = 10 (even)
Median = (5th term + 6th term)/2 = (46+49)/2 = 47.5
|xᵢ − M|: 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Σ|xᵢ − M| = 11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5 = 70
M.D.(M) = 70/10 = 7
Q5. Find the mean deviation about the mean for the following data:
n = Σfᵢ = 2+3+4+3+2 = 14 (but actually n = 14, however let us check: total = 14)
Wait — n = 2+3+4+3+2 = 14. But Σfᵢxᵢ = 4+12+24+24+20 = 84
Mean x̄ = 84/14 = 6
|xᵢ − x̄|: 4, 2, 0, 2, 4
fᵢ|xᵢ − x̄|: 8, 6, 0, 6, 8
Σfᵢ|xᵢ − x̄| = 8+6+0+6+8 = 28
M.D.(x̄) = 28/14 = 2
Q6. Find the mean deviation about the mean for the following data:
n = Σfᵢ = 4+24+28+16+8 = 80
Σfᵢxᵢ = 40+720+1400+1120+720 = 4000
Mean x̄ = 4000/80 = 50
|xᵢ − x̄|: 40, 20, 0, 20, 40
fᵢ|xᵢ − x̄|: 160, 480, 0, 320, 320
Σfᵢ|xᵢ − x̄| = 160+480+0+320+320 = 1280
M.D.(x̄) = 1280/80 = 16
Q7. Find the mean deviation about the median for the following data:
n = Σfᵢ = 8+6+2+2+2+6 = 26
Cumulative frequencies: 8, 14, 16, 18, 20, 26
n/2 = 13 → Median lies in the class where cf first exceeds 13 → Median = 7
|xᵢ − M|: 2, 0, 2, 3, 5, 8
fᵢ|xᵢ − M|: 16, 0, 4, 6, 10, 48
Σfᵢ|xᵢ − M| = 16+0+4+6+10+48 = 84
M.D.(M) = 84/26 = 42/13 ≈ 3.23
Q8. Find the mean deviation about the median for the following data:
n = Σfᵢ = 3+5+6+7+8 = 29
Cumulative frequencies: 3, 8, 14, 21, 29
n/2 = 14.5 → Median = value at 15th observation → cf first exceeding 14.5 is 21 → Median = 30
|xᵢ − M|: 15, 9, 3, 0, 5
fᵢ|xᵢ − M|: 45, 45, 18, 0, 40
Σfᵢ|xᵢ − M| = 45+45+18+0+40 = 148
M.D.(M) = 148/29 = 148/29 ≈ 5.1
Q9. Find the mean deviation about the mean for the following data:
Mid-values (xᵢ): 5, 15, 25, 35, 45
n = Σfᵢ = 5+8+15+16+6 = 50
Σfᵢxᵢ = 25+120+375+560+270 = 1350
Mean x̄ = 1350/50 = 27
|xᵢ − x̄|: 22, 12, 2, 8, 18
fᵢ|xᵢ − x̄|: 110, 96, 30, 128, 108
Σfᵢ|xᵢ − x̄| = 110+96+30+128+108 = 472
M.D.(x̄) = 472/50 = 9.44
Q10. Find the mean deviation about the mean for the following data:
Mid-values (xᵢ): 15, 25, 35, 45, 55, 65, 75
n = Σfᵢ = 2+3+8+14+8+3+2 = 40
Σfᵢxᵢ = 30+75+280+630+440+195+150 = 1800
Mean x̄ = 1800/40 = 45
|xᵢ − x̄|: 30, 20, 10, 0, 10, 20, 30
fᵢ|xᵢ − x̄|: 60, 60, 80, 0, 80, 60, 60
Σfᵢ|xᵢ − x̄| = 60+60+80+0+80+60+60 = 400
M.D.(x̄) = 400/40 = 10
Q11. Find the mean deviation about the median for the following data:
n = Σfᵢ = 6+7+15+16+4+2 = 50
Cumulative frequencies: 6, 13, 28, 44, 48, 50
n/2 = 25 → Median class = 20–30 (cf before = 13, f = 15)
Median = 20 + [(25−13)/15] × 10 = 20 + (12/15)×10 = 20 + 8 = 28
Mid-values: 5, 15, 25, 35, 45, 55
|xᵢ − M|: 23, 13, 3, 7, 17, 27
fᵢ|xᵢ − M|: 138, 91, 45, 112, 68, 54
Σfᵢ|xᵢ − M| = 138+91+45+112+68+54 = 508
M.D.(M) = 508/50 = 10.16
Q12. Calculate the mean deviation about the median for the following frequency distribution:
n = Σfᵢ = 40
Cumulative frequencies: 2, 5, 13, 27, 35, 38, 40
n/2 = 20 → Median class = 40–50 (cf before = 13, f = 14)
Median = 40 + [(20−13)/14] × 10 = 40 + 5 = 45
Mid-values: 15, 25, 35, 45, 55, 65, 75
|xᵢ − M|: 30, 20, 10, 0, 10, 20, 30
fᵢ|xᵢ − M|: 60, 60, 80, 0, 80, 60, 60
Σfᵢ|xᵢ − M| = 400
M.D.(M) = 400/40 = 10
FAQs – Class 11 Maths Chapter 13 Exercise 13.1 Statistics
Q1. What is mean deviation?
Mean deviation is a measure of dispersion that tells us how spread out the data values are around a central value (mean or median). It is the average of the absolute deviations of all observations from the central value.
Q2. What is the formula for mean deviation about the mean?
M.D.(x̄) = Σ|xᵢ − x̄| / n for ungrouped data, and M.D.(x̄) = Σfᵢ|xᵢ − x̄| / Σfᵢ for grouped data.
Q3. What is the formula for mean deviation about the median?
M.D.(M) = Σ|xᵢ − M| / n for ungrouped data, and M.D.(M) = Σfᵢ|xᵢ − M| / Σfᵢ for grouped data.
Q4. Why do we use absolute values in mean deviation?
Without absolute values, positive and negative deviations would cancel each other, giving a sum of zero. Taking absolute values ensures all deviations contribute positively to the measure of spread.
Q5. Why is Exercise 13.1 important for board exams?
Mean deviation questions are directly asked in CBSE board exams. Exercise 13.1 covers all types — individual data, discrete frequency distributions, and continuous frequency distributions — making it essential for complete exam preparation.
Q6. How can students prepare effectively for Exercise 13.1?
Students should practise calculating mean and median first, then apply the mean deviation formula step by step. Making a table with columns for xᵢ, fᵢ, fᵢxᵢ, |xᵢ − x̄|, and fᵢ|xᵢ − x̄| helps organise calculations clearly.