CBSE X Maths — Statistics: Notes, Practice Questions & Worked Solutions

CBSE Class X Mathematics (041/241) · Unit VII: Statistics and Probability

Statistics — Mean, Median, and Mode of Grouped Data

For grouped data, we don’t know individual values — only how many fall in each class interval. The mean, median, and mode all have specific formulas to handle this.

1. Mean — Direct, Assumed Mean, and Step-Deviation Methods

Direct: Mean = Σfᵢxᵢ / Σfᵢ

Assumed mean: Mean = A + Σfᵢdᵢ/Σfᵢ, where dᵢ=xᵢ−A

Step-deviation: Mean = A + h×Σfᵢuᵢ/Σfᵢ, where uᵢ=(xᵢ−A)/h

xᵢ is the midpoint (class mark) of each class interval.

Worked example — mean by the direct method Classes 0-10, 10-20, 20-30, 30-40, 40-50 have frequencies 5, 8, 15, 16, 6 (midpoints 5, 15, 25, 35, 45).
Σf = 50. Σfx = 25+120+375+560+270 = 1350.
Mean = 1350/50 = 27

2. Median of Grouped Data

Median = l + [(n/2 − cf)/f] × h

l = lower boundary of the median class, cf = cumulative frequency before the median class, f = frequency of the median class, h = class width. The median class is the first class whose cumulative frequency is ≥ n/2.

Worked example — median Classes 0-10…50-60 have frequencies 5,10,18,12,9,6 (n=60). Cumulative frequencies: 5,15,33,45,54,60. n/2=30.
The median class is 20-30 (first cf ≥ 30, with cf=33). l=20, cf=15, f=18, h=10.
Median = 20 + [(30−15)/18]×10 = 20 + 8.33 = 28.33

3. Mode of Grouped Data

Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h

l = lower boundary of the modal class (the class with the highest frequency), f₁ = frequency of the modal class, f₀ = frequency of the class before it, f₂ = frequency of the class after it.

Worked example — mode Using the same data as above, the modal class is 20-30 (highest frequency, 18). l=20, f₁=18, f₀=10, f₂=12, h=10.
Mode = 20 + [(18−10)/(36−10−12)]×10 = 20 + 5.71 = 25.71

Common mistakes to watch for

Using the frequency of the median/modal class as “n” in the formula — n is the total of all frequencies, not just one class’s.
Picking the wrong median class — it’s the first class where the cumulative frequency reaches or exceeds n/2, not the class containing n/2 itself as a raw frequency.
Forgetting bimodal data (two equally highest frequencies) is avoided in this syllabus — such datasets won’t appear in this topic’s problems.

Key Vocabulary Recap

Class mark (xᵢ) — the midpoint of a class interval

Cumulative frequency — running total of frequencies up to a class

Modal class — the class interval with the highest frequency

Syllabus Reference

This note covers CBSE Class X Mathematics Unit VII, Statistics: mean, median, and mode of grouped data, with bimodal situations avoided.

Practice questions

CBSE Class X Mathematics · Unit VII: Statistics — Practice Question Bank (no answers)

Statistics — Practice

Covers: mean (direct, assumed mean, step-deviation), median, and mode of grouped data. 10 questions.

Find the mean of the following data using the direct method: classes 0-10, 10-20, 20-30, 30-40, 40-50 with frequencies 5, 8, 15, 16, 6.

Using the same data as Q1, verify your answer using the assumed mean method (assumed mean = 25).

Find the median of the grouped data: classes 0-10 to 50-60 with frequencies 5, 10, 18, 12, 9, 6.

Find the mode of the data given in Q3.

Find the mean using the step-deviation method for classes 10-20 to 60-70 with frequencies 8, 6, 12, 11, 9, 4 (assumed mean = 45).

Explain what “bimodal” data means, and why it is avoided in this syllabus.

State the formula for the mean using the direct method, and identify what fᵢ and xᵢ represent.

A survey of 50 students’ heights (in cm) gives classes 150-155, 155-160, 160-165, 165-170, 170-175, 175-180 with frequencies 4, 7, 15, 12, 8, 4. Find the modal class and the mean height.

Explain the difference between mean, median, and mode as measures of central tendency.

Q10

Find the mean of the first 10 natural numbers.

More Practice

Q11

Lifetimes (hours) of 225 electrical components are grouped into classes 0-20,20-40,40-60,60-80,80-100,100-120 with frequencies 10,35,52,61,38,29. Find the modal lifetime.

Q12

Find the mean of the lifetime data in Q11, using the direct method.

Q13

The median of a distribution is 525, with total frequency 100. Given classes 0-100 to 900-1000 with frequencies 2,5,x,12,17,20,y,9,7,4 (in order), find x and y.

Q14

Find the mean using the assumed mean method for classes 0-20,20-40,40-60,60-80,80-100 with frequencies 6,8,10,4,2 (assumed mean = 50).

Q15

If the mean of the data — classes 0-10,10-20,20-30,30-40,40-50 with frequencies 7,f,8,10,5 — is 25, find the missing frequency f.

Q16

Find the mode of the data: classes 25-30,30-35,35-40,40-45,45-50,50-55 with frequencies 25,34,50,42,38,14.

Q17

Ages of patients admitted to a hospital: classes 5-15,15-25,25-35,35-45,45-55,55-65 with frequencies 6,11,21,23,14,5. Find the mode.

Q18

Find the median for the data in Q17.

Q19

Find the mean for the data in Q17 using the direct method.

Q20

Explain, using algebra, why the assumed mean method always gives the same result as the direct method.

Worked solutions

Show worked solutions

CBSE Class X Mathematics · Unit VII: Statistics — Solutions

Statistics — Practice: Solutions

Mean by direct method.

Midpoints 5,15,25,35,45. Σf=50, Σfx=25+120+375+560+270

Σfx=1350

✓Mean = 1350÷50 = 27

Verify via assumed mean method (A=25).

d=x−A: −20,−10,0,10,20. Σfd=5(−20)+8(−10)+15(0)+16(10)+6(20)

Σfd=−100−80+0+160+120=100

✓Mean = 25 + 100/50 = 27 (matches Q1)

Median, freqs 5,10,18,12,9,6.

Cumulative freq: 5,15,33,45,54,60. n=60, n/2=30. Median class = 20-30 (cf=33 first ≥30)

l=20,cf=15,f=18,h=10

Median = l+[(n/2−cf)/f]×h

20+[(30−15)/18]×10

✓Median ≈ 28.33

Mode of same data.

Modal class=20-30 (highest freq=18). l=20,f₁=18,f₀=10,f₂=12,h=10

Mode=20+[(18−10)/(36−10−12)]×10

✓Mode ≈ 25.71

Step-deviation mean, A=45,h=10.

Midpoints 15,25,35,45,55,65. u=(x−45)/10: −3,−2,−1,0,1,2

Σfu=8(−3)+6(−2)+12(−1)+11(0)+9(1)+4(2)=−31

✓Mean = 45 + 10×(−31/50) = 45 − 6.2 = 38.8

Explain bimodal data.

✓Bimodal data has two different values (or class intervals) tied for the highest frequency. It’s avoided in this syllabus because the standard mode formula assumes a single, clearly-defined modal class — bimodal situations would need special handling not covered here.

State the direct-method mean formula.

✓Mean = Σfᵢxᵢ / Σfᵢ, where xᵢ is the midpoint (class mark) of the iᵗʰ class interval, and fᵢ is the frequency of that class.

Heights survey, find modal class and mean.

Midpoints 152.5,157.5,162.5,167.5,172.5,177.5. Σf=50

Σfx=610+1102.5+2437.5+2010+1380+710=8250

✓Modal class = 160-165 (highest frequency, 15); Mean = 8250÷50 = 165 cm

Difference between mean, median, mode.

✓Mean is the average value, using every data point equally. Median is the middle value when data is arranged in order — it isn’t affected by extreme values. Mode is the most frequently occurring value or class — useful for categorical or heavily-repeated data.

Q10

Mean of first 10 natural numbers.

Mean of first n natural numbers = (n+1)/2

(10+1)/2

✓5.5

More Practice — Solutions

Q11

Modal lifetime, freqs 10,35,52,61,38,29.

Modal class=60-80 (highest freq 61). l=60,f₁=61,f₀=52,f₂=38,h=20

Mode=60+[(61−52)/(122−52−38)]×20

✓≈ 65.63 hours

Q12

Mean of lifetime data.

Midpoints 10,30,50,70,90,110. Σf=225

Σfx=100+1050+2600+4270+3420+3190=14630

✓Mean ≈ 65.02 hours

Q13

Median 525, n=100. Find x, y.

Known frequencies sum to 76, so x+y=24. Median class is 500-600, l=500,f=20,h=100

525=500+[(50−(36+x))/20]×100

Solve for x

25=5(14−x) → x=9

✓x = 9, y = 24−9 = 15

Q14

Assumed mean A=50, freqs 6,8,10,4,2.

Midpoints 10,30,50,70,90. d=x−A: −40,−20,0,20,40

Σfd=6(−40)+8(−20)+10(0)+4(20)+2(40)=−240−160+0+80+80=−240

✓Mean = 50 + (−240/30) = 42

Q15

Mean=25, freqs 7,f,8,10,5. Find f.

Midpoints 5,15,25,35,45. Σfx=810+15f, Σf=30+f

(810+15f)/(30+f)=25

Solve

810+15f=750+25f → 60=10f

✓f = 6

Q16

Mode, freqs 25,34,50,42,38,14.

Modal class=35-40 (highest 50). l=35,f₁=50,f₀=34,f₂=42,h=5

Mode=35+[(50−34)/(100−34−42)]×5

✓≈ 38.33

Q17

Ages mode, freqs 6,11,21,23,14,5.

Modal class=35-45 (highest 23). l=35,f₁=23,f₀=21,f₂=14,h=10

Mode=35+[(23−21)/(46−21−14)]×10

✓≈ 36.82

Q18

Median for same ages data (n=80).

Cumulative freq: 6,17,38,61,75,80. n/2=40, median class=35-45(cf=61 first≥40)

l=35,cf=38,f=23,h=10

✓Median = 35+[(40−38)/23]×10 ≈ 35.87

Q19

Mean for same ages data, direct method.

Midpoints 10,20,30,40,50,60. Σf=80

Σfx=60+220+630+920+700+300=2830

✓Mean = 2830÷80 = 35.375

Q20

Why does assumed mean = direct method?

✓Mean = A + Σf(x−A)/Σf = A + (Σfx − AΣf)/Σf = A + Σfx/Σf − A = Σfx/Σf — which is exactly the direct-method formula. The two methods are algebraically identical, just different arithmetic paths to the same answer.

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