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Statistics — CBSE Class 10 Maths Important Questions

13 CBSE Class 10 Maths practice questions on Statistics, each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 30 September 2026

Last updated 25 October 2026

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Statistics — CBSE Class 10 Maths Important Questions

Mean, Median, Mode — No Mix-Ups

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Quick answer

For grouped data: Mean =a+∑fidi∑fi=a+\dfrac{\sum f_id_i}{\sum f_i}=a+(Σ f_id_i)/(Σ f_i) (assumed-mean method), Median =l+N2−cff×h=l+\dfrac{\frac{N}{2}-cf}{f}\times h=l+N/2-cff× h, Mode =l+f1−f02f1−f0−f2×h=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times h=l+(f₁-f₀)/(2f₁-f₀-f₂)× h. The empirical relation is Mode=3 Median−2 Mean\text{Mode}=3\,\text{Median}-2\,\text{Mean}Mode=3 Median-2 Mean.

Syllabus reference (2026–27): NCERT (National Council of Educational Research and Training) · CBSE (Central Board of Secondary Education)

About Statistics

This chapter is about the three measures of central tendency for grouped (continuous) data: mean, median and mode. It is a formula-driven chapter, and questions often include a missing frequency to find. Identify the correct class first (median class = the class where cumulative frequency first reaches N/2N/2N/2; modal class = the class with the highest frequency) before substituting into the formula.

Mean of grouped data (direct, assumed-mean, step-deviation)Median of grouped dataMode of grouped dataEmpirical relationship: Mode = 3 Median minus 2 MeanFinding missing frequencies; identifying median and modal class

Key concepts & formulas

Mean (assumed-mean method)

Take an assumed mean aaa (a convenient class mark), compute di=xi−ad_i=x_i-ad_i=x_i-a, then xˉ=a+∑fidi∑fi\bar{x}=a+\dfrac{\sum f_id_i}{\sum f_i}x̄=a+(Σ f_id_i)/(Σ f_i). The direct method xˉ=∑fixi∑fi\bar{x}=\dfrac{\sum f_ix_i}{\sum f_i}x̄=(Σ f_ix_i)/(Σ f_i) gives the same answer, where xix_ix_i is the class mark.

Median of grouped data

Median=l+N2−cff×h\text{Median}=l+\dfrac{\frac{N}{2}-cf}{f}\times hMedian=l+N/2-cff× h, where l=l=l= lower limit of the median class, N=∑fiN=\sum f_iN=Σ f_i, cf=cf=cf= cumulative frequency of the class before the median class, f=f=f= frequency of the median class, h=h=h= class width.

Mode of grouped data

Mode=l+f1−f02f1−f0−f2×h\text{Mode}=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times hMode=l+(f₁-f₀)/(2f₁-f₀-f₂)× h, where l=l=l= lower limit of the modal class (highest-frequency class), f1=f_1=f₁= its frequency, f0=f_0=f₀= frequency of the preceding class, f2=f_2=f₂= frequency of the following class, h=h=h= class width.

Empirical relationship

For a moderately skewed distribution, Mode=3 Median−2 Mean\text{Mode}=3\,\text{Median}-2\,\text{Mean}Mode=3 Median-2 Mean. If any two are known, the third can be estimated.

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Important questions with answers

Try each on paper first, then reveal the model answer to check your method.

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The class mark of the class interval 151515–353535 is:

  1. (a)

    252525

  2. (b)

    202020

  3. (c)

    505050

  4. (d)

    101010

Show model answer

Answer: (a) 252525

Class mark =lower limit+upper limit2=15+352=25.=\dfrac{\text{lower limit}+\text{upper limit}}{2}=\dfrac{15+35}{2}=25.=(lower limit+upper limit)/2=(15+35)/2=25.

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Q2MCQEasy1 mark

For a distribution, the median is 150150150 and the mean is 148148148. Using the empirical relationship, the mode is:

  1. (a)

    154154154

  2. (b)

    152152152

  3. (c)

    150150150

  4. (d)

    146146146

Show model answer

Answer: (a) 154154154

Mode=3 Median−2 Mean=3(150)−2(148)=450−296=154.\text{Mode}=3\,\text{Median}-2\,\text{Mean}=3(150)-2(148)=450-296=154.Mode=3 Median-2 Mean=3(150)-2(148)=450-296=154.

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Q3MCQModerate1 mark

In the step-deviation method for a grouped distribution, a=45a=45a=45, h=10h=10h=10, ∑fi=60\sum f_i=60Σ f_i=60 and ∑fiui=−12\sum f_iu_i=-12Σ f_iu_i=-12. The mean is:

  1. (a)

    434343

  2. (b)

    474747

  3. (c)

    44.844.844.8

  4. (d)

    45.245.245.2

Show model answer

Answer: (a) 434343

xˉ=a+h×∑fiui∑fi=45+10×−1260=45−2=43.\bar{x}=a+h\times\dfrac{\sum f_iu_i}{\sum f_i}=45+10\times\dfrac{-12}{60}=45-2=43.x̄=a+h× (Σ f_iu_i)/(Σ f_i)=45+10× -12/60=45-2=43.

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Q4MCQHOTS1 mark

For the distribution below, the sum of the lower limits of the median class and the modal class is:

Class0–55–1010–1515–2020–25
Frequency101512209
  1. (a)

    252525

  2. (b)

    151515

  3. (c)

    303030

  4. (d)

    353535

Show model answer

Answer: (a) 252525

N=66, N2=33.N=66,\ \dfrac{N}{2}=33.N=66, N/2=33. Cumulative frequencies: 10,25,37,57,66.10,25,37,57,66.10,25,37,57,66. The c.f. first reaches 333333 in class 101010–151515, so the median class is 101010–151515 (lower limit 101010). The highest frequency 202020 is in 151515–202020, so the modal class is 151515–202020 (lower limit 151515). Sum =10+15=25.=10+15=25.=10+15=25.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): In the formula for the mode of grouped data, f0f_0f₀ is the frequency of the class just before the modal class.

Reason (R): The modal class is the class with the highest cumulative frequency.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

Show model answer

Answer: (c) A is true but R is false.

The modal class is the class with the highest frequency. The highest cumulative frequency always belongs to the last class, so it cannot be used to find the modal class.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

The mean of 555 observations is 181818. If one observation is removed, the mean of the remaining observations becomes 161616. Find the value of the removed observation.

Show model answer

Sum of 555 observations =5×18=90.=5\times 18=90.=5× 18=90.

Sum of remaining 444 observations =4×16=64.=4\times 16=64.=4× 16=64.

Removed observation =90−64=26.=90-64=26.=90-64=26.

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Q7Very ShortModerate2 marks

For the following distribution, write the median class and the modal class.

Class0–1010–2020–3030–4040–50
Frequency581575
Show model answer

N=40, N2=20.N=40,\ \dfrac{N}{2}=20.N=40, N/2=20. Cumulative frequencies: 5,13,28,35,40.5,13,28,35,40.5,13,28,35,40. The c.f. first reaches 202020 in class 202020–303030, so the median class is 202020–303030.

The highest frequency is 151515 (class 202020–303030), so the modal class is also 202020–303030.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Find the mean daily wage from the following data.

Daily wage (Rs)100–120120–140140–160160–180180–200
No. of workers12148610
Show model answer

Class marks xix_ix_i: 110,130,150,170,190.110,130,150,170,190.110,130,150,170,190.

∑fixi=110(12)+130(14)+150(8)+170(6)+190(10)\sum f_ix_i=110(12)+130(14)+150(8)+170(6)+190(10)Σ f_ix_i=110(12)+130(14)+150(8)+170(6)+190(10)
=1320+1820+1200+1020+1900=7260.=1320+1820+1200+1020+1900=7260.=1320+1820+1200+1020+1900=7260.

N=∑fi=50.N=\sum f_i=50.N=Σ f_i=50. Mean =726050=Rs 145.20.=\dfrac{7260}{50}=\text{Rs }145.20.=7260/50=Rs 145.20.

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Q9Short AnswerModerate3 marks

Find the mode of the following distribution of the ages of patients admitted to a hospital.

Age (years)0–1010–2020–3030–4040–50
No. of patients81636346
Show model answer

The highest frequency is 363636, so the modal class is 202020–303030.

Here l=20, f1=36, f0=16, f2=34, h=10.l=20,\ f_1=36,\ f_0=16,\ f_2=34,\ h=10.l=20, f₁=36, f₀=16, f₂=34, h=10.

Mode=l+f1−f02f1−f0−f2×h=20+36−1672−16−34×10=20+2022×10=20+9.09=29.09 years.\text{Mode}=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times h=20+\dfrac{36-16}{72-16-34}\times 10=20+\dfrac{20}{22}\times 10=20+9.09=29.09\ \text{years}.Mode=l+(f₁-f₀)/(2f₁-f₀-f₂)× h=20+(36-16)/(72-16-34)× 10=20+20/22× 10=20+9.09=29.09 years.

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Q10Short AnswerHOTS3 marks

The median of the following distribution is 28.528.528.5 and the total frequency is 606060. Find the missing frequencies xxx and yyy.

Class0–1010–2020–3030–4040–5050–60
Frequency5xxx2015yyy5
Show model answer

Total: 5+x+20+15+y+5=60⇒x+y=15.5+x+20+15+y+5=60\Rightarrow x+y=15.5+x+20+15+y+5=60⇒ x+y=15.

N2=30\dfrac{N}{2}=30N/2=30; since the median 28.528.528.5 lies in 202020–303030, that is the median class (l=20, f=20, h=10l=20,\ f=20,\ h=10l=20, f=20, h=10, cf=5+xcf=5+xcf=5+x).

Median=l+N2−cff×h⇒28.5=20+30−(5+x)20×10.\text{Median}=l+\dfrac{\frac{N}{2}-cf}{f}\times h\Rightarrow 28.5=20+\dfrac{30-(5+x)}{20}\times 10.Median=l+N/2-cff× h⇒ 28.5=20+(30-(5+x))/20× 10.

8.5=25−x2⇒25−x=17⇒x=8.8.5=\dfrac{25-x}{2}\Rightarrow 25-x=17\Rightarrow x=8.8.5=(25-x)/2⇒ 25-x=17⇒ x=8. Then y=15−8=7.y=15-8=7.y=15-8=7.

x=8, y=7.x=8,\ y=7.x=8, y=7.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

For the following frequency distribution, find the mean and the median. Hence, using the empirical relationship, estimate the mode.

Class0–1010–2020–3030–4040–50
Frequency5815166
Show model answer

Mean: class marks 5,15,25,35,45.5,15,25,35,45.5,15,25,35,45. ∑fixi=25+120+375+560+270=1350, N=50.\sum f_ix_i=25+120+375+560+270=1350,\ N=50.Σ f_ix_i=25+120+375+560+270=1350, N=50. Mean =135050=27.=\dfrac{1350}{50}=27.=1350/50=27.

Median: N2=25.\dfrac{N}{2}=25.N/2=25. Cumulative frequencies: 5,13,28,44,50.5,13,28,44,50.5,13,28,44,50. Median class =20=20=20–303030 (l=20, cf=13, f=15, h=10l=20,\ cf=13,\ f=15,\ h=10l=20, cf=13, f=15, h=10).

Median=20+25−1315×10=20+1215×10=20+8=28.\text{Median}=20+\dfrac{25-13}{15}\times 10=20+\dfrac{12}{15}\times 10=20+8=28.Median=20+(25-13)/15× 10=20+12/15× 10=20+8=28.

Mode (empirical): Mode=3 Median−2 Mean=3(28)−2(27)=84−54=30.\text{Mode}=3\,\text{Median}-2\,\text{Mean}=3(28)-2(27)=84-54=30.Mode=3 Median-2 Mean=3(28)-2(27)=84-54=30.

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Q12Long AnswerHOTS5 marks

The mean of the following distribution is 62.862.862.8 and the sum of all frequencies is 505050. Find the missing frequencies f1f_1f₁ and f2f_2f₂.

Class0–2020–4040–6060–8080–100100–120
Frequency5f1f_1f₁10f2f_2f₂78
Show model answer

Total: 5+f1+10+f2+7+8=50⇒f1+f2=20.5+f_1+10+f_2+7+8=50\Rightarrow f_1+f_2=20.5+f₁+10+f₂+7+8=50⇒ f₁+f₂=20. …(1)

Class marks: 10,30,50,70,90,110.10,30,50,70,90,110.10,30,50,70,90,110.

∑fixi=5(10)+f1(30)+10(50)+f2(70)+7(90)+8(110)=2060+30f1+70f2.\sum f_ix_i=5(10)+f_1(30)+10(50)+f_2(70)+7(90)+8(110)=2060+30f_1+70f_2.Σ f_ix_i=5(10)+f₁(30)+10(50)+f₂(70)+7(90)+8(110)=2060+30f₁+70f₂.

Mean =∑fixi50=62.8⇒2060+30f1+70f2=3140⇒3f1+7f2=108.=\dfrac{\sum f_ix_i}{50}=62.8\Rightarrow 2060+30f_1+70f_2=3140\Rightarrow 3f_1+7f_2=108.=(Σ f_ix_i)/50=62.8⇒ 2060+30f₁+70f₂=3140⇒ 3f₁+7f₂=108. …(2)

From (1), f1=20−f2.f_1=20-f_2.f₁=20-f₂. Substituting: 3(20−f2)+7f2=108⇒60+4f2=108⇒f2=12.3(20-f_2)+7f_2=108\Rightarrow 60+4f_2=108\Rightarrow f_2=12.3(20-f₂)+7f₂=108⇒ 60+4f₂=108⇒ f₂=12. Then f1=8.f_1=8.f₁=8.

f1=8, f2=12.f_1=8,\ f_2=12.f₁=8, f₂=12.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

A company tested the life (in hours) of 200200200 electric bulbs. The results are shown below.

Life (hours)300–400400–500500–600600–700700–800
No. of bulbs1456604030

(i) Write the modal class.

(ii) Write the median class.

(iii) Find the median life of the bulbs.

Show model answer

(i) Highest frequency is 606060 (class 500500500–600600600), so the modal class is 500500500–600600600.

(ii) N=200, N2=100.N=200,\ \dfrac{N}{2}=100.N=200, N/2=100. Cumulative frequencies: 14,70,130,170,200.14,70,130,170,200.14,70,130,170,200. The c.f. first reaches 100100100 in 500500500–600600600, so the median class is 500500500–600600600.

(iii) l=500, cf=70, f=60, h=100.l=500,\ cf=70,\ f=60,\ h=100.l=500, cf=70, f=60, h=100.
Median=500+100−7060×100=500+3060×100=500+50=550 hours.\text{Median}=500+\dfrac{100-70}{60}\times 100=500+\dfrac{30}{60}\times 100=500+50=550\ \text{hours}.Median=500+(100-70)/60× 100=500+30/60× 100=500+50=550 hours.

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All CBSE Class 10 Maths Chapters

Real NumbersPolynomialsPair of Linear Equations in Two VariablesQuadratic EquationsArithmetic ProgressionsTrianglesCoordinate GeometryIntroduction to TrigonometrySome Applications of TrigonometryCirclesAreas Related to CirclesSurface Areas and VolumesStatisticsProbability

Related study guide

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Trigonometry Class 10 CBSE: Understand Every Formula in 1 Hour
Sin, cos, tan, identities — all the Class 10 trigonometry formulas explained the way a real tutor would teach them, not memorised.

Frequently asked questions

  • What are the most important questions from Statistics for Class 10?
    Finding the mean, median and mode of grouped data (using the direct, assumed-mean or step-deviation method for the mean, and the standard formulas for median and mode from a frequency distribution table) accounts for most marks in this chapter. Drawing and reading an ogive (cumulative frequency curve) to find the median graphically is the other question type that comes up regularly.
  • What are the important questions in Statistics Class 10?
    The mean by the direct, assumed-mean or step-deviation method, the median and modal classes, missing frequencies when the mean or median is given, and the empirical relation between the three measures. This page has 13 such questions with answers, finishing with a case study on the life of 200200200 electric bulbs.
  • What are the formulas for the mean, median and mode of grouped data?
    Mean (assumed-mean method) =a+∑fidi∑fi=a+\dfrac{\sum f_id_i}{\sum f_i}=a+(Σ f_id_i)/(Σ f_i); median =l+N2−cff×h=l+\dfrac{\frac N2-cf}{f}\times h=l+( N2-cf)/f× h; mode =l+f1−f02f1−f0−f2×h=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times h=l+(f₁-f₀)/(2f₁-f₀-f₂)× h. In the median formula lll belongs to the median class; in the mode formula it belongs to the modal class.
  • How do you use the empirical relationship between mean, median and mode?
    Mode=3 Median−2 Mean\text{Mode}=3\,\text{Median}-2\,\text{Mean}Mode=3 Median-2 Mean. With a median of 150150150 and a mean of 148148148, the mode is 450−296=154450-296=154450-296=154. Use it to check a calculated value or to estimate the third measure when two are given.
  • How do you find ui and the mean by the step-deviation method?
    Take an assumed mean aaa and class width hhh, set ui=xi−ahu_i=\dfrac{x_i-a}{h}u_i=(x_i-a)/h, then xˉ=a+h⋅∑fiui∑fi\bar x=a+h\cdot\dfrac{\sum f_iu_i}{\sum f_i}x=a+h· (Σ f_iu_i)/(Σ f_i). With a=45a=45a=45, h=10h=10h=10, ∑fi=60\sum f_i=60Σ f_i=60 and ∑fiui=−12\sum f_iu_i=-12Σ f_iu_i=-12: xˉ=45+10×(−0.2)=43\bar x=45+10\times(-0.2)=43x=45+10×(-0.2)=43.
  • Where can I find Statistics Class 10 notes?
    Key concepts & formulas, at the top of this page, gives Statistics notes on one screen: the assumed-mean method, the median and mode of grouped data and the empirical relationship. NCERT Exercises 13.1 to 13.3 are solved with full tables on the chapter's NCERT Solutions page.
  • Where can I get Statistics Class 10 extra questions?
    Beyond NCERT, this page sets a mean daily wage, the mode of hospital patients' ages, missing frequencies from a median of 28.528.528.5 and from a mean of 62.862.862.8, a mean and median followed by the mode from the empirical relation, and a bulb-life case study. Model answers show every column of the table.
  • Where can I practise Statistics Class 10 MCQs?
    Five objective questions open the page: the class mark of 151515–353535, a mode from the empirical relation, a step-deviation mean, the lower limits of the median and modal classes, and an assertion–reason item on f0f_0f₀. Reveal each answer after choosing. Create Test in a free account makes timed Statistics quizzes.

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