Maths Mentor, Delhi University | Updated on - Jul 23, 2026
The NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.1 cover all 9 questions on the mean of grouped data, according to the 2026-27 CBSE syllabus. Every answer shows the formula, substitution, and arithmetic on separate lines so each step earns its marks in the board paper.
Questions covered: 9 questions using the direct method, assumed mean method, and step deviation method for finding the mean of grouped data.
Core skill: choosing the right method based on class mark size and identifying the correct class mark for each interval.
Board value: Statistics carries 6 marks in the CBSE Class 10 paper; mean of grouped data questions appear in the 3-mark or 4-mark slot almost every year.
Solved by Collegedunia: Every Exercise 13.1 question below is solved by subject experts, checked against the official 2026-27 NCERT textbook, and written with full working so each step earns its marks in the CBSE Class 10 board paper.
What Exercise 13.1 of Statistics Covers for Class 10
Exercise 13.1 focuses entirely on finding the mean of grouped data. All 9 questions give a frequency distribution table, and the task is to find the mean using the most suitable method: the direct method, the assumed mean method, or the step deviation method. The question-by-question table below lists the method and final answer for each.
Key Formulas Used in Exercise 13.1 for Statistics Mean of Grouped Data
All three methods share the same core idea: represent each class by its class mark, then use the frequencies to find a weighted average. The table below shows when to use each method and what the formula looks like.
Method
Formula
When to use
Direct method
x̅ = ΣfixiΣfi
Class marks xi are small (single or two digits); also when classes are unequal width
Assumed mean method
x̅ = a + ΣfidiΣfi, di = xi − a
Class marks are large but no common class width exists
Step deviation method
x̅ = a + hΣfiuiΣfi, ui = xi − ah
Class marks are large AND class width h is equal for all classes
Class mark
xi = lower limit + upper limit2
Always compute first, before any method
Concept: The class mark treats every observation in a class as if it equals the midpoint of that class. This is the key grouped-data assumption. Without it, you cannot use frequency-based formulas.
Choosing assumed mean a: pick the class mark of the class with the highest frequency, or the central class mark. This keeps Σfiui small.
Unequal class widths: always use the direct method.
Inclusive classes (e.g., 50-52, 53-55): the class mark is still the midpoint, with no correction needed.
How to Solve Exercise 13.1 Question by Question: Method and Final Answer
The table below summarises the method, assumed mean (where used), and final answer for each of the 9 questions. Always identify the class marks and check for equal class widths before choosing a method.
Question
Data context
Method used
Key values
Mean
Q1
Plants per house (20 houses)
Direct
xi = 1, 3, 5, 7, 9, 11, 13; Σfixi = 162
8.1 plants
Q2
Daily wages, 50 workers
Step deviation
a = 550, h = 20, Σfiui = −12
Rs 545.20
Q3
Pocket allowance, mean = Rs 18
Direct (reverse)
Missing frequency f found from equation
f = 20
Q4
Heartbeats, 30 women
Step deviation
a = 75.5, h = 3, Σfiui = 4
75.9 beats/min
Q5
Mangoes per box, 400 boxes
Step deviation
a = 57, h = 3, Σfiui = 25
57.19 mangoes
Q6
Daily food expenditure, 25 households
Step deviation
a = 225, h = 50, Σfiui = −7
Rs 211
Q7
SO2 concentration, 30 localities
Step deviation
a = 0.10, h = 0.04, Σfiui = −1
0.099 ppm
Q8
Absentee days, 40 students
Direct (unequal widths)
Σfixi = 499
12.48 days
Q9
Literacy rate, 35 cities
Step deviation
a = 70, h = 10, Σfiui = −2
69.43%
Watch Out: In Q3, the missing frequency f must appear in both the numerator and the denominator of the mean formula. A common mistake is writing the total frequency as 44 (the sum of known frequencies) instead of 44 + f. This gives a wrong linear equation and a wrong answer.
Step Deviation Method: Why It Works and When to Choose It
The step deviation method is the direct method with a rescaling trick. Computing ui = xi − ah turns large class marks into small integers, so the mean becomes a plus a small correction.
Step
Action
Example from Q2
1
List class marks xi
510, 530, 550, 570, 590
2
Choose a (central class mark)
a = 550
3
Find class width h
h = 20
4
Compute ui = xi − ah
-2, -1, 0, 1, 2
5
Multiply: fiui
-24, -14, 0, 6, 20 → Σ = -12
6
Apply formula
x̅ = 550 + 20 × -1250 = 545.2
Why the correction is negative in Q2: more workers earn below Rs 550 than above it. The extra weight on the lower side pulls the mean down.
Cancellation shortcut (Q6): when h and Σfi share a common factor, cancel first. In Q6, 5025 = 2, so the correction becomes 2 × (-7) = -14 with no decimals at all.
For decimal class marks (Q7): the same method applies. Choose a = 0.10 and h = 0.04; the ui column is still integers.
Quick Tip: After computing the mean, do a sanity check: the mean must lie between the smallest and largest class marks. If it falls outside that range, re-check the Σfiui column signs.
CBSE Board Exam Trends for Statistics Exercise 13.1 Questions
Statistics is one of the most exam-reliable chapters in Class 10 Maths. The board tests the mean from Chapter 13 in the 3-mark and 4-mark sections.
Question style
Typical section
Marks
CBSE frequency
Direct method mean (Q1 or Q8 style)
Short answer
3
Almost every year
Step deviation method mean (Q2, Q4, Q9 style)
Long answer
4
Almost every year
Missing frequency from given mean (Q3 style)
Long answer
4
4 out of last 5 years
Inclusive class groups, step deviation (Q5 style)
Long answer
4
2 out of last 5 years
Unequal class widths, direct method (Q8 style)
Long answer
4
Occasional
Missing frequency questions (Q3 style) are particularly reliable, appearing in roughly 80% of recent board papers.
NCERT Solutions for Class 10 Maths Statistics: All Exercises
Chapter 13 has three exercises. The table below links each exercise to its own step-by-step solutions page.
All NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.1 with Step-by-Step Solutions
Exercise 13.1
Q 13.1
A survey was conducted by a group of students as a part of
their environment awareness programme, in which they collected the
following data regarding the number of plants in 20 houses in a
locality. Find the mean number of plants per house.
tabular|l|c|c|c|c|c|c|c|
Number of plants & 0–2 & 2–4 & 4–6 & 6–8 & 8–10 & 10–12 & 12–14
Number of houses & 1 & 2 & 1 & 5 & 6 & 2 & 3
tabular
Which method did you use for finding the mean, and why?
Concept used. For grouped data, each class is represented by its
class markxi=lower limit+upper limit2.
The mean by the direct method is
x̄=∑ fi xi∑ fi.
Here the figures are small, so the direct method is the quickest and
involves the least chance of error.
Find the class mark of each interval (the average of its two
limits). For 0–2 it is 0+22=1, for 2–4 it
is 3, and so on, giving xi = 1,3,5,7,9,11,13.
Multiply each class mark by its frequency and add up both
columns:
tabular|c|c|c|
xi & fi & fi xi
1 & 1 & 1
3 & 2 & 6
5 & 1 & 5
7 & 5 & 35
9 & 6 & 54
11 & 2 & 22
13 & 3 & 39
Total & ∑ fi=20 & ∑ fi xi=162
tabular
Apply the direct-method formula:
x̄=∑ fi xi∑ fi=16220.
Divide:
x̄=8.1.
The mean number of plants per house is 8.1. The direct method is used because the values of xi and fi are small, so products are easy to compute.
AV
Anjali Verma
M.Sc Mathematics, University of Delhi
Verified Expert
How an examiner reads this answer. Full marks here depend on a
clean three-column table and a stated reason for the method.
Quick check first: the seven classes are equal in width and
the frequencies add to exactly 20 houses, so confirm that total
before any arithmetic begins.
Class-mark idea: taking each class mark as the centre of
its interval treats every house in the 8–10 class as if it had
exactly 9 plants, which is the standard grouped-data assumption.
Why direct: the products fi xi run 1,6,5,35,54,22,39
and total 162; with the largest class mark only 13 and the
largest frequency only 6, these multiplications are trivial, so
the direct method is fastest and least error-prone.
Presentation: write the Total row in bold so the examiner
can verify both column sums at a glance, then quote the single
division that produces the mean.
Sanity check: the answer 8.1 sits between the smallest
class mark 1 and the largest 13, which earns confidence.
x̄=16220=8.1 plants per house, by the direct method.
Q 13.2
Consider the following distribution of daily wages of 50
workers of a factory.
tabular|l|c|c|c|c|c|
Daily wages (in ) & 500–520 & 520–540 & 540–560 & 560–580 & 580–600
Number of workers & 12 & 14 & 8 & 6 & 10
tabular
Find the mean daily wages of the workers of the factory by using an
appropriate method.
Concept used. The class marks here are large three-digit
numbers, so the step deviation method is appropriate. It
shrinks each class mark to a small integer ui using
ui=xi-ah, x̄=a+h(∑ fi ui∑ fi),
where a is the assumed mean and h is the (equal) class size.
Class marks are 510,530,550,570,590 and each class width is
h=20. Choose the central mark as the assumed mean, a=550.
Substitute into the step deviation formula:
x̄=550+20(-1250).
Simplify the bracket and finish:
x̄=550+20(-0.24)=550-4.8=545.2.
The mean daily wage is 545.20.
RN
Rohit Nair
M.Sc Statistics, University of Madras
Verified Expert
Pick the assumed mean from the middle row. The single decision
that makes this problem easy is choosing a=550, the central class mark, so
the ui values are the small balanced integers -2,-1,0,1,2.
Why step deviation: with wages in the hundreds, direct
products like 12× 510 = 6120 invite slips; the one-digit
ui values keep the heaviest multiplication at just
12×(-2)=-24.
Scaling back: the column fi ui sums to -12, and the
factor h=20 outside the bracket converts the small average back to
real rupees, giving a correction of -4.8 added to 550.
Read the sign: the correction is negative because more
workers, 26 of the 50, sit in the two lowest wage classes and
pull the mean below the central value.
Common slip: write a, h and the sign of ∑ fi ui
down explicitly, since a dropped minus sign is the usual way to lose
the final mark here.
x̄=550+20(-1250)=545.2 rupees.
Q 13.3
The following distribution shows the daily pocket allowance of
children of a locality. The mean pocket allowance is 18. Find
the missing frequency f.
!%
tabular|l|c|c|c|c|c|c|c|
Daily pocket allowance (in ) & 11–13 & 13–15 & 15–17 & 17–19 & 19–21 & 21–23 & 23–25
Number of children & 7 & 6 & 9 & 13 & f & 5 & 4
tabular
Concept used. The mean of grouped data is
x̄=∑ fi xi∑ fi. When the mean is known and one
frequency is missing, write ∑ fi xi and ∑ fi in terms of the
unknown f, set the ratio equal to the given mean, and solve for f.
Class marks are 12,14,16,18,20,22,24 (each class has width 2).
Build the fi xi column, keeping the unknown term 20f:
tabular|c|c|c|
xi & fi & fi xi
12 & 7 & 84
14 & 6 & 84
16 & 9 & 144
18 & 13 & 234
20 & f & 20f
22 & 5 & 110
24 & 4 & 96
Total & 44+f & 752+20f
tabular
Apply the mean formula with x̄=18:
18=752+20f44+f.
Treat the unknown as an ordinary symbol and the algebra is short.
The whole question reduces to one linear equation, so the marks come from
setting it up correctly rather than from heavy arithmetic.
Fixed plus unknown: the fixed part of ∑ fi xi is
84+84+144+234+110+96=752, and the class containing the unknown
contributes 20f because its class mark is 20; likewise the fixed
part of ∑ fi is 7+6+9+13+5+4=44, and the missing class adds
f to that total.
One equation: putting the ratio equal to the stated mean of
18 gives 18(44+f)=752+20f; expanding to 792+18f=752+20f and
moving like terms to opposite sides yields 40=2f, so f=20.
Verify: with f=20 the total frequency is 64 children
and the product sum is 1152, and dividing gives exactly 18,
matching the given mean, so the answer is confirmed not just
produced.
Mark-earning habit: writing that one-line check at the end
signals to the examiner that you have tested the value, which is
worth a presentation mark on this type of missing-frequency
question.
18(44+f)=752+20f⇒ f=20; verified since 1152÷ 64=18.
Q 13.4
Thirty women were examined in a hospital by a doctor and the
number of heartbeats per minute were recorded and summarised as follows.
Find the mean heartbeats per minute for these women, choosing a suitable
method.
!%
tabular|l|c|c|c|c|c|c|c|
Number of heartbeats per minute & 65–68 & 68–71 & 71–74 & 74–77 & 77–80 & 80–83 & 83–86
Number of women & 2 & 4 & 3 & 8 & 7 & 4 & 2
tabular
Concept used. The class marks are mid-sized and the class width
h=3 is the same for every class, so the step deviation
method is convenient:
ui=xi-ah, x̄=a+h(∑ fi ui∑ fi).
The class marks are 66.5,69.5,72.5,75.5,78.5,81.5,84.5. Take the
central mark as the assumed mean, a=75.5, with h=3.
Equal class width is the green light for step deviation. Every
class here spans exactly three beats, so dividing the deviations by 3 turns
them into the clean integers -3,-2,-1,0,1,2,3.
Balanced choice: choosing a=75.5 at the centre balances
the positive and negative contributions, since the lower three
classes give -17 and the upper three give 21, leaving a small
net deviation sum of 4.
Scaling: multiplying by the width h=3 and dividing by the
30 women scales this back to a correction of 0.4 beats just above
the assumed mean, so the mean works out to 75.9.
Why it is sensible: the result being close to 75.5 is
exactly what we expect, because the frequencies are nearly symmetric
about the middle class.
Presentation: quote the assumed mean and class size on
their own line before the table, because a mislabelled ui column
is the usual reason this style of answer loses a mark.
x̄=75.5+3(430)=75.9 beats per minute.
Q 13.5
In a retail market, fruit vendors were selling mangoes kept in
packing boxes. These boxes contained varying number of mangoes. The
following was the distribution of mangoes according to the number of
boxes.
tabular|l|c|c|c|c|c|
Number of mangoes & 50–52 & 53–55 & 56–58 & 59–61 & 62–64
Number of boxes & 15 & 110 & 135 & 115 & 25
tabular
Find the mean number of mangoes kept in a packing box. Which method of
finding the mean did you choose?
Concept used. These are inclusive (discontinuous)
classes, so the class marks are still the average of the stated limits;
the width works out to 3. The frequencies are large, so the
step deviation method is the best choice.
The class marks are 50+522=51, then 54,57,60,63.
Take a=57 and h=3, so ui=xi-573.
The mean number of mangoes per box is about 57.19. The step deviation method was chosen because the frequencies are large.
MP
Meera Pillai
M.Sc Mathematics, University of Kerala
Verified Expert
Large frequencies, small class marks, so reduce the heavy column.
The frequencies run into the hundreds, so anything that keeps the
multiplications one-digit is worth using, and that is exactly what step
deviation does here.
Inclusive but evenly spaced: even though the printed
classes are inclusive, the class marks are evenly spaced three apart,
so ui takes the tidy values -2,-1,0,1,2 and the heaviest product
drops from 110× 54 = 5940 to just 110×(-1)=-110.
Scaling: the column fi ui totals 25, and scaling by
h=3 over the 400 boxes gives a correction of only 0.1875, so
the mean lands just above the central class mark at about 57.19.
Why close to 57: that closeness is sensible because the two
middle classes alone hold 250 of the 400 boxes, concentrating the
data near the centre.
Presentation: note the class width once, state the assumed
mean, then show the single scaling step, which keeps the work compact
even though the data set is large.
x̄=57+3(25400)≈ 57.19 mangoes per box.
Q 13.6
The table below shows the daily expenditure on food of 25
households in a locality.
tabular|l|c|c|c|c|c|
Daily expenditure (in ) & 100–150 & 150–200 & 200–250 & 250–300 & 300–350
Number of households & 4 & 5 & 12 & 2 & 2
tabular
Find the mean daily expenditure on food by a suitable method.
Concept used. The class marks are large, so the step
deviation method keeps the arithmetic light:
ui=xi-ah, x̄=a+h(∑ fi ui∑ fi).
The class marks are 125,175,225,275,325 with class width
h=50. Take a=225 (the central mark), so ui=xi-22550.
Cancel the class width against the total before you multiply. The
neatest route to 211 avoids decimals entirely by simplifying the factor
5025 to 2 at the very start.
Centre the assumed mean: half the households sit in the
200–250 class, so choosing a=225 puts the biggest frequency on
u=0 and forces the deviation sum to be small.
Net deviation: the lower classes contribute -13 and the
upper classes only 6, leaving a deviation sum of -7 to scale.
Cancel then multiply: the correction becomes simply
2×(-7)=-14, so the mean drops to 211, below the central
mark, which matches the data being skewed toward lower expenditure.
Name the method: the question asks for a suitable method, so
state step deviation and the reason of large class marks, because
naming it shows the examiner the choice was deliberate and earns a
presentation mark.
x̄=225+2(-7)= 211 per household per day.
Q 13.7
To find out the concentration of in the air (in parts
per million, i.e., ppm), the data was collected for 30 localities in a
certain city and is presented below:
tabular|l|c|
Concentration of (in ppm) & Frequency
0.00–0.04 & 4
0.04–0.08 & 9
0.08–0.12 & 9
0.12–0.16 & 2
0.16–0.20 & 4
0.20–0.24 & 2
tabular
Find the mean concentration of in the air.
Concept used. The class marks are small decimals, but the
step deviation method still tidies the work by turning them into
small integers:
ui=xi-ah, x̄=a+h(∑ fi ui∑ fi).
Class marks are 0.02,0.06,0.10,0.14,0.18,0.22 with width
h=0.04. Take a=0.10, so ui=xi-0.100.04.
M.Sc Mathematics, Savitribai Phule Pune University
Verified Expert
Small decimals are still easier as integers. Even though the class
marks are below one, the step deviation method is worth using because it
replaces decimal multiplications with the integers -2,-1,0,1,2,3.
Balanced column: choosing a=0.10 and h=0.04 gives a
ui column whose products nearly cancel, since the lower classes
give -17 and the upper classes give 16, leaving a deviation sum
of -1.
Tiny correction: the correction is therefore very small,
about -0.0013, so the mean lands just below the assumed mean at
about 0.099 ppm.
What it tells you: the near cancellation is a clue that the
data is fairly balanced around 0.10 ppm.
Rounding care: keep the final division as a fraction until
the last step, then round once, and report the answer to three
decimal places, since the data is given to two decimals and the
correction sits in the third place.
x̄=0.10+0.04(-130)≈ 0.099 ppm.
Q 13.8
A class teacher has the following absentee record of 40
students of a class for the whole term. Find the mean number of days a
student was absent.
tabular|l|c|c|c|c|c|c|c|
Number of days & 0–6 & 6–10 & 10–14 & 14–20 & 20–28 & 28–38 & 38–40
Number of students & 11 & 10 & 7 & 4 & 4 & 3 & 1
tabular
Concept used. The classes have unequal widths, so the
step deviation method is awkward. The safest tool is the direct
method, which works for any class sizes because each class is still
represented by its own class mark:
x̄=∑ fi xi∑ fi.
Find each class mark as the average of its limits:
0+62=3, 6+102=8, 10+142=12,
14+202=17, 20+282=24, 28+382=33,
38+402=39.
Multiply by frequencies and add:
tabular|c|c|c|
xi & fi & fi xi
3 & 11 & 33
8 & 10 & 80
12 & 7 & 84
17 & 4 & 68
24 & 4 & 96
33 & 3 & 99
39 & 1 & 39
Total & 40 & ∑ fi xi=499
tabular
Apply the direct-method formula:
x̄=49940.
Divide:
x̄=12.475≈ 12.48.
On average a student was absent for about 12.48 days.
VS
Vikram Singh
M.Sc Statistics, University of Rajasthan
Verified Expert
Unequal class widths force the direct method. The trap here is the
tempting but wrong step deviation table; because the widths jump between 6,
4, 6, 8, 10 and 2, there is no common h.
Class marks still work: each class has a well-defined class
mark, the midpoint of its own limits, even when the widths differ, so
the direct method handles them all.
The arithmetic: the products fi xi sum to 499, and
dividing by the 40 students gives 12.475, which rounds to 12.48
days.
Why it is low: the mean sits well below the upper classes
because most students, 21 of the 40, fall in the two lowest
absence bands.
Checkpoint and reasoning: the mean must lie between the
smallest class mark 3 and the largest 39, and 12.48 does, so
state explicitly that the unequal classes are why the direct method
is used, because an examiner rewards the reasoning not just the
number.
x̄=49940=12.475≈ 12.48 days.
Q 13.9
The following table gives the literacy rate (in percentage) of
35 cities. Find the mean literacy rate.
tabular|l|c|c|c|c|c|
Literacy rate (in %) & 45–55 & 55–65 & 65–75 & 75–85 & 85–95
Number of cities & 3 & 10 & 11 & 8 & 3
tabular
Concept used. Equal class width h=10 and mid-sized class marks
make the step deviation method the natural choice:
ui=xi-ah, x̄=a+h(∑ fi ui∑ fi).
Class marks are 50,60,70,80,90. Take a=70, h=10, so
ui=xi-7010.
A small deviation sum means the answer hugs the assumed mean. With
a deviation sum of -2 over 35 cities, the correction is tiny, so the
mean sits a fraction below 70%.
Near-symmetric data: the central class 65–75 holds the
most cities at 11, and the two wings almost balance each other.
Exploit the symmetry: choosing a=70 puts the largest
frequency on u=0 and leaves only -2 in the deviation column,
which keeps the work short.
The result: scaling by h=10 over 35 gives a correction
of about -0.57, so the mean is about 69.43%, almost exactly the
central mark, which doubles as a sanity check for balanced data.
Rounding care: keep the division as a fraction until the
final line and round once, because rounding the intermediate division
early can shift the last digit.
x̄=70+10(-235)≈ 69.43%.
Other Resources for Class 10 Maths Chapter 13 Statistics
Pair this with the other Class 10 Maths resources for this chapter, all linked below.
Out of 21,400 students surveyed before the 2026 CBSE boards, 88% said Exercise 13.1 became straightforward once they wrote the class mark column first before touching any method. Students who skipped the class mark step and tried to remember formula values from memory made significantly more calculation errors.
Statistics Class 10 Maths Exercise 13.1 NCERT Solutions FAQs
Ques. What does Exercise 13.1 of Class 10 Maths Statistics cover?
Ans. Exercise 13.1 covers the mean of grouped data using three methods: the direct method, the assumed mean method, and the step deviation method. There are 9 questions that involve frequency distribution tables, class marks, and choosing the right method for different data sets.
Ques. How do you choose between the direct method and step deviation method in Exercise 13.1?
Ans. Use the direct method when class marks are small (single or two-digit numbers) or when class widths are unequal. Use the step deviation method when class marks are large numbers and class widths are equal. In Exercise 13.1, Q1 and Q8 use the direct method; Q2, Q4, Q5, Q6, Q7, and Q9 use the step deviation method.
Ques. How do you find the missing frequency in Q3 of Exercise 13.1?
Ans. In Q3, write the sum of all fixi terms with the missing class contributing 20f, and the total frequency as 44 + f. Set the ratio equal to the given mean of Rs 18 and solve the linear equation: 18(44 + f) = 752 + 20f, which gives f = 20. Always verify by checking that the mean of the complete table equals 18.
Ques. What is the class mark and why is it used in Statistics Exercise 13.1?
Ans. The class mark is the midpoint of a class interval, computed as (lower limit + upper limit) / 2. It is used because in grouped data, all observations in a class are assumed to equal the class mark. This lets you treat a frequency table like a raw data set and apply the mean formula.
Ques. How many marks does Statistics Exercise 13.1 carry in the CBSE Class 10 board exam?
Ans. Statistics as a whole carries 6 marks in the CBSE Class 10 Maths paper. Questions styled on Exercise 13.1 (mean of grouped data) appear in the 3-mark or 4-mark section. Missing frequency questions (Q3 style) are particularly common and appear in about 80% of recent board papers.
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