Class 11 Maths NCERT Solutions Chapter 13 Ex 13.1 – Statistics | Boundless Maths
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Chapter 13 · Statistics

Class 11 Maths NCERT Solutions Chapter 13 Ex 13.1

Complete step-by-step solutions for Exercise 13.1 of Statistics — mean deviation about the mean and about the median, for ungrouped data, discrete frequency distributions, and continuous frequency distributions. Every question is worked with full, properly laid-out tables, in exam-ready detail as per the CBSE 2026-27 syllabus.

12Questions Solved
Ex 13.1Statistics
2026-27CBSE Syllabus

Class 11 Maths NCERT Solutions Chapter 13 Ex 13.1 — All 12 Questions

1

Find the mean deviation about the mean for the data: 4,7,8,9,10,12,13,17.

Easy +
Solution

Step 1: Find the mean

\bar{x}=\dfrac{4+7+8+9+10+12+13+17}{8}=\dfrac{80}{8}=10

Step 2: Tabulate the deviations and their absolute values

Deviations from the mean (x̄ = 10)
xi478910121317Total
xi − x̄−6−3−2−10237—
|xi − x̄|6321023724

Step 3: Compute the mean deviation

\text{M.D.}(\bar{x})=\dfrac{1}{n}\sum|x_i-\bar{x}|=\dfrac{24}{8}=3

Mean deviation about the mean = 3.
2

Find the mean deviation about the mean for the data: 38,70,48,40,42,55,63,46,54,44.

Easy +
Solution

Step 1: Find the mean

\bar{x}=\dfrac{38+70+48+40+42+55+63+46+54+44}{10}=\dfrac{500}{10}=50

Step 2: Tabulate the deviations and their absolute values

Deviations from the mean (x̄ = 50)
xi38704840425563465444Total
xi − x̄−1220−2−10−8513−44−6—
|xi − x̄|1220210851344684

Step 3: Compute the mean deviation

\text{M.D.}(\bar{x})=\dfrac{1}{n}\sum|x_i-\bar{x}|=\dfrac{84}{10}=8.4

Mean deviation about the mean = 8.4.
3

Find the mean deviation about the median for the data: 13,17,16,14,11,13,10,16,11,18,12,17.

Medium +
Solution

Step 1: Arrange the data in ascending order

10,\,11,\,11,\,12,\,13,\,13,\,14,\,16,\,16,\,17,\,17,\,18

Step 2: Find the median

There are n=12 observations (even), so the median is the mean of the 6th and 7th observations:

\text{M}=\dfrac{13+14}{2}=\dfrac{27}{2}=13.5

Step 3: Tabulate the absolute deviations from the median

Deviations from the median (M = 13.5)
xi101111121313141616171718Total
|xi − M|3.52.52.51.50.50.50.52.52.53.53.54.528

Step 4: Compute the mean deviation

\text{M.D.}(\text{M})=\dfrac{1}{n}\sum|x_i-\text{M}|=\dfrac{28}{12}=\dfrac{7}{3}\approx2.33

Mean deviation about the median = 7/3 ≈ 2.33.
4

Find the mean deviation about the median for the data: 36,72,46,42,60,45,53,46,51,49.

Medium +
Solution

Step 1: Arrange the data in ascending order

36,\,42,\,45,\,46,\,46,\,49,\,51,\,53,\,60,\,72

Step 2: Find the median

There are n=10 observations (even), so the median is the mean of the 5th and 6th observations:

\text{M}=\dfrac{46+49}{2}=\dfrac{95}{2}=47.5

Step 3: Tabulate the absolute deviations from the median

Deviations from the median (M = 47.5)
xi36424546464951536072Total
|xi − M|11.55.52.51.51.51.53.55.512.524.570

Step 4: Compute the mean deviation

\text{M.D.}(\text{M})=\dfrac{1}{n}\sum|x_i-\text{M}|=\dfrac{70}{10}=7

Mean deviation about the median = 7.
5

Find the mean deviation about the mean for the data:
xi510152025
fi74635

Medium +
Solution

Step 1: Set up the working table and find the mean

Table 1 — Computing the mean and mean deviation
xififixi|xi − x̄|fi|xi − x̄|
5735963
10440416
1569016
20360618
2551251155
TotalN = 25350—158

\bar{x}=\dfrac{1}{\text{N}}\sum f_ix_i=\dfrac{350}{25}=14

(The |x_i-\bar{x}| column above uses this mean of 14 — e.g. for x_i=5: |5-14|=9.)

Step 2: Compute the mean deviation

\text{M.D.}(\bar{x})=\dfrac{1}{\text{N}}\sum f_i|x_i-\bar{x}|=\dfrac{158}{25}=6.32

Mean deviation about the mean = 158/25 = 6.32.
6

Find the mean deviation about the mean for the data:
xi1030507090
fi42428168

Medium +
Solution

Step 1: Set up the working table and find the mean

Table 1 — Computing the mean and mean deviation
xififixi|xi − x̄|fi|xi − x̄|
1044040160
302472020480
5028140000
7016112020320
90872040320
TotalN = 804000—1280

\bar{x}=\dfrac{1}{\text{N}}\sum f_ix_i=\dfrac{4000}{80}=50

Step 2: Compute the mean deviation

\text{M.D.}(\bar{x})=\dfrac{1}{\text{N}}\sum f_i|x_i-\bar{x}|=\dfrac{1280}{80}=16

Mean deviation about the mean = 16.
7

Find the mean deviation about the median for the data:
xi579101215
fi862226

Medium +
Solution

Step 1: Build the cumulative frequency table

The observations are already in ascending order.

Table 1 — Cumulative frequencies
xi579101215
fi862226
c.f.81416182026

Step 2: Find the median

Here \text{N}=26 is even, so the median is the mean of the 13th and 14th observations. Both lie in the cumulative frequency 14, corresponding to x_i=7:

\text{M}=\dfrac{7+7}{2}=7

Step 3: Tabulate the absolute deviations from the median

Table 2 — Mean deviation about the median (M = 7)
xifi|xi − M|fi|xi − M|
58216
7600
9224
10236
122510
156848
TotalN = 26—84

Step 4: Compute the mean deviation

\text{M.D.}(\text{M})=\dfrac{1}{\text{N}}\sum f_i|x_i-\text{M}|=\dfrac{84}{26}=\dfrac{42}{13}\approx3.23

Mean deviation about the median = 42/13 ≈ 3.23.
8

Find the mean deviation about the median for the data:
xi1521273035
fi35678

Medium +
Solution

Step 1: Build the cumulative frequency table

Table 1 — Cumulative frequencies
xi1521273035
fi35678
c.f.38142129

Step 2: Find the median

Here \text{N}=29 is odd, so the median is the \left(\dfrac{29+1}{2}\right)^{\text{th}}=15^{\text{th}} observation. The cumulative frequency just greater than or equal to 15 is 21, corresponding to x_i=30:

\text{M}=30

Step 3: Tabulate the absolute deviations from the median

Table 2 — Mean deviation about the median (M = 30)
xifi|xi − M|fi|xi − M|
1531545
215945
276318
30700
358540
TotalN = 29—148

Step 4: Compute the mean deviation

\text{M.D.}(\text{M})=\dfrac{1}{\text{N}}\sum f_i|x_i-\text{M}|=\dfrac{148}{29}\approx5.10

Mean deviation about the median = 148/29 ≈ 5.10.
9

Find the mean deviation about the mean for the data:
Income per day in ₹0-100100-200200-300300-400400-500500-600600-700700-800
Number of persons489107543

Hard +
Solution

Step 1: Find the mid-point of each class and set up the working table

Table 1 — Computing the mean and mean deviation
ClassfiMid-point xifixi|xi − x̄|fi|xi − x̄|
0-1004502003081232
100-200815012002081664
200-30092502250108972
300-400103503500880
400-5007450315092644
500-60055502750192960
600-700465026002921168
700-800375022503921176
TotalN = 50—17900—7896

\bar{x}=\dfrac{1}{\text{N}}\sum f_ix_i=\dfrac{17900}{50}=358

(The |x_i-\bar{x}| column uses this mean of 358 — e.g. for the mid-point 50: |50-358|=308.)

Step 2: Compute the mean deviation

\text{M.D.}(\bar{x})=\dfrac{1}{\text{N}}\sum f_i|x_i-\bar{x}|=\dfrac{7896}{50}=157.92

Mean deviation about the mean = 7896/50 = 157.92.
10

Find the mean deviation about the mean for the data:
Height in cms95-105105-115115-125125-135135-145145-155
Number of boys91326301210

Hard +
Solution

Step 1: Find the mid-point of each class and set up the working table

Table 1 — Computing the mean and mean deviation
ClassfiMid-point xifixi|xi − x̄|fi|xi − x̄|
95-105910090025.3227.7
105-11513110143015.3198.9
115-1252612031205.3137.8
125-1353013039004.7141
135-14512140168014.7176.4
145-15510150150024.7247
TotalN = 100—12530—1128.8

\bar{x}=\dfrac{1}{\text{N}}\sum f_ix_i=\dfrac{12530}{100}=125.3

(The |x_i-\bar{x}| column uses this mean of 125.3 — e.g. for the mid-point 100: |100-125.3|=25.3.)

Step 2: Compute the mean deviation

\text{M.D.}(\bar{x})=\dfrac{1}{\text{N}}\sum f_i|x_i-\bar{x}|=\dfrac{1128.8}{100}=11.288

Mean deviation about the mean = 1128.8/100 = 11.288.
11

Find the mean deviation about median for the following data :
Marks0-1010-2020-3030-4040-5050-60
Number of Girls68141642

Hard +
Solution

Step 1: Build the cumulative frequency table with mid-points

Table 1 — Cumulative frequencies and mid-points
Classfic.f.Mid-point xi
0-10665
10-2081415
20-30142825
30-40164435
40-5044845
50-6025055

Step 2: Find the median

Here \text{N}=50, so \dfrac{\text{N}}{2}=25. The class whose cumulative frequency is just greater than or equal to 25 is 20\text{-}30 — this is the median class, with l=20, \text{C}=14 (c.f. of the preceding class), f=14, h=10.

\text{Median}=l+\dfrac{\dfrac{\text{N}}{2}-\text{C}}{f}\times h=20+\dfrac{25-14}{14}\times10=20+\dfrac{110}{14}=\dfrac{195}{7}\approx27.86

Step 3: Tabulate the absolute deviations from the median

Table 2 — Mean deviation about the median (M ≈ 27.86)
ClassfiMid-point xi|xi − M|fi|xi − M|
0-106522.86137.1
10-2081512.86102.9
20-3014252.8640
30-4016357.14114.3
40-5044517.1468.6
50-6025527.1454.3
TotalN = 50——517.1

Step 4: Compute the mean deviation

\text{M.D.}(\text{M})=\dfrac{1}{\text{N}}\sum f_i|x_i-\text{M}|=\dfrac{1}{50}\times\dfrac{3620}{7}=\dfrac{362}{35}\approx10.34

Mean deviation about the median = 362/35 ≈ 10.34.
12

Calculate the mean deviation about median age for the age distribution of 100 persons given below:
Age (in years)16-2021-2526-3031-3536-4041-4546-5051-55
Number5612142612169
[Hint: Convert the given data into a continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.]

Hard +
Solution

Step 1: Convert the data into a continuous frequency distribution

Following the hint, 0.5 is subtracted from every lower limit and added to every upper limit:

Table 1 — Converting to continuous class intervals
Original classContinuous classfic.f.Mid-point xi
16-2015.5-20.55518
21-2520.5-25.561123
26-3025.5-30.5122328
31-3530.5-35.5143733
36-4035.5-40.5266338
41-4540.5-45.5127543
46-5045.5-50.5169148
51-5550.5-55.5910053

Step 2: Find the median

Here \text{N}=100, so \dfrac{\text{N}}{2}=50. The class whose cumulative frequency is just greater than or equal to 50 is 35.5\text{-}40.5 — this is the median class, with l=35.5, \text{C}=37, f=26, h=5.

\text{Median}=l+\dfrac{\dfrac{\text{N}}{2}-\text{C}}{f}\times h=35.5+\dfrac{50-37}{26}\times5=35.5+2.5=38

Step 3: Tabulate the absolute deviations from the median

Table 2 — Mean deviation about the median (M = 38)
ClassfiMid-point xi|xi − M|fi|xi − M|
15.5-20.551820100
20.5-25.56231590
25.5-30.5122810120
30.5-35.51433570
35.5-40.5263800
40.5-45.51243560
45.5-50.5164810160
50.5-55.595315135
TotalN = 100——735

Step 4: Compute the mean deviation

\text{M.D.}(\text{M})=\dfrac{1}{\text{N}}\sum f_i|x_i-\text{M}|=\dfrac{735}{100}=7.35

Mean deviation about the median age = 7.35 years.

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Common Questions

Class 11 Maths NCERT Solutions Chapter 13 Ex 13.1 — FAQs

How many questions are there in Exercise 13.1?
Exercise 13.1 has 12 questions on mean deviation — about the mean for Q1, 2, 5, 6, 9, 10, and about the median for Q3, 4, 7, 8, 11, 12 — covering ungrouped data, discrete frequency distributions, and continuous frequency distributions.
What is the formula for mean deviation about the mean?
For ungrouped data, mean deviation about the mean is M.D.(x̄) = (1/n) Σ|xi − x̄|, the average of the absolute deviations of each observation from the mean. For frequency data, it is M.D.(x̄) = (1/N) Σ fi|xi − x̄|, where N is the sum of the frequencies.
Where can I find the official NCERT textbook for this chapter?
Statistics is Chapter 13 of the NCERT Class 11 Mathematics textbook, published by the National Council of Educational Research and Training (NCERT) and prescribed by CBSE. You can download the official textbook PDF directly from ncert.nic.in, NCERT's official website — the solutions on this page follow the exercise exactly as it appears there.

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