Statistics NCERT Solutions Exercise 13.1 Class 11 Math PDF Free Download

Anand Classes provides detailed NCERT Solutions for Class 11 Maths Chapter 13 Statistics Exercise 13.1, aimed at helping students understand the fundamental concepts of measures of dispersion, mean deviation, and variance. These step-by-step solutions are prepared as per the latest CBSE and NCERT syllabus, ensuring accuracy and conceptual clarity. Each question is solved in a systematic and easy-to-understand manner to help students build a strong foundation in statistical analysis and data interpretation. Click the print button to download study material and notes.

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Problems Based on Mean Deviation about Mean

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NCERT Question 1: Find the mean deviation about the mean for the data:
$$4, 7, 8, 9, 10, 12, 13, 17$$

Solution:

We make the following table from the given data:

$x_i$	$x_i – \bar{x}$	$\lvert x_i – \bar{x} \rvert$
4	$-6$	$6$
7	$-3$	$3$
8	$-2$	$2$
9	$-1$	$1$
10	$0$	$0$
12	$2$	$2$
13	$3$	$3$
17	$7$	$7$

$$
\sum |x_i – \bar{x}| = 24, \quad n = 8
$$

We know,
$$
\bar{x} = \frac{\sum x_i}{n} = \frac{80}{8} = 10
$$

Mean Deviation about Mean is given by:
$$
\text{M.D.}(\bar{x}) = \frac{\sum |x_i – \bar{x}|}{n} = \frac{24}{8} = 3
$$

$$
\boxed{\text{M.D. about mean} = 3}
$$

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NCERT Question 2: Find the mean deviation about the mean for the data:
$$38, 70, 48, 40, 42, 55, 63, 46, 54, 44$$

Solution:

Here,
$$
n = 10, \quad \bar{x} = \frac{\sum x_i}{n}
$$

$$
\sum x_i = 38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44 = 500
$$

$$
\bar{x} = \frac{500}{10} = 50
$$

Now we prepare the table:

$x_i$	$x_i – \bar{x}$	$\lvert x_i – \bar{x} \rvert$
38	$-12$	$12$
70	$20$	$20$
48	$-2$	$2$
40	$-10$	$10$
42	$-8$	$8$
55	$5$	$5$
63	$13$	$13$
46	$-4$	$4$
54	$4$	$4$
44	$-6$	$6$

$$
\sum |x_i – \bar{x}| = 84
$$

Mean Deviation about Mean is given by:
$$
\text{M.D.}(\bar{x}) = \frac{\sum |x_i – \bar{x}|}{n} = \frac{84}{10} = 8.4
$$

$$
\boxed{\text{M.D. about mean} = 8.4}
$$

For more detailed NCERT Class 11 Statistics solutions and JEE, NDA, and CUET preparation materials, download comprehensive study notes from Anand Classes.

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Problems Based on Mean Deviation about Median

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NCERT Question 3: Find the mean deviation about the median for the data:
$$13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$$

Solution:

Arranging the data in ascending order, we have:
$$10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18$$

Number of observations,
$$n = 12 \ (\text{even})$$

Hence, the median is the mean of the 6th and 7th terms:
$$
M = \frac{13 + 14}{2} = 13.5
$$

Now we prepare the following table:

$x_i$	$x_i – M$	$\lvert x_i – M \rvert$
10	$-3.5$	$3.5$
11	$-2.5$	$2.5$
11	$-2.5$	$2.5$
12	$-1.5$	$1.5$
13	$-0.5$	$0.5$
13	$-0.5$	$0.5$
14	$0.5$	$0.5$
16	$2.5$	$2.5$
16	$2.5$	$2.5$
17	$3.5$	$3.5$
17	$3.5$	$3.5$
18	$4.5$	$4.5$

$$
\sum |x_i – M| = 28
$$

Mean deviation about the median is given by:
$$
\text{M.D.}(M) = \frac{\sum |x_i – M|}{n} = \frac{28}{12} = 2.33
$$

$$
\boxed{\text{M.D. about median} = 2.33}
$$

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NCERT Question 4: Find the mean deviation about the median for the data:
$$36, 72, 46, 42, 60, 45, 53, 46, 51, 49$$

Solution:

Arranging the data in ascending order, we have:
$$36, 42, 45, 46, 46, 49, 51, 53, 60, 72$$

Number of observations,
$$n = 10 \ (\text{even})$$

Hence, the median is the mean of the 5th and 6th terms:
$$
M = \frac{46 + 49}{2} = 47.5
$$

Now we prepare the following table:

$x_i$	$x_i – M$	$\lvert x_i – M \rvert$
36	$-11.5$	$11.5$
42	$-5.5$	$5.5$
45	$-2.5$	$2.5$
46	$-1.5$	$1.5$
46	$-1.5$	$1.5$
49	$1.5$	$1.5$
51	$3.5$	$3.5$
53	$5.5$	$5.5$
60	$12.5$	$12.5$
72	$24.5$	$24.5$

$$
\sum |x_i – M| = 70
$$

Mean deviation about the median is given by:
$$
\text{M.D.}(M) = \frac{\sum |x_i – M|}{n} = \frac{70}{10} = 7
$$

$$
\boxed{\text{M.D. about median} = 7}
$$

For complete NCERT Class 11 Statistics explanations and JEE, NDA, and CUET preparation material, explore in-depth concept notes and solutions by Anand Classes.

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Problems Based on Mean Deviation about Mean using Frequency Table

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NCERT Question 5: Find the mean deviation about the mean for the data:

$x_i$	5	10	15	20	25
$f_i$	7	4	6	3	5

Solution:

We prepare the following table:

$x_i$	$f_i$	$f_i x_i$	$\lvert x_i – \bar{x} \rvert$	$f_i \lvert x_i – \bar{x} \rvert$
5	7	35	9	63
10	4	40	4	16
15	6	90	1	6
20	3	60	6	18
25	5	125	11	55

$$
\sum f_i = 25, \quad \sum f_i x_i = 350, \quad \sum f_i |x_i – \bar{x}| = 158
$$

We know,
$$
\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{350}{25} = 14
$$

Mean deviation about the mean is given by:
$$
\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i – \bar{x}|}{\sum f_i} = \frac{158}{25} = 6.32
$$

$$
\boxed{\text{M.D. about mean} = 6.32}
$$

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NCERT Question 6: Find the mean deviation about the mean for the data:

$x_i$	10	30	50	70	90
$f_i$	4	24	28	16	8

Solution:

We prepare the following table:

$x_i$	$f_i$	$f_i x_i$	$\lvert x_i – \bar{x} \rvert$	$f_i \lvert x_i – \bar{x} \rvert$
10	4	40	40	160
30	24	720	20	480
50	28	1400	0	0
70	16	1120	20	320
90	8	720	40	320

$$
\sum f_i = 80, \quad \sum f_i x_i = 4000, \quad \sum f_i |x_i – \bar{x}| = 1280
$$

We know,
$$
\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4000}{80} = 50
$$

Mean deviation about the mean is given by:
$$
\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i – \bar{x}|}{\sum f_i} = \frac{1280}{80} = 16
$$

$$
\boxed{\text{M.D. about mean} = 16}
$$

For complete NCERT Class 11 Statistics solutions and expert guidance for JEE, NDA, and CUET preparation, refer to comprehensive study notes by Anand Classes.

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Problems Based on Mean Deviation about Median using Frequency Table

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NCERT Question.7 : Find the Mean Deviation about the Median for the following data:
Given Data:
$x_i : 5, 7, 9, 10, 12, 15$
$f_i : 8, 6, 2, 2, 2, 6$

Solution:

The given observations (values of $x$) are already in ascending order.

We first prepare the following table:

$x_i$	$f_i$	C.F.	$\lvert x_i – M \rvert$	$f_i \lvert x_i – M \rvert$
5	8	8	2	16
7	6	14	0	0
9	2	16	2	4
10	2	18	3	6
12	2	20	5	10
15	6	26	8	48
Total	26			84

Here,
$$N = \sum f_i = 26$$

Since $N$ is even, the median is the mean of the $\dfrac{N}{2}$th and $\left(\dfrac{N}{2} + 1\right)$th observations.

That is,
$$
M = \dfrac{x_{\frac{N}{2}} + x_{\frac{N}{2}+1}}{2}
= \dfrac{x_{13} + x_{14}}{2}
$$

From the cumulative frequency table, both the 13th and 14th observations correspond to $x = 7$.

Hence,
$$
M = 7
$$

Now,
$$
\sum f_i \lvert x_i – M \rvert = 84
$$

Therefore, the Mean Deviation about the Median is given by:

$$
\text{M.D.}(M) = \frac{\sum f_i \lvert x_i – M \rvert}{\sum f_i}
= \frac{84}{26} = 3.23
$$

$$
\boxed{\text{Mean Deviation about Median} = 3.23}
$$

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NCERT Question.8 : Find the Mean Deviation about the Median for the following data:
Given Data:
$x_i : 15, 21, 27, 30, 35$
$f_i : 3, 5, 6, 7, 8$

Solution:
The given observations are already in ascending order.

We now prepare the following table:

$x_i$	$f_i$	C.F.	$\lvert x_i – M \rvert$	$f_i \lvert x_i – M \rvert$
15	3	3	15	45
21	5	8	9	45
27	6	14	3	18
30	7	21	0	0
35	8	29	5	40
Total	29			148

Here,
$$N = \sum f_i = 29$$

Since $N$ is odd, the median is the $\left(\dfrac{N + 1}{2}\right)^{th}$ observation.

$$
\text{Median} = \left(\frac{29 + 1}{2}\right)^{th} = 15^{th} \text{ observation}
$$

From the cumulative frequency table, the 15th observation corresponds to $x = 30$.

Hence,
$$
M = 30
$$

Now,
$$
\sum f_i \lvert x_i – M \rvert = 148
$$

Therefore, the Mean Deviation about the Median is:

$$
\text{M.D.}(M) = \frac{\sum f_i \lvert x_i – M \rvert}{\sum f_i}
= \frac{148}{29} = 5.10
$$

$$
\boxed{\text{Mean Deviation about Median} = 5.10}
$$

For more detailed NCERT Statistics Solutions for Class 11 Maths, explore Anand Classes — helping students master Mean, Median, Mode, and Dispersion concepts for JEE, NDA, and CUET preparation.

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NCERT Question 9 : Find the mean deviation about the mean for the following data:

Income (per day)	Number of persons ($f_i$)
0–100	4
100–200	8
200–300	9
300–400	10
400–500	7
500–600	5
600–700	4
700–800	3

Solution:
Let the mid-values of each class be $x_i$.

We take the assumed mean $a = 350$ and class width $h = 100$, 100, form
the following table :

Class Interval	$f_i$	$x_i$	$u_i = \dfrac{x_i – 350}{100}$	$f_i u_i$	$\lvert x_i – \bar{x} \rvert$	$f_i \lvert x_i – \bar{x} \rvert$	$f_i x_i$
0–100	4	50	–3	–12	308	1232	200
100–200	8	150	–2	–16	208	1664	1200
200–300	9	250	–1	–9	108	972	2250
300–400	10	350	0	0	8	80	3500
400–500	7	450	1	7	92	644	3150
500–600	5	550	2	10	192	960	2750
600–700	4	650	3	12	292	1168	2600
700–800	3	750	4	12	392	1176	2250
Total	50			4		7896	17900

Calculation of Mean

$$
\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h
$$

Substituting the values:

$$
\bar{x} = 350 + \frac{4}{50} \times 100 = 350 + 8 = 358
$$

Hence,
$$
\boxed{\bar{x} = 358}
$$

Calculation of Mean Deviation

We know,
$$
\text{M.D.}(\bar{x}) = \frac{\sum f_i \lvert x_i – \bar{x} \rvert}{\sum f_i}
$$

From the table,
$$
\sum f_i \lvert x_i – \bar{x} \rvert = 7896, \quad \sum f_i = 50
$$

Therefore,
$$
\text{M.D.}(\bar{x}) = \frac{7896}{50} = 157.92
$$

$$
\boxed{\text{Mean Deviation about Mean} = 157.92}
$$

Result:
Mean = 358
Mean Deviation about Mean = 157.92

For detailed NCERT solutions and dispersion examples, visit Anand Classes — your trusted guide for JEE, NDA, and CUET Maths preparation.

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NCERT Question 10 : Find the mean deviation about the mean for the following data.

Height (in cms) and Number of boys

Height (in cms)	Number of boys $f_i$
95–105	9
105–115	13
115–125	26
125–135	30
135–145	12
145–155	10

Solution.
Take mid-values $x_i$ of the class-intervals and let the assumed mean $a = 120$, class width $h = 10$ and form the following table:

$x_i$	$f_i$	$u_i=\dfrac{x_i-120}{10}$	$f_i u_i$	$\lvert x_i – \bar{x}\rvert$	$f_i \lvert x_i – \bar{x}\rvert$
100	9	$-2$	$-18$	$25.3$	$227.7$
110	13	$-1$	$-13$	$15.3$	$198.9$
120	26	$0$	$0$	$5.3$	$137.8$
130	30	$1$	$30$	$4.7$	$141.0$
140	12	$2$	$24$	$14.7$	$176.4$
150	10	$3$	$30$	$24.7$	$247.0$
Total	100		53		1128.8

Here,
$$N=\sum f_i=100,\qquad \sum f_i u_i = 53.$$

Using the assumed mean formula,
$$
\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i}\times h
=120 + \frac{53}{100}\times 10 = 120 + 5.3 = 125.3.
$$

Mean deviation about the mean:
$$
\text{M.D.}(\bar{x})=\frac{\sum f_i\lvert x_i-\bar{x}\rvert}{\sum f_i}
= \frac{1128.8}{100} = 11.288\ \text{cm} \approx 11.29\ \text{cm}.
$$

$$\boxed{\text{Mean} = 125.3\ \text{cm},\qquad \text{M.D. about mean} = 11.29\ \text{cm}}$$

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NCERT Question 11 : Find the mean deviation about the median for the following data.

Marks and Number of girls

Class (marks)	Number of girls $f_i$
0–10	6
10–20	8
20–30	14
30–40	16
40–50	4
50–60	2

Solution.
Compute cumulative frequencies and mid-points $x_i$ and form the following table :

Class	$f_i$	C.F.	$x_i$	$\lvert x_i – M \rvert$	$f_i \lvert x_i – M \rvert$
0–10	6	6	5	$22.86$	$137.16$
10–20	8	14	15	$12.86$	$102.88$
20–30	14	28	25	$2.86$	$40.04$
30–40	16	44	35	$7.14$	$114.24$
40–50	4	48	45	$17.14$	$68.56$
50–60	2	50	55	$27.14$	$54.28$
Total	50				517.16

Here $N=\sum f_i=50$.

The median class is the class whose cumulative frequency just exceeds $N/2 = 25$.
From the C.F. column the 25th observation lies in class 20–30 (C.F. = 28). Thus median class is $20$–$30$.

Median by formula:
$$
M = l + \frac{\dfrac{N}{2} – C}{f}\times h,
$$
where $l=20$ (lower limit of median class), $C=14$ (c.f. before median class), $f=14$, $h=10$.

$$
M = 20 + \frac{25-14}{14}\times 10 = 20 + \frac{11}{14}\times 10 = 20 + 7.86 = 27.86.
$$

Now mean deviation about median:
$$
\text{M.D.}(M) = \frac{\sum f_i\lvert x_i – M\rvert}{\sum f_i}
= \frac{517.16}{50} = 10.3432 \approx 10.34\ \text{marks}.
$$

$$\boxed{\text{Median} = 27.86,\qquad \text{M.D. about median} = 10.34\ \text{marks}}$$

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NCERT Question 12 : Calculate the mean deviation about median age for the age distribution of 100 persons:

Age (years)	Number
16–20	5
21–25	6
26–30	12
31–35	14
36–40	26
41–45	12
46–50	16
51–55	9

Solution.
First make the class intervals continuous (since they were given discontinuous): subtract $0.5$ ($=( 21-20)/2 = 0.5$) from lower limits and add $0.5$ to upper limits. So the continuous classes are:

15.5–20.5, 20.5–25.5, 25.5–30.5, 30.5–35.5, 35.5–40.5, 40.5–45.5, 45.5–50.5, 50.5–55.5.

Compute cumulative frequencies:

Class (cont.)	$f_i$	C.F.	$x_i$ (mid)	$\lvert x_i – M \rvert$	$f_i\lvert x_i – M\rvert$
15.5–20.5	5	5	18	$20$	$100$
20.5–25.5	6	11	23	$15$	$90$
25.5–30.5	12	23	28	$10$	$120$
30.5–35.5	14	37	33	$5$	$70$
35.5–40.5	26	63	38	$0$	$0$
40.5–45.5	12	75	43	$5$	$60$
45.5–50.5	16	91	48	$10$	$160$
50.5–55.5	9	100	53	$15$	$135$
Total	100				735

Here $N=\sum f_i=100$. The median position is $N/2 = 50$th observation. The c.f. just ≥50 is 63, so the median class is 35.5–40.5. Thus:

$$l = 35.5,\quad h = 5,\quad f = 26,\quad C = 37.$$

Median:
$$
M = l + \frac{\dfrac{N}{2} – C}{f}\times h $$
$$M = 35.5 + \frac{50-37}{26}\times 5 $$
$$M = 35.5 + \frac{13}{26}\times 5 $$
$$M = 35.5 + 2.5 = 38.
$$

Now compute mean deviation about median:
$$
\sum f_i\lvert x_i – M\rvert = 735.
$$

So
$$
\text{M.D.}(M) = \frac{735}{100} = 7.35.
$$

$$\boxed{\text{Median age} = 38\ \text{years},\qquad \text{M.D. about median} = 7.35\ \text{years}}$$