NCERT Class 11 Mathematics Chapter 13 Statistics

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NCERT Class 11 Mathematics Chapter 13 Statistics

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Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 11 Mathematics Chapter 13 Statistics Textual Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 11 Mathematics Chapter 13 Statistics Solutions for All Subject, You can practice these here.

Statistics

Chapter – 13

Exercise 13.1

Find the mean deviation about the mean for the data in Exercises 1 and 2. 

1. 4, 7, 8, 9, 10, 12, 13, 17

Ans: The given data is

4, 7, 8, 9, 10, 12, 13, 17

Mean of the data,

The deviations of the respective observations from the mean X̅, i.e. xᵢ – X̅ , are

– 6, – 3, – 2, – 1, 0, 2, 3, 7 

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The absolute values of the deviations, i.e. | xᵢ – X̅ | are

6, 3, 2, 1, 0, 2, 3, 7 

The required mean deviation about the mean is

2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Ans: The given data is 

38, 70, 48, 40, 42, 55, 63, 46, 54, 44 

Mean of the given data,

The deviations of the respective observations from the mean X̅ , i.e. xᵢ – X̅ , are

– 12,20, – 2,- 10,- 8,5,13,- 4,4,- 6 

The absolute values of the deviations, i.e.| xᵢ – X̅ | are

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

The required mean deviation about the mean is

Find the mean deviation about the median for the data in Exercises 3 and 4.

3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.

Ans: The given data is

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

Here, the numbers of observations are 12, which is even.

Arranging the data in ascending order, we obtain

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

The deviations of the respective observations from the median, i.e. xᵢ – M, are

– 3.5, – 2.5, – 2.5, – 1.5, – 0.5, – 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The absolute values of the deviations, |xᵢ – M|, are 

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The required mean deviation about the median is

4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.

Ans: The given data is 

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

Here, the number of observations is 10, which is even. 

Arranging the data in ascending order, we obtain

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

The deviations of the respective observations from the median, i.e. xᵢ – M, are

– 11.5 , – 5.5 ,- 2.5, – 1.5 – 1.5, 1.5, 3.5, 5.5, 12.5, 24.5 

The absolute values of the deviations, |xᵢ – M|, are 

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5 

Thus, the required mean deviation about the median is 

Find the mean deviation about the mean for the data in Exercises 5 and 6.

5.

xᵢ510152025
fᵢ74635

Ans: 

xᵢfᵢfᵢ xᵢ|xᵢ – X̅|fᵢ|xᵢ – X̅|
5735963
10440416
1569016
20360618
2551251155
25350158

6.

xᵢ1030507090
fᵢ42428168

Ans: 

xᵢfᵢfᵢ xᵢ|xᵢ – X̅|fᵢ|xᵢ – X̅|
1044040160
302472020480
5028140000
7016112020320
90872040320
8040001280

Find the mean deviation about the median for the data in Exercises 7 and 8.

7.

xᵢ579101215
fᵢ862226

Ans: The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

xᵢfᵢc.f.
588
7614
9216
10218
12220
15626

Here, N = 26, which is even.

Median is the mean of 13ᵗʰ and 14ᵗʰ observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

The absolute values of the deviations from median, i.e. |xᵢ – M|, are

|xᵢ — M|202358
fᵢ862226
fᵢ|xᵢ — M|160461048

8.

xᵢ1521273035
fᵢ35678

Ans: The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

xᵢfᵢc.f.
1533
2158
27614
30721
35829

Here, N = 29, which is odd.

observation = 15ᵗʰ observation

This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.

.:  Median = 30

The absolute values of the deviations from median, i.e.|xᵢ – M|, are

|xᵢ – M|159305
fᵢ35678
fᵢ|xᵢ – M|454518040

Find the mean deviation about the mean for the data in Exercises 9 and 10.

9.

Income per day in ₹0 -100100 – 200200 – 300300 – 400400 – 500500 – 600600 – 700700 – 800
Number of persons 489107543

Ans: The following table is formed.

Income per day in Number of persons fᵢMid – point xᵢfᵢ xᵢ|xᵢ – X̅|fᵢ|xᵢ – X̅|
0 -1004502003081232
100 – 200815012002081664
200 – 30092502250108972
300 – 400103503500880
400 – 5007450315092644
500 – 60055502750192960
600 – 700465026002921168
700 – 800375022503921176
50179007896

10. 

Height in cms95 – 105105 – 115115 – 125125 – 135135 – 145145 – 155
Number of boys 91326301210

Ans: The following table is formed.

Height in cmsNumber of boys fᵢMid – point xᵢfᵢ xᵢ|xᵢ – X̅|fᵢ|xᵢ – X̅|
95 – 105910090025.3227.7
105 – 11513110143015.3198.9
115 – 1252612031205.3137.8
125 – 1353013039004.7141
135 – 14512140168014.7176.4
145 – 15510150150024.7247

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

Marks 0 – 1010 – 2020 – 3030-4040-5050-60
Number of girls 68141642

Ans: 

Marks fᵢCumulative frequencyxᵢ|xᵢ – M|fᵢ|xᵢ – M|
0 – 1066522.86137.16
10 – 208141512.86102.88
20 – 301428252.8640.04
30 – 401644357.14114.24
40 – 504484517.1468.56
50 – 602505527.1454.28
Total50517.32

Since, 25 occurs in the cumulative frequency 25. 

So, the median class is 20 – 30.

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

Ans: The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by 

subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is formed as follows.

AgeNumber fᵢCumulative frequency(c.f.)Mid – point xᵢ|xᵢ – Med.|fᵢ|xᵢ – Med.|
15.5 – 20.5551820100
20.5 – 25.5611231590
25.5 – 30.512232810120
30.5 – 35.5143733570
35.5 – 40.526633800
40.5 – 45.5127543560
45.5 – 50.516914810160
50.5 – 55.591005315135
100735

The class interval containing the Nᵗʰ/2 or 50ᵗʰ item is 35.5 – 40.5.

Therefore, 35.5 – 40.5 is the median class.

It is known that,

Here, /= 35.5, C = 37, f = 26, h = 5, and N = 100

Thus, mean deviation about the median is given by,

Exercise 13.2

Find the mean and variance for each of the data in Exercise 1 to 5.

1. 6, 7, 10, 12, 13, 4, 8, 12.

Ans: 6, 7, 10, 12, 13, 4, 8, 12

Mean,

The following table is obtained.

xᵢ(xᵢ – X̅)(xᵢ – X̅)²
6– 39
7– 24
10-11
1239
13416
4– 525
8– 11
1239
74

2. First n natural numbers.

Ans: The mean of first n natural numbers is calculated as follows.

3. First 10 multiples of 3.

Ans: The first 10 multiples of 3 are 

3, 6, 9, 12, 15, 18, 21, 24, 27, 30 

Here, number of observations, n = 10

The following table is obtained.

xᵢ(xᵢ – X̅)(xᵢ – X̅)²
3– 13.5182.25
6-10.5110.25
9– 7.556.25
12– 4.520.25
15– 1.52.25
181.52.25
214.520.25
247.556.25
2710.5110.25
3013.5182.25
742.5

4.

xᵢ6101418242830
fᵢ24712843

Ans: The data is obtained in tabular form as follows.

xᵢfᵢfᵢxᵢxᵢ – X̅(xᵢ – X̅)²fᵢ(xᵢ – X̅)²
6212-13169338
10440-981324
14798-525175
1812216-1112
248192525200
284112981324
3039011121363
407601736

5.

xᵢ92939798102104109
fᵢ3232633

Ans: The data is obtained in tabular form as follows.

xᵢfᵢfᵢxᵢxᵢ – X̅(xᵢ – X̅)²fᵢ(xᵢ – X̅)²
923276-864192
932186-74998
973291-3927
982196-248
10266122424
104331241648
1093327981243
222200640

6. Find the mean and standard deviation using short-cut method.

xᵢ606162636465666768
fᵢ21122925121045

Ans: The data is obtained in tabular form as follows.

xᵢfᵢyᵢ²fᵢyᵢfᵢyᵢ²
602-416-832
611-39-39
6212-24-2448
6329-11-2929
64250000
6512111212
6610242040
674391236
6854162080
1002200286

Find the mean and variance for the following frequency distributions in Exercises 7 and 8.

7.

Classes0 – 3030 – 6060 – 9090 – 120120 – 150150 – 180180 – 210
Frequencies23510352

Ans: 

ClassFrequencyfᵢMid – point xᵢyᵢ²fyᵢfyᵢ²
0 – 30215-39-618
30 – 60345-24-612
60 – 90575-11-55
90 – 120101050000
120 – 15031351133
150 – 1805165241020
180 – 210219539618
30276

8.

Classes0 – 1010 – 2020 – 3030 – 4040 – 50
Frequencies5815166

Ans: 

ClassFrequencyfᵢMid – point xᵢyᵢ²fyᵢfyᵢ²
0 – 1055-24-1020
10 – 20815-11-88
20 – 3015250000
30 – 401635111616
40 – 50645241224
501068

9. Find the mean, variance and standard deviation using short-cut method

Heightin cms 70-7575-8080-8585-90 90-95 95-100 100-105105-110110-115
No. ofchildren3477159663

Ans: 

Class IntervalFrequencyfᵢMid – point xᵢyᵢ²fyᵢfyᵢ²
70 – 75372.5-416-1248
75 – 80477.5-39-1236
80 – 85782.5-24-1428
85 – 90 787.5-11-77
90 – 95 1592.50000
95 -100 997.51199
100 -1056102.5241224
105 -1106107.5391854
110 – 1153112.54161248
606254

10. The diameters of circles (in mm) drawn in a design are given below:

Diameters33 – 3637 – 4041 – 4445 – 4849 – 52
No. of circles1517212225

Ans:

Class IntervalFrequencyfᵢMid – point xᵢfᵢ²fyᵢfyᵢ²
32.5 – 36.51534.5-24-3060
36.5 – 40.51738.5-11-1717
40.5 – 44.52142.50000
44.5 – 48.52246.5112222
48.5 – 52.52550.52450100
10025199

Here, N = 100, h = 4

Let the assumed mean, A, be 42.5.

Miscellaneous Exercise on Chapter 12

1. The mean and variance of eight observations are 9 and 9.25, respectively. If six

of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations. 

Ans: Let the remaining two observations be x and y.

Therefore, the observations are 6, 7, 10, 12, 12, 13, х, у.

⇒ 60 + x + y = 72 

⇒ x + y = 12                             …(1)

⇒ x² + y² = 80                               …(2)

From (1), we obtain 

x² + y² + 2xy =144 -| (3) 

From (2) and (3), we obtain 

2xy = 64 -| (4) 

Subtracting (4) from (2), we obtain 

x² + y² – 2xy = 80 – 64 = 16 

⇒ x – y= A ± 4-|(5)

Therefore, from (1) and (5), we obtain

x = 8 and Y = 4 when x – Y = 4

x = 4 and y = 8 when x – y = – 4

Thus, the remaining observations are 4 and 8.

2. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations. 

Ans: Let the remaining two observations be x and y.

The observations are 2, 4, 10, 12, 14, x, y.

⇒ 56 = 42 + x + y

⇒x + y = 14                                            ….(1)

⇒ x² + y² = 112 – 12 = 100 

x² + y² = 100                                           …..(2)

From (1), we obtain 

x² + y² + 2xy = 196 -| (3) 

From (2) and (3), we obtain 

2xy =196 – 100 

⇒ 2xy = 96 -| (4) 

Subtracting (4) from (2), we obtain

x² + y² – 2xy = 100 – 96 

⇒ (x – y)² = 4 

⇒ x – y= A ± 2 -| (5) 

Therefore, from (1) and (5), we obtain 

x = 8 and y = 6 when x – y = 2 

x = 6 and y = 8 when x – y = – 2 

Thus, the remaining observations are 6 and 8.

3. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Ans: Let the observations be X₁ , X₂ , X₃ , X₄ ,X₅ , and X₆ .

It is given that mean is 8 and standard deviation is 4.

If each observation is multiplied by 3 and the resulting observations are y, then

= 3 × 8                                          ….[Using(1)]

= 24

Standard deviation,

From (1) and (2), it can be observed that,

Substituting the values of xᵢ and X̅ in (2), we obtain.

Therefore, variance of new observations =

Hence, the standard deviation of new observations is √144 = 12

4. Given that X̅ is the mean and Ïf² is the variance of n observations x₁, x₂ -| xₙ.Prove that the mean and variance of the observations ax₁, ax₂, ax₃ -axₙ are aX̅ and a² Ïf², respectively, ( a ≠ 0).

Ans: The given n observations are x₁ ,x₂ -| xₙ.

Mean = X̅

Variance = Ïf²

If each observation is multiplied by a and the new observations are y, then

Therefore, mean of the observations, ax₁, ax₂ -| axₙ, is aX̅.

Substituting the values of xand X̅ in (1), we obtain

Thus, the variance of the observations, ax₁,ax₂ -| axₙ, is a² Ïf² .

Q5: The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

(i) If the wrong item is omitted.

Ans: Number of observations (n) = 20

Incorrect mean = 10

Incorrect standard deviation = 2

That is, incorrect sum of observations = 200 

Correct sum of observations = 200 – 8 = 192

∴Correct mean

(ii) If it is replaced by 12.

Ans: When 8 is replaced by 12,

Incorrect sum of observations = 200

∴ Correct sum of observations = 200 – 8 + 12 = 204

6. The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

Ans: Number of observations (n) = 100 

Incorrect mean ( X̅ )=20 

Incorrect standard deviation (σ) = 3

∴ Incorrect sum of observations = 2000

⇒ Correct sum of observations = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940

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