NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1

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Chapter 15 Ex.15.1 Question 1

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

Solution

The given data is \(4,7,8,9,10,12,13,17\)

Mean of the data

\[\begin{align}\bar x &= \frac{4 + 7 + 8 + 9 + 10 + 12 + 13 + 17}{8}\\&= \frac{80}{8}\\&= 10\end{align}\]

The deviations of the respective observations from the mean \(\bar x\), i.e., \({x_i} - \bar x\) are

\[ - 6, - 3, - 2, - 1,0,2,3,7\]

The absolute values of the deviations, \(\left| {{x_i} - \bar x} \right|\) are \(6,3,2,1,0,2,3,7\)

The required mean deviation about the mean is

\[\begin{align}M.D.\left( {\bar x} \right)& = \frac{{\sum\limits_{i = 1}^8 {\left| {x_i - \bar x} \right|} }}{8}\\&= \frac{6 + 3 + 2 + 1 + 0 + 2 + 3 + 7}{8}\\&= \frac{24}{8}\\&= 3\end{align}\]

Chapter 15 Ex.15.1 Question 2

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

Solution

The given data is \(38,70,48,40,42,55,63,46,54,44\)

Mean of the data

\[\begin{align}\bar x &= \frac{38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44}{10}\\&= \frac{500}{10}\\&= 50\end{align}\]

The deviations of the respective observations from the mean \(\bar x\), i.e., \({x_i} - \bar x\) are

\[ - 12,20, - 2, - 10, - 8,5,13, - 4,4, - 6\]

The absolute values of the deviations, \(\left| {x_i - \bar x} \right|\) are

\[12,20,2,10,8,5,13,4,4,6\]

The required mean deviation about the mean is

\[\begin{align}M.D.\left( {\bar x} \right) &= \frac{{\sum\limits_{i = 1}^{10} {\left| {x_i - \bar x} \right|} }}{10}\\&= \frac{12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6}{10}\\&= \frac{84}{10}\\&= 8.4\end{align}\]

Chapter 15 Ex.15.1 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

The given data is \(13,17,16,14,11,13,10,16,11,18,12,17\)

Here, the numbers of observations are \({\rm{12}}\), i.e., even.

Arranging the above data in ascending order, we obtain \(10,11,11,12,13,13,14,16,16,17,17,18\)

Median of the data

\[\begin{align}M &= \frac{{\left( {\frac{12}{2}} \right)^{th}observation + {\left( {\frac{12}{2} + 1} \right)^{th}}observation}}{2}\\&= \frac{{6^{th}observation + 7^{th}observation}}{2}\\&= \frac{13 + 14}{2}\\&= \frac{27}{2}\\&= 13.5\end{align}\]

The deviations of the respective observations from the median, i.e., \({x_i} - M\) 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, \(\left| {{x_i} - M} \right|\) 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

\[\begin{align}M.D.\left( M \right) &= \frac{{\sum\limits_{i = 1}^{12} {\left| {{x_i} - M} \right|} }}{{12}}\\&= \frac{3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5}{12}\\&= \frac{28}{12}\\&= 2.33\end{align}\]

Chapter 15 Ex.15.1 Question 4

Find the mean deviation about the median for the data \(36,72,46,42,60,45,53,46,51,49\)

Solution

The given data is \(36,72,46,42,60,45,53,46,51,49\)

Here, the numbers of observations are \({\rm{1}}0\), hence even.

Arranging the above data in ascending order, we obtain \(36,42,45,46,46,49,51,53,60,72\)

Median of the data

\[\begin{align}M &= \frac{{{{\left( {\frac{10}{2}} \right)}^{th}}observation + {\left( {\frac{{10}}{2} + 1} \right)^{th}}observation}}{2}\\&= \frac{{5^{th}observation \;+\; 6^{th}observation}}{2}\\&= \frac{46 + 49}{2}\\&= \frac{95}{2}\\&= 47.5\end{align}\]

The deviations of the respective observations from the median, i.e., \({x_i} - M\) 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, \(\left| {{x_i} - M} \right|\)are

\[11.5,5.5,2.5,1.5,1.5,1.5,3.5,5.5,12.5,24.5\]

The required mean deviation about the median is

\[\begin{align}M.D.\left( M \right) &= \frac{{\sum\limits_{i = 1}^{10} {\left| {{x_i} - M} \right|} }}{{10}}\\&= \frac{11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5}{10}\\&= \frac{70}{10}\\&= 7\end{align}\]

Chapter 15 Ex.15.1 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

\({x_i}\)

\({f_i}\)

\({f_i}{x_i}\)

\(\left| {{x_i} - \bar x} \right|\)

\({f_i}\left| {{x_i} - \bar x} \right|\)

\(5\)

\(7\)

35

9

63

\(10\)

\(4\)

\(40\)

\(4\)

\(16\)

\(15\)

\(6\)

\(90\)

\(1\)

\(6\)

\(20\)

\(3\)

\(60\)

\(6\)

\(18\)

\(25\)

\(5\)

\(125\)

\(11\)

\(55\)

\(\)

\(25\)

\(350\)

\(\)

\(158\)

\(N = \sum\limits_{i = 1}^5 {{f_i}} = 25\) and \(\sum\limits_{i = 1}^5 {{f_i}{x_i} = 350} \)

Therefore,

\[\begin{align}\overline x &= \frac{1}{N}\sum\limits_{i = 1}^5 {{f_i}{x_i}} \\&= \frac{1}{{25}} \times 350\\&= 14\end{align}\]

Mean deviation about the mean

\[\begin{align}M.D.\left( {\bar x} \right) &= \frac{1}{N}\left| {{x_i} - \bar x} \right|\\&= \frac{1}{{25}} \times 158\\&= 6.32\end{align}\]

Chapter 15 Ex.15.1 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

\({x_i}\)

\({f_i}\)

\({f_i}{x_i}\)

\(\left| {{x_i} - \overline x } \right|\)

\({f_i}\left| {{x_i} - \overline x } \right|\)

\(10\)

\(4\)

\(40\)

\(40\)

\(160\)

\(30\)

\(24\)

\(720\)

\(20\)

\(480\)

\(50\)

\(28\)

\(1400\)

\(0\)

\(0\)

\(70\)

\(16\)

\(1120\)

\(20\)

\(320\)

\(90\)

\(8\)

\(720\)

\(40\)

\(320\)

 

\(80\)

\(4000\)

 

\(1280\)

\(N = \sum\limits_{i = 1}^5 {{f_i}} = 80\) and \(\sum\limits_{i = 1}^5 {{f_i}{x_i} = 4000} \)

Therefore,

\[\begin{align}\bar x &= \frac{1}{N}\sum\limits_{i = 1}^5 {{f_i}{x_i}} \\&= \frac{1}{{80}} \times 4000\\&= 50\end{align}\]

Mean deviation about the mean

\[\begin{align}M.D.\left( {\overline x } \right) &= \frac{1}{N}\sum\limits_{i = 1}^5 {{f_i}\left| {{x_i} - \overline x } \right|} \\&= \frac{1}{{80}} \times 1280\\&= 16\end{align}\]

Chapter 15 Ex.15.1 Question 7

Find the mean deviation about the median for the data

\({x_i}\)

\(5\)

\(7\)

\(9\)

\(10\)

\(12\)

\(15\)

\({f_i}\)

\(8\)

\(6\)

\(2\)

\(2\)

\(2\)

\(6\)

Solution

The given observations are in ascending order.

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

\({x_i}\)

\({f_i}\)

\(c.f\)

\(5\)

\(8\)

\(8\)

\(7\)

\(6\)

\(14\)

\(9\)

\(2\)

\(16\)

\(10\)

\(2\)

\(18\)

\(12\)

\(2\)

\(20\)

\(15\)

\(6\)

26

Here, \(N = 26\), which is even.

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

\[\begin{align} \text{Median}&= \frac{{{13^{th}}\,observation + {14^{th}}observation}}{2}\\& = \frac{7 + 7}{2}\\&= 7 \end{align}\]

The absolute values of the deviations, \(\left| {{x_i} - M} \right|\)are

\(\left| {{x_i} - M} \right|\)

2

0

2

3

5

8

\({f_i}\)

8

6

2

2

2

6

\({f_i}\left| {{x_i} - M} \right|\)

16

0

4

6

10

48

\(\sum\limits_{i = 1}^6 {{f_i}} = 26\) and \(\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - M} \right|} = 84\)

Hence, the mean deviation about the median

\[\begin{align}M.D.\left( M \right) &= \frac{1}{N}\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - M} \right|} \\&= \frac{1}{26} \times 84\\&= 3.23\end{align}\]

Chapter 15 Ex.15.1 Question 8

Find the mean deviation about the median for the data

\({x_i}\)

\(15\)

\(21\)

\(27\)

\(30\)

\(35\)

\({f_i}\)

\(3\)

\(5\)

\(6\)

\(7\)

\(8\)

Solution

The given observations are in ascending order.

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

\({x_i}\)

\({f_i}\)

\(c.f\)

\(15\)

\(3\)

\(3\)

\(21\)

\(5\)

\(8\)

\(27\)

\(6\)

\(14\)

\(30\)

\(7\)

\(21\)

\(35\)

\(8\)

\(29\)

Here, \(N = 29\), which is odd.

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

This observation lies in the cumulative frequency \({\rm{21}}\), for which the corresponding observation is\(30\).

Therefore, Median = \(30\)

The absolute values of the deviations, \(\left| {{x_i} - M} \right|\)are

\(\left| {{x_i} - M} \right|\)

\(15\)

\(9\)

\(3\)

\(0\)

\(5\)

\({f_i}\)

\(3\)

\(5\)

\(6\)

\(7\)

\(8\)

\({f_i}\left| {{x_i} - M} \right|\)

\(45\)

\(45\)

\(18\)

\(0\)

\(40\)

Here, \(\sum\limits_{i = 1}^5 {{f_i}} = 29\) and \(\sum\limits_{i = 1}^5 {{f_i}\left| {{x_i} - M} \right|} = 148\)

\(M.D.\left( M \right) = \frac{1}{N}\sum\limits_{i = 1}^5 {{f_i}\left| {{x_i} - M} \right|} = \frac{1}{{29}} \times 148 = 5.1\)

Chapter 15 Ex.15.1 Question 9

Find the mean deviation about the mean for the data

Income per day

Number of persons

\(0 - 100\)

\(4\)

\(100 - 200\)

\(8\)

\(200 - 300\)

\(9\)

\(300 - 400\)

\(10\)

\(400 - 500\)

\(7\)

\(500 - 600\)

\(5\)

\(600 - 700\)

\(4\)

\(700 - 800\)

\(3\)

Solution

The following table is formed.

Income per day

Number of persons \({f_i}\)

Mid-point \({x_i}\)

\({f_i}{x_i}\)

\(\left| {{x_i} - \overline x } \right|\)

\({f_i}\left| {{x_i} - \overline x } \right|\)

\(0 - 100\)

\(4\)

\(50\)

\(200\)

\(308\)

\(1232\)

\(100 - 200\)

\(8\)

\(150\)

\(1200\)

\(208\)

\(1664\)

\(200 - 300\)

\(9\)

\(250\)

\(2250\)

\(108\)

\(972\)

\(300 - 400\)

\(10\)

\(350\)

\(3500\)

\(8\)

\(80\)

\(400 - 500\)

\(7\)

\(450\)

\(3150\)

\(92\)

\(644\)

\(500 - 600\)

\(5\)

\(550\)

\(2750\)

\(192\)

\(960\)

\(600 - 700\)

\(4\)

\(650\)

\(2600\)

\(292\)

\(1168\)

\(700 - 800\)

\(3\)

\(750\)

\(2250\)

\(392\)

\(1176\)

Total

\(50\)

\(17900\)

\(7896\)

Here, \(N = \sum\limits_{i = 1}^8 {{f_i}} = 50\) and \(\sum\limits_{i = 1}^8 {{f_i}{x_i} = 17900} \)

Therefore,

\[\begin{align}\overline x &= \frac{1}{N}\sum\limits_{i = 1}^8 {{f_i}{x_i}} \\ &= \frac{1}{{50}} \times 17900\\ &= 358\end{align}\]

Hence, the mean deviation about the mean

\[\begin{align}M.D.\left( {\bar x} \right) &= \frac{1}{N}\sum\limits_{i = 1}^8 {{f_i}\left| {{x_i} - \bar x} \right|} \\&= \frac{1}{{50}} \times 7896\\&= 157.92\end{align}\]

Chapter 15 Ex.15.1 Question 10

Find the mean deviation about the mean for the data

Height in cms

\(95-105\)

\(105-115\)

\(115-125\)

\(125-135\)

\(135-145\)

\(135-145\)

Number of boys

\(9\)

13

26

30

12

10

Solution

The following table is formed.

Height in cms

Number of boys\({f_i}\)

Mid-point \({x_i}\)

\({f_i}{x_i}\)

\(\left| {{x_i} - \overline x } \right|\)

\({f_i}\left| {{x_i} - \overline x } \right|\)

\(95 - 105\)

\(9\)

\(100\)

\(900\)

\(25.3\)

\(227.7\)

\(105 - 115\)

\(13\)

\(110\)

\(1430\)

\(15.3\)

\(198.9\)

\(115 - 125\)

\(26\)

\(120\)

\(3120\)

\(5.3\)

\(137.8\)

\(125 - 135\)

\(30\)

\(130\)

\(3900\)

\(4.7\)

\(141\)

\(135 - 145\)

\(12\)

\(140\)

\(1680\)

\(14.7\)

\(176.4\)

\(145 - 155\)

\(10\)

\(150\)

\(1500\)

\(24.7\)

\(247\)

Total

\(100\)

\(12530\)

\(1128.8\)

Here, \(N = \sum\limits_{i - 1}^6 {f_i = 100} \) and \(\sum\limits_{i = 1}^6 {{f_i}{x_i} = 12530} \)

Therefore,

\[\begin{align}\bar x &= \frac{1}{N}\sum\limits_{i = 1}^6 {{f_i}{x_i}} \\&= \frac{1}{{100}} \times 12530\\&= 125.3\end{align}\]

Hence, the mean deviation about the mean

\[\begin{align}M.D.\left( {\bar x} \right) &= \frac{1}{N}\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - \bar x} \right|} \\&= \frac{1}{{100}} \times 1128.8\\&= 11.28\end{align}\]

Chapter 15 Ex.15.1 Question 11

Find the mean deviation about median for the following data:

Marks

\(0 - 10\)

\(10 - 20\)

\(20 - 30\)

\(30 - 40\)

\(40 - 50\)

\(50 - 60\)

No. of girls

6

8

14

16

4

2

Solution

Marks

No. of girls

c.f.

Mid-point \(\left( {{x_i}} \right)\)

\(\left| {{x_i} - M} \right|\)

\({f_i}\left| {{x_i} - M} \right|\)

\(0 - 10\)

\(6\)

\(6\)

\(5\)

\(22.85\)

\(137.1\)

\(10 - 20\)

\(8\)

\(14\)

\(15\)

\(12.85\)

\(120.8\)

\(20 - 30\)

\(14\)

\(28\)

\(25\)

\(2.85\)

\(39.9\)

\(30 - 40\)

\(16\)

\(44\)

\(35\)

\(7.15\)

\(114.4\)

\(40 - 50\)

\(4\)

\(48\)

\(45\)

\(17.15\)

\(68.6\)

\(50 - 60\)

\(2\)

\(50\)

\(55\)

\(27.15\)

\(54.3\)

 

\(50\)

 

 

 

\(517.1\)

Here, \(\sum\limits_{i = 1}^6 {{f_i}} = 50\) and \(\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - M} \right|} = 517.1\)

\(l = 20,C = 14,f = 14,h = 10,N = 50\)

Median,

\[\begin{align}&= l + \frac{{\frac{N}{2} - C}}{f} \times h\\&= 20 + \frac{25 - 14}{14} \times 10\\&= 20 + 7.85\\&= 27.85\end{align}\]

Hence, the mean deviation about median

\[\begin{align}M.D.\left( M \right) &= \frac{1}{N}\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - M} \right|} \\&= \frac{1}{50} \times 517.1\\&= 10.34\end{align}\]

Chapter 15 Ex.15.1 Question 12

Calculate the mean deviation about median age for the age distribution of \(100\) persons given below:

Age

(in years)

\(16 - 20\)

\(21 - 25\)

\(26 - 30\)

\(31 - 35\)

\(36 - 40\)

\(41 - 45\)

\(46 - 50\)

\(51 - 55\)

Number

5

6

12

14

26

12

16

9

[Hint: Convert the given data into continuous frequency distribution by subtracting \(0.5\) from the lower limit and adding \(0.5\) to the upper limit of each class interval]

Solution

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

Age

(in years)

Number of persons \({f_i}\)

\(c.f\)

Mid-point

\({x_i}\)

\(\left| {{x_i} - M} \right|\)

\({f_i}\left| {{x_i} - M} \right|\)

\(15.5 - 20.5\)

\(5\)

\(5\)

\(18\)

\(20\)

\(100\)

\(20.5 - 25.5\)

\(6\)

\(11\)

\(23\)

\(15\)

\(90\)

\(25.5 - 30.5\)

\(12\)

\(23\)

\(28\)

\(10\)

\(120\)

\(30.5 - 35.5\)

\(14\)

\(37\)

\(33\)

\(5\)

\(70\)

\(35.5 - 40.5\)

\(26\)

\(63\)

\(38\)

\(0\)

\(0\)

\(40.5 - 45.5\)

\(12\)

\(75\)

\(43\)

\(5\)

\(60\)

\(45.5 - 50.5\)

\(16\)

\(91\)

\(48\)

\(10\)

\(160\)

\(50.5 - 55.5\)

\(9\)

\(100\)

\(53\)

\(15\)

\(135\)

 

\(100\)

 

 

 

\(735\)

The class interval containing the \({\left( {\frac{N}{2}} \right)^{th}}\)or \({50^{th}}\) item is \(35.5 - 40.5\)

Therefore, \(35.5 - 40.5\) is the median class.

It is known that,

Median \( = l + \left( {\frac{{\frac{N}{2} - C}}{f}} \right) \times h\)

Here, \(l = 35.5,C = 37,f = 26,h = 5,N = 100\)

Median,

\[\begin{align}M &= 35.5 + \frac{{50 - 37}}{{26}} \times 5\\&= 35.5 + \frac{{13 \times 5}}{{26}}\\&= 35.5 + 2.5\\&= 38\end{align}\]

Thus, mean deviation about the median is given by,

\[\begin{align}M.D.\left( M \right) = \frac{1}{N}\sum\limits_{i = 1}^8 {{f_i}\left| {{x_i} - M} \right|} \\= \frac{1}{{100}} \times 735\\= 7.35\end{align}\]

  
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