NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.2

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Chapter 15 Ex.15.2 Question 1

Find the mean and variance for the data \(6,7,10,12,13,4,8,12\).

Solution

The given data is \(6,7,10,12,13,4,8,12\)

Mean of the data

\[\begin{align}\bar x &= \frac{{\sum\limits_{i = 1}^8 {{x_i}} }}{n}\\&= \frac{{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}}{8}\\&= \frac{{72}}{8}\\&= 9\end{align}\]

The following table is obtained from the given above data

\(x_i\)

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

\({\left( {x_i - \bar x} \right)^2}\)

\(6\)

\( - 3\)

\(9\)

\(7\)

\( - 2\)

\(4\)

\(10\)

\(1\)

\(1\)

\(12\)

\(3\)

\(9\)

\(13\)

\(4\)

\(16\)

\(4\)

\( - 5\)

\(25\)

\(8\)

\( - 1\)

\(1\)

\(12\)

\(3\)

\(9\)

 

 

\(74\)

Variance of the data

\[\begin{align}\left( {{\sigma ^2}} \right)& = \frac{1}{n}\sum\limits_{i = 1}^8 {{{\left( {{x_i} - \bar x} \right)}^2}} \\&= \frac{1}{8} \times 74\\&= 9.25\end{align}\]

Chapter 15 Ex.15.2 Question 2

Find the mean and variance for the first \(n\) natural numbers.

Solution

The mean of first \(n\) natural numbers is calculated as follows.

\[\frac{\text{Sum of all observations}}{\text{Number of observations}} = \frac{{\frac{{n\left( {n + 1} \right)}}{2}}}{n} = \frac{n + 1}{2}\]

Variance,

\[\begin{align}\left( {{\sigma ^2}} \right) &= \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} \\&= \frac{1}{n}\sum\limits_{i = 1}^n {{{\left[ {{x_i} - \left( {\frac{{n + 1}}{2}} \right)} \right]}^2}} \\&= \frac{1}{n}\sum\limits_{i = 1}^n {{x_i}^2 - } \frac{1}{n}\sum\limits_{i = 1}^n {2\left( {\frac{{n + 1}}{n}} \right){x_i} + } \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {\frac{{n + 1}}{2}} \right)}^2}} \\&= \frac{1}{n}\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \left( {\frac{{n + 1}}{n}} \right)\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right] + {\frac{{\left( {n + 1} \right)}}{{4n}}^2} \times n\\&= \frac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \frac{{{{\left( {n + 1} \right)}^2}}}{2} + \frac{{{{\left( {n + 1} \right)}^2}}}{4}\\&= \frac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \frac{{{{\left( {n + 1} \right)}^2}}}{4}\\&= \left( {n + 1} \right)\left[ {\frac{{4n + 2 - 3n - 3}}{{12}}} \right]\\&= \frac{{\left( {n + 1} \right)\left( {n - 1} \right)}}{{12}}\\&= \frac{{{n^2} - 1}}{{12}}\end{align}\]

Chapter 15 Ex.15.2 Question 3

Find the mean and variance for the first \({\rm{1}}0\) multiples of \({\rm{3}}\).

Solution

The first ten multiples of \({\rm{3}}\) are \(3,6,9,12,15,18,21,24,27,30\)

Here, the number of observations, \(n = 10\)

Mean of the data

\[\begin{align}\bar x &= \frac{{\sum\limits_{i = 1}^{10} {{x_i}} }}{{10}}\\&= \frac{{165}}{{10}}\\&= 16.5\end{align}\]

The following table is obtained for the first multiples of \({\rm{3}}\)

\({x_i}\)

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

\({\left( {x_i - \bar x} \right)^2}\)

\(3\)

\(-13.5\)

\(182.25\)

\(6\)

\(-10.5\)

\(110.25\)

\(9\)

\( - 7.5\)

\(56.25\)

\(12\)

\( - 4.5\)

\(20.25\)

\(15\)

\( - 1.5\)

\(2.25\)

\(1.8\)

\(1.5\)

\(2.25\)

\(21\)

\(4.5\)

\(20.25\)

\(24\)

\(7.5\)

\(56.25\)

\(27\)

\(10.5\)

\(110.25\)

\(30\)

\(13.5\)

\(182.25\)

 

 

\(742.5\)

Variance

\[\begin{align}\left( {{\sigma ^2}} \right) &= \frac{1}{n}\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - \bar x} \right)}^2}} \\& = \frac{1}{{10}} \times 742.5\\& = 74.25\end{align}\]

Chapter 15 Ex.15.2 Question 4

Find the mean and variance for the data

\({x_i}\)

\(6\)

\(10\)

\(14\)

\(18\)

\(24\)

\(28\)

\(30\)

\({f_i}\)

\(2\)

\(4\)

\(7\)

\(12\)

\(8\)

\(4\)

\(3\)

Solution

\({x_i}\)

\({f_i}{x_i}\)

\({f_i}{x_i}\)

\({x_i} - \bar x\)

\({\left( {{x_i} - \bar x} \right)^{^2}}\)

\({f_i}{\left( {{x_i} - \bar x} \right)^2}\)

\(6\)

\(2\)

\(12\)

\( - 13\)

\(169\)

\(338\)

\(10\)

\(4\)

\(40\)

\( - 9\)

\(81\)

\(324\)

\(14\)

\(7\)

\(98\)

\( - 5\)

\(25\)

\(175\)

\(18\)

\(12\)

\(216\)

\( - 1\)

\(1\)

\(12\)

\(24\)

\(8\)

\(192\)

\(5\)

\(25\)

\(200\)

\(28\)

\(4\)

\(112\)

\(9\)

\(81\)

\(324\)

\(30\)

\(3\)

\(90\)

\(11\)

\(121\)

\(363\)

 

\(40\)

\(760\)

 

 

\(1736\)

Here,

\[\begin{align}N &= 40,\sum\limits_{i = 1}^7 {{f_1}{x_1} = 760} \\\therefore \overline x &= \frac{{\sum\limits_{i = 1}^7 {{f_1}{x_1}} }}{N} = \frac{{760}}{{40}} = 19\end{align}\]

Variance, \(\left( {{\sigma ^2}} \right) = \frac{1}{n}\sum\limits_{i = 1}^7 {{{({x_i} - \overline x )}^2} = \frac{1}{{40}} \times 1736 = 43.4} \)

Chapter 15 Ex.15.2 Question 5

Find the mean and variance for the data

\(x_i\)

\(92\)

\(93\)

\(97\)

\(98\)

\(102\)

\(104\)

\(109\)

\({f_i}\)

\(3\)

\(2\)

\(3\)

\(2\)

\(6\)

\(3\)

\(3\)

Solution

\(x_i\)

\(f_i\)

\({f_i}{x_i}\)

\({x_i} - \bar x\)

\({\left( {{x_i} - \bar x} \right)^{^2}}\)

\({f_i}{\left( {{x_i} - \overline x } \right)^2}\)

\(92\)

\(3\)

\(276\)

\(-8\)

\(64\)

\(192\)

\(93\)

\(2\)

\(186\)

\( - 7\)

\(49\)

\(98\)

\(97\)

\(3\)

\(291\)

\( - 3\)

\(9\)

\(27\)

\(98\)

\(2\)

\(196\)

\( - 2\)

\(4\)

\(8\)

\(102\)

\(6\)

\(612\)

\(2\)

\(4\)

\(24\)

\(104\)

\(3\)

\(312\)

\(4\)

\(16\)

\(48\)

\(109\)

\(3\)

\(327\)

\(9\)

\(81\)

\(243\)

 

 

\(2200\)

 

 

\(640\)

Here, \(N = 22\) and \(\sum\limits_{i = 1}^7 {{f_1}{x_1} = 2200} \)

Therefore,

\[\begin{align}\bar x& = \frac{1}{n}\sum\limits_{i = 1}^7 {{f_i}{x_i}} \\&= \frac{1}{{22}} \times 2200\\&= 100\end{align}\]

Variance

\[\begin{align}\left( {{\sigma ^2}} \right) &= \frac{1}{N}\sum\limits_{i = 1}^7 {{{\left( {{x_i} - \bar x} \right)}^2}} \\&= \frac{1}{{22}} \times 640\\&= 29.09\end{align}\]

Chapter 15 Ex.15.2 Question 6

Find the mean and standard deviation using shortcut method.

\({x_i}\)

\(60\)

\(61\)

\(62\)

\(63\)

\(64\)

\(65\)

\(66\)

\(67\)

\(68\)

\({f_i}\)

\(2\)

\(1\)

\(12\)

\(29\)

\(25\)

\(12\)

\(10\)

\(4\)

\(5\)

Solution

\({x_i}\)

\({f_i}\)

\({f_i} = \frac{{{x_i} - 64}}{1}\)

\({y_1}^2\)

\({f_1}{y_1}\)

\({f_1}{y_1}^{^2}\)

\(60\)

\(2\)

\( - 4\)

\(16\)

\( - 8\)

\( 32\)

\(61\)

\(1\)

\( - 3\)

\(9\)

\( - 3\)

\(9\)

\(62\)

\(12\)

\( - 2\)

\(4\)

\( - 24\)

\(48\)

\(63\)

\(29\)

\( - 1\)

\(1\)

\( - 29\)

\(29\)

\(64\)

\(25\)

\(0\)

\(0\)

\(0\)

\(0\)

\(65\)

\(12\)

\(1\)

\(1\)

\(12\)

\(12\)

\(66\)

\(10\)

\(2\)

\(4\)

\(20\)

\(40\)

\(67\)

\(4\)

\(3\)

\(9\)

\(12\)

\(36\)

\(68\)

\(5\)

\(4\)

\(16\)

\(20\)

\(80\)

 

\(100\)

\(220\)

 

\(0\)

\(286\)

Mean,

\[\begin{align}\bar x& = A + \frac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{N} \times h\\&= 64 + \frac{0}{{100}} \times 1\\&= 64\end{align}\]

Variance,

\[\begin{align}{\sigma ^2} &= \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^9 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}} } \right)}^2}} } \right]\\&= \frac{1}{{{{100}^2}}}\left[ {100 \times 286 - 0} \right]\\&= 2.86\end{align}\]

Standard deviation,

\[\begin{align}\left( \sigma \right) &= \sqrt {2.86} \\&= 1.69\end{align}\]

Chapter 15 Ex.15.2 Question 7

Find the mean and variance for the following frequency distribution.

Classes

\(0 -30\)

\(30 -60\)

\(60 -90\)

\(90 -120\)

\(120 -150\)

\(150 -180\)

\(180 -210\)

Frequencies

\( - 2\)

\(3\)

\(5\)

\(10\)

\(3\)

\(5\)

\(2\)

Solution

Class

Frequency \({f_i}\)

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

\({y_i} = \frac{{{x_i} - 105}}{{30}}\)

\({y_i}^2\)

\({f_i}{y_i}\)

\({f_i}{y_i}^2\)

\(0 - {\rm{3}}0\)

\(2\)

\(15\)

\( - 3\)

\(9\)

\( - 6\)

\(18\)

\(30 - 60\)

\(3\)

\(45\)

\(-2\)

\(4\)

\( - 6\)

\(12\)

\(60 - 90\)

\(5\)

\(75\)

\( - 1\)

\(1\)

\( - 5\)

\(5\)

\(90 - 120\)

\(10\)

\(105\)

\(0\)

\(0\)

\(0\)

\(0\)

\(120 - 150\)

\(3\)

\(135\)

\(1\)

\(1\)

\(3\)

\(3\)

\(150 - 180\)

\(5\)

\(165\)

\(2\)

\(4\)

\(10\)

\(20\)

\(180 - 210\)

\(2\)

\(195\)

\(3\)

\(9\)

\(6\)

\(18\)

 

 

 

 

 

\(2\)

\(76\)

Mean,

\[\begin{align}\bar x &= A + \frac{{\sum\limits_{i = 1}^7 {{f_i}{y_i}} }}{N} \times h\\&= 105 + \frac{2}{{30}} \times 30\\&= 105 + 2\\&= 107\end{align}\]

Variance,

\[\begin{align}{\sigma ^2} &= \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^9 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}} } \right)}^2}} } \right]\\&= \frac{{{{\left( {30} \right)}^2}}}{{{{\left( {30} \right)}^2}}}\left[ {30 \times 76 - {{\left( 2 \right)}^2}} \right]\\&= 2280 - 4\\&= 2276\end{align}\]

Chapter 15 Ex.15.2 Question 8

Find the mean and variance for the following frequency distribution.

Classes

\(0 - 10\)

\(10 - 20\)

\(20 - 30\)

\(30 - 40\)

\(40 - 50\)

Frequencies

\(5\)

\(8\)

\(15\)

\(16\)

\(6\)

Solution

Class

Frequency \({f_i}\)

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

\({y_i} = \frac{{{x_i} - 25}}{10}\)

\({y_i}^2\)

\({f_i}{y_i}\)

\({f_i}{y}_{i}^{2}\)

\(0-10\)

\(5\)

\(-2\)

\( 4\)

\( - 10\)

\(20\)

\(10 - 20\)

\(8\)

\(15\)

\( - 1\)

\(1\)

\( - 8\)

\(8\)

\(20 - 30\)

\(15\)

\(25\)

\(0\)

\(0\)

\(0\)

\(0\)

\(30 - 40\)

\(16\)

\(35\)

\(1\)

\(1\)

\(16\)

\(16\)

\(40 - 50\)

\(6\)

\(45\)

\(2\)

\(4\)

\(12\)

\(24\)

 

\(50\)

 

 

 

\(10\)

\(68\)

Mean,

\[\begin{align}\bar x&= A + \frac{{\sum\limits_{i = 1}^5 {{f_i}{y_i}} }}{N} \times h\\&= 25 + \frac{{10}}{{50}} \times 10\\&= 25 + 2\\&= 27\end{align}\]

Variance,

\[\begin{align}\left( {{\sigma ^2}} \right) &= \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^5 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^5 {{f_i}{y_i}} } \right)}^2}} } \right]\\&= \frac{{{{\left( {10} \right)}^2}}}{{{{\left( {50} \right)}^2}}}\left[ {50 \times 68 - {{\left( {10} \right)}^2}} \right]\\&= \frac{1}{{25}}\left[ {3400 - 100} \right]\\&= \frac{{3300}}{{25}}\\&= 132\end{align}\]

Chapter 15 Ex.15.2 Question 9

Find the mean, variance and standard deviation using shortcut method.

Height in cms

Number of children

\(70 - 75\)

\(3\)

\(75 - 80\)

\(4\)

\(80 - 85\)

\(7\)

\(85 - 90\)

\(7\)

\(95 - 100\)

\(9\)

\(100 - 105\)

\(6\)

\(105 - 110\)

\(6\)

\(110 - 115\)

\(3\)

Solution

Class

Frequency \({f_i}\)

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

\(y_i = \frac{{{x_i} - 92.5}}{5}\)

\(y_i^2\)

\({f_i}{y_i}\)

\({f_i}{y_i}^2\)

\(70 - 75\)

\(3\)

\(72.5\)

\(-4\)

\(16\)

\( - 12\)

\(48\)

\(75 - 80\)

\(4\)

\(77.5\)

\( - 3\)

\(9\)

\( - 12\)

\(36\)

\(80 - 85\)

\(7\)

\(82.5\)

\( - 2\)

\(4\)

\( - 14\)

\(28\)

\(85 - 95\)

\(7\)

\(87.5\)

\( - 1\)

\(1\)

\( - 7\)

\(7\)

\(90 - 95\)

\(15\)

\(92.5\)

\(0\)

\(0\)

\(0\)

\(0\)

\(95-100\)

\(9\)

\(97.5\)

\(1\)

\(1\)

\(9\)

\(9\)

\(100-105\)

\(6\)

\(102.5\)

\(2\)

\(4\)

\(12\)

\(24\)

\(105 - 110\)

\(6\)

\(107.5\)

\(3\)

\(9\)

\(18\)

\(54\)

\(110-115\)

\(3\)

\(112.5\)

\(4\)

\(16\)

\(12\)

\(48\)

 

\(60\)

 

 

 

\(6\)

\(254\)

Mean,

\[\begin{align}\bar x &= A + \frac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{N} \times h\\&= 92.5 + \frac{6}{{60}} \times 5\\&= 92.5 + 0.5\\&= 93\end{align}\]

Variance,

\[\begin{align}\left( {{\sigma ^2}} \right)& = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^9 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}} } \right)}^2}} } \right]\\&= \frac{{{{\left( 5 \right)}^2}}}{{{{\left( {60} \right)}^2}}}\left[ {60 \times 254 - {{\left( 6 \right)}^2}} \right]\\&= \frac{{25}}{{3600}} \times 15204\\&= 105.58\end{align}\]

Standard deviation,

\[\begin{align}\left( \sigma \right)&= \sqrt {105.58} \\&= 10.27\end{align}\]

Chapter 15 Ex.15.2 Question 10

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

Diameters

No. of circles

\(33 - 36\)

\(15\)

\(37 - 40\)

\(17\)

\(41 - 44\)

\(21\)

\(45-48\)

22

\(49 - 52\)

\(25\)

Calculate the standard deviation and mean diameter of the circles.

[Hint: First make the data continuous by making the classes as \(32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5 - 48.5, 48.5 - 52.5\) and then proceed.]

Solution

Class interval

Frequency \({f_i}\)

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

\({y_i} = \frac{{{x_i} - 42.5}}{4}\)

\({y_i}^2\)

\({f_i}{y_i}\)

\({f_i}{y_i}^2\)

\(33 - 36\)

\(15\)

\(34.5\)

\(-2\)

\(4\)

\(-30\)

\(60\)

\(37 - 40\)

\(17\)

\(38.5\)

\( - 1\)

\(1\)

\( - 17\)

\(17\)

\(41 - 44\)

\(21\)

\(42.5\)

\(0\)

\(0\)

\(0\)

\(0\)

\(45 - 48\)

\(22\)

\(46.5\)

\(1\)

\(1\)

\(22\)

\(22\)

\(49-52\)

\(25\)

\(50.5\)

\(2\)

\(4\)

\(50\)

\(100\)

 

\(100\)

 

 

 

\(25\)

\(199\)

Here, \(N = 100\), \(h = 4\)

Let the assumed mean, A be \(42.5\)

Mean,

\[\begin{align}\overline x &= A + \frac{{\sum\limits_{i = 1}^5 {{f_i}{y_i}} }}{N} \times h\\&= 42.5 + \frac{{25}}{{100}} \times 4\\&= 42.5 + 1\\&= 43.5\end{align}\]

Variance,

\[\begin{align}{\sigma ^2}& = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^5 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^5 {{f_i}{y_i}} } \right)}^2}} } \right]\\&= \frac{{16}}{{10000}}\left[ {100 \times 199 - {{\left( {25} \right)}^2}} \right]\\&= \frac{{16}}{{10000}}\left[ {19900 - 625} \right]\\&= \frac{{16}}{{10000}} \times 19275\\&= 30.84\end{align}\]

Standard deviation,

\[\begin{align}\left( \sigma \right) &= \sqrt {30.84} \\&= 5.55\end{align}\]

  
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