# NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.2

## 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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