# Miscellaneous Exercise Statistics - NCERT Class 11

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## Chapter 15 Ex.15.ME Question 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.

### Solution

Let the remaining two observations be $$x$$ and $$y$$.

Therefore, the observations are $$6,7,10,12,12,13,x,y$$

Mean,

\begin{align}\overline x &= \frac{{6 + 7 + 10 + 12 + 12 + 13 + x + y}}{8}\\9 &= \frac{{60 + x + y}}{8}\\60 + x + y& = 72\\x + y &= 12 \qquad \quad \ldots \left( 1 \right)\end{align}

Variance,

\begin{align}9.25 &= \frac{1}{n}{\sum\limits_{i = 1}^8 {\left( {{x_i} - \bar x} \right)} ^2}\\9.25 &= \frac{1}{8}\left[ {{{\left( { - 3} \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + {x^2} + {y^2} - 2 \times 9\left( {x + y} \right) + 2 \times {{\left( 9 \right)}^2}} \right]\\9.25 &= \frac{1}{8}\left[ {9 + 4 + 1 + 9 + 9 + 16 + {x^2} + {y^2} - 18\left( {12} \right) + 162} \right]\\9.25 &= \frac{1}{8}\left[ {48 + {x^2} + {y^2} - 216 + 162} \right]\\9.25& = \frac{1}{8}\left[ {{x^2} + {y^2} - 6} \right]\\{x^2} + {y^2}& = 80 \qquad \quad \ldots \left( 2 \right)\end{align}

From $$\left( 1 \right)$$, we obtain

${x^2} + {y^2} + 2xy = 144 \qquad \quad \ldots \left( 3 \right)$

From $$\left( 2 \right)$$and $$\left( 3 \right)$$, we obtain

$2xy = 64 \qquad \quad \ldots \left( 4 \right)$

Subtracting $$\left( 4 \right)$$from $$\left( 2 \right)$$, we obtain

\begin{align}{x^2} + {y^2} - 2xy &= 16\\x - y&= \pm 4 \quad \quad \ldots \left( 5 \right)\end{align}

Therefore, from $$\left( 1 \right)$$and $$\left( 5 \right)$$, 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 $$8$$.

## Chapter 15 Ex.15.ME Question 2

The mean and variance of seven observations are $$8$$ and $$16$$, respectively. If six of the observations are $$2,4,10,12$$ and $$14$$, find the remaining two observations.

### Solution

Let the remaining two observations be $$x$$ and $$y$$.

Therefore, the observations are $$2,4,10,12,14,x,y$$

Mean,

\begin{align}\overline x &= \frac{{2 + 4 + 10 + 12 + 14 + x + y}}{7}\\8& = \frac{{42 + x + y}}{7}\\42 + x + y& = 56\\x + y &= 14 \qquad \quad \ldots \left( 1 \right)\end{align}

Variance,

\begin{align}16 &= \frac{1}{n}{\sum\limits_{i = 1}^7 {\left( {{x_i} - \overline x } \right)} ^2}\\16 &= \frac{1}{7}\left[ {{{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 6 \right)}^2} + {x^2} + {y^2} - 2 \times 8\left( {x + y} \right) + 2 \times {{\left( 8 \right)}^2}} \right]\\16 &= \frac{1}{7}\left[ {36 + 16 + 4 + 16 + 36 + {x^2} + {y^2} - 16\left( {14} \right) + 2\left( {64} \right)} \right]\\16 &= \frac{1}{7}\left[ {108 + {x^2} + {y^2} - 224 + 128} \right]\\16& = \frac{1}{7}\left[ {12 + {x^2} + {y^2}} \right]\\ &\Rightarrow {x^2} + {y^2} = 100\dots\dots\left( 2 \right)\end{align}

From $$\left( 1 \right)$$, we obtain

$${x^2} + {y^2} + 2xy = 196\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)$$

From $$\left( 2 \right)$$and $$\left( 3 \right)$$, we obtain

\begin{align}2xy &= 196 - 100\\2xy &= 96\qquad \ldots \left( 4 \right)\end{align}

Subtracting $$\left( 4 \right)$$from $$\left( 2 \right)$$, we obtain

\begin{align}{x^2} + {y^2} - 2xy& = 100 - 96\\{\left( {x - y} \right)^2} &= 4\\x - y &= \pm 2\qquad \quad \ldots \left( 5 \right) \end{align}

Therefore, from $$\left( 1 \right)$$ and $$\left( 5 \right)$$, 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 $$8$$.

## Chapter 15 Ex.15.ME Question 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.

### Solution

Let the observations be $${x_1},{x_2},{x_3},{x_4},{x_5}$$ and $${x_6}$$.

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

Mean,

$$\bar x = \frac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = 8 \qquad \quad \ldots \left( 1 \right)$$

If each observation is multiplied by $${\rm{3}}$$ and the resulting observations are $${y_i}$$, then

$${y_i} = 3{x_i}$$ i.e., $${x_i} = \frac{1}{3}{y_i}$$, for $$i = 1$$to $$6$$

Therefore, new mean,

\begin{align}\bar y& = \frac{{{y_1} + {y_2} + {y_3} + {y_4} + {y_5} + {y_6}}}{6}\\&= \frac{{3\left( {{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}} \right)}}{6}\\&= 3 \times 8 \qquad \quad \ldots \left( {{\rm{from }}1} \right)\\&= 24\end{align}

Standard deviation,

\begin{align}\left( \sigma \right) &= \sqrt {\frac{1}{n}\sum\limits_{i = 1}^6 {{{\left( {{x_1} - \overline x } \right)}^2}} } \\{\left( 4 \right)^2} &= \frac{1}{6}\sum\limits_{i = 1}^6 {{{\left( {{x_1} - \overline x } \right)}^2}} \\\sum\limits_{i = 1}^6 {{{\left( {{x_1} - \overline x } \right)}^2}} &= 96 \qquad \quad \ldots \left( 2 \right)\end{align}

From $$\left( 1 \right)$$and $$\left( 2 \right)$$, it can be observed that,

$$\bar y = 3\bar x$$ and $$\bar x = \frac{1}{3}\bar y$$

Substituting the values of $${x_1}$$ and $$\bar x$$ in $$\left( 2 \right)$$, we obtain

\begin{align}\sum\limits_{i = 1}^6 {{{\left( {\frac{1}{3}{y_i} - \frac{1}{3}\bar y} \right)}^2}} &= 96 \\\sum\limits_{i = 1}^6 {{{\left( {{y_i} - \bar y} \right)}^2}} &= 864 \end{align}

Therefore, variance of new observations is $$\left( {\frac{1}{6} \times 864} \right) = 144$$

Hence, the standard deviation of new observations is $$\sqrt {144} = 12$$

## Chapter 15 Ex.15.ME Question 4

Given that $$\bar x$$ is the mean and $${\sigma ^2}$$ is the variation of n observations $${x_1},{x_2}, \ldots ,{x_n}$$, Prove that the mean and variance of the observations $$a{x_1},a{x_2}, \ldots ,a{x_n}$$ are $$a\bar x$$ and $${a^2}{\sigma ^2}$$, respectively $$\left( {a \ne 0} \right)$$.

### Solution

The given n observations are $${x_1},{x_2}, \ldots ,{x_n}$$

Mean $$= \bar x$$

Variance=$$= {\sigma ^2}$$

Therefore,

$${\sigma ^2} = \frac{1}{n}\sum\limits_{i = 1}^n {{y_i}} {\left( {{x_i} - \bar x} \right)^2} \qquad \quad \ldots \left( 1 \right)$$

If each observation is multiplied by $$a$$ and the new observations are$${y_i}$$, then

$${y_i} = a{x_i}$$ i.e., $${x_i} = \frac{1}{a}{y_i}$$

Hence,

\begin{align} \bar{y}&=\frac{1}{n}\sum\limits_{i=1}^{n}{{{y}_{i}}} \\ & =\frac{1}{n}\sum\limits_{i=1}^{n}{a{{x}_{i}}} \\ & =\frac{a}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}} \\ & =a\bar{x}\ \ \ \ \ \ \ \ \ \ \ \ \ \left( \because \overline{x}=\frac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}} \right) \end{align}

Therefore, mean of the observations, $$a{x_1},a{x_2}, \ldots ,a{x_n}$$ is $$a\bar x$$

Substituting the values of $${x_i}$$ and $$\bar x$$ in $$\left( 1 \right)$$, we obtain

\begin{align}{\sigma ^2} &= \frac{1}{n}{\sum\limits_{i = 1}^n {\left( {\frac{1}{a}{y_i} - \frac{1}{a}\bar y} \right)} ^2}\\{a^2}{\sigma ^2}& = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{y_i} - \bar y} \right)}^2}} \end{align}

Thus, the variance of the observations, $$a{x_1},a{x_2},............a{x_n},$$ is $${a^2}{\sigma ^2}$$

## Chapter 15 Ex.15.ME Question 5

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 wrong item is omitted.

(ii) If it is replaced by $$12$$.

### Solution

(i) Number of observations $$\left( n \right) = 20$$

Incorrect mean $$= 10$$

Incorrect standard deviation $$= 2$$

\begin{align}\overline x &= \frac{1}{n}\sum\limits_{i = 1}^{20} {{x_i}} \\10 &= \frac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}} \\\sum\limits_{i = 1}^{20} {{x_i}} &= 200\end{align}

That is, incorrect sum of observations $$= 200$$

Correct sum of observations $$= 200 - 8 = 192$$

Therefore, correct mean $$= \frac{{{\rm{correct sum}}}}{{19}} = \frac{{192}}{{19}} = 10.1$$

Standard deviation,

\begin{align}\sigma &= \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {x_i^2 - } \frac{1}{{{n^2}}}{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}^2}} = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {x_i^2 - } {{\left( {\overline x } \right)}^2}} \\2 &= \sqrt {\frac{1}{{20}}\text{Incorrect}\sum\limits_{i = 1}^n {x_i^2 - } {{\left( {10} \right)}^2}} \\4 &= \frac{1}{{20}}\text{Incorrect}\sum\limits_{i = 1}^n {x_i^2 - } 100\\&\text{Incorrect}\sum\limits_{i = 1}^n {x_i^2 = 2080} \end{align}

Hence, $$Correct\sum\limits_{i = 1}^n {x_i^2 =\text{Incorrect}} \sum\limits_{i = 1}^n {x_i^2 - {{\left( 8 \right)}^2}}$$

\begin{align}&= 2080 - 64\\&= 2016\end{align}

Correct standard deviation $$= \sqrt {\frac{{Correct\sum\limits_{i = 1}^n {x_i^2} }}{n} - {{\left( {\text{correct}{\text{ mean}}} \right)}^2}}$$

\begin{align}&= \sqrt {\frac{{2016}}{{19}} - {{\left( {\frac{{192}}{{19}}} \right)}^2}} \\&= \sqrt {\frac{{1440}}{{361}}} \\&= \sqrt {3.988} \\&= 1.99\end{align}

(ii) When $${\rm{8}}$$ is replaced by $$12$$,

Incorrect sum of observations $$= 200$$

Correct sum of observations $$= 200 - 8 + 12 = 204$$

Hence, Correct mean $$= \frac{{correct{\rm{ }}sum}}{{20}} = \frac{{204}}{{20}} = 10.2$$

Standard deviation,

\begin{align}\sigma &= \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {x_i^2 - } \frac{1}{{{n^2}}}{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}^2}} = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {x_i^2 - } {{\left( {\overline x } \right)}^2}} \\2 &= \sqrt {\frac{1}{{20}}\text{Incorrect}\sum\limits_{i = 1}^n {x_i^2 - } {{\left( {10} \right)}^2}} \\4& = \frac{1}{{20}}\text{Incorrect}\sum\limits_{i = 1}^n {x_i^2 - } 100\\&\text{Incorrect}\sum\limits_{i = 1}^n {x_i^2 = 2080}\end{align}

Therefore, $$Correct\sum\limits_{i = 1}^n {x_i^2 = \text{Incorrect}} \sum\limits_{i = 1}^n {x_i^2 - {{\left( 8 \right)}^2}} + {\left( {12} \right)^2}$$

\begin{align}&= 2080 - 64 + 144\\&= 2160\end{align}

Correct standard deviation $$= \sqrt {\frac{{Correct\sum {{x_i}^2} }}{n} - {{\left( {\text{correct}{\text{ mean}}} \right)}^2}}$$

\begin{align}&= \sqrt {\frac{{2160}}{{20}} - {{\left( {10.2} \right)}^2}} \\&= \sqrt {108 - 104.04} \\&= \sqrt {3.96} \\&= 1.98\end{align}

## Chapter 15 Ex.15.ME Question 6

The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:

 Subject Mathematics Physics Chemistry Mean $$42$$ $$32$$ $$40.9$$ Standard deviation $$12$$ $$15$$ $$20$$

Which of the three subjects shows the highest variability in marks and which shows the lowest?

### Solution

Standard deviation of mathematics $$= 12$$

Standard deviation of Physics $$= 15$$

Standard deviation of Chemistry $$= 20$$

The coefficient of variation (C.V) is given by $$\frac{{\text{standard}-{\text{deviation }}}}{{mean}} \times 100$$

$$C.V\left( {Mathematics} \right) = \frac{{12}}{{42}} \times 100 = 28.57$$

$$C.V\left( {Physics} \right) = \frac{{15}}{{32}} \times 100 = 46.87$$

$$C.V\left( {Chemistry} \right) = \frac{{20}}{{40.9}} \times 100 = 48.89$$

The subject with greater C.V is more variable than others.

Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is in Mathematics.

## Chapter 15 Ex.15.ME Question 7

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.

### Solution

Number of observations $$= 100$$

Incorrect mean $$\left( {\bar x} \right) = 20$$

Incorrect standard deviation $$\left( \sigma \right) = 3$$

\begin{align}20 &= \frac{1}{{100}}\sum\limits_{i = 1}^{100} {{x_i}} \\\sum\limits_{i = 1}^{100} {{x_i}} &= 20 \times 100\\&= 2000\end{align}

Incorrect sum of observations $$= 2000$$

Correct sum of observations $$= 2000 - 21 - 21 - 18 = 2000 - 60 = 1940$$

Therefore, Correct mean $$= \frac{{correct{\rm{ }}sum}}{{100 - 3}} = \frac{{1940}}{{97}} = 20$$

Standard deviation $$\left( \sigma \right) = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {{x_i} - \frac{1}{{{n^2}}}{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}^2}} } = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {{x_i}^2} - {{\left( {\overline x } \right)}^2}}$$

\begin{align}&\Rightarrow 3 = \sqrt {\frac{1}{{100}} \times \text{Incorrect}\sum {{x_i}^2} - {{\left( {20} \right)}^2}} \\&\Rightarrow \text{Incorrect}\sum {{x_i}^2 = 100} \left( {9 + 400} \right) = 40900\end{align}

\begin{align}\text{Correct}\sum\limits_{i = 1}^n {x_i^2} &= \text{Incorrect}\sum\limits_{i = 1}^n {x_i^2 - {{\left( {21} \right)}^2} - } {\left( {21} \right)^2} - {\left( {18} \right)^2}\\A&= 40900 - 441 - 441 - 324\\&= 39694\end{align}

Correct standard deviation

\begin{align}&= \sqrt {\frac{{\text{Correct}\sum {{x_i}^2} }}{n} - {{\left( {\text{Correct}{\text{ mean}}} \right)}^2}} \\&= \sqrt {\frac{{39694}}{{97}} - {{\left( {20} \right)}^2}} \\&= \sqrt {409.216 - 400} \\&= \sqrt {9.216} \\&= 3.036\end{align}

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