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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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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