NCERT Solutions for Class 11 Maths Chapter 13 Statistics Miscellaneous Exercise

Miscellaneous Exercise

1. The mean and variance of eight observations are ${\text{9}}$ and ${\text{9}}{\text{.25}}$ respectively. If six of the observations are \[{\text{6,}}\,{\text{7,10,}}\,{\text{12,}}\,{\text{12}}\] and \[{\text{13}}\], find the remaining two observations.

Ans: Consider the remaining two observations to be $x$ and $y$. Thus the eight observations are \[6,\,7,10,\,12,\,12,\,13,\,x,\,y\].

Therefore, from the mean of the data it can be obtained that,

\[\overline x  = \dfrac{{6 + 7 + \,10 + 12 + \,12 + 13 + x + y}}{8} = 9\] 

\[ \Rightarrow 60 + x + y = 72\]

\[ \Rightarrow x + y = 12\] ......(i)

Again, from the variance of the data it can be obtained that,

\[{\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^8 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{N} = 9.25\]

\[\therefore \dfrac{1}{8}\left[ {{{\left( { - 3} \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + {x^2} + {y^2} - (2)(9)\left( {x + y} \right) + \left( {2{{\left( 9 \right)}^2}} \right)} \right] = 9.25\]

\[ \Rightarrow \dfrac{1}{8}\left[ {9 + 4 + 9 + 16 + {x^2} + {y^2} - (18)\left( {12} \right) + 162} \right] = 9.25\]   [By, using (i)]

\[ \Rightarrow \dfrac{1}{8}\left[ {{x^2} + {y^2} - 6} \right] = 9.25\]

\[ \Rightarrow {x^2} + {y^2} = 80\] ......(ii)

Observe that form equation (i), it can be obtained that,

${x^2} + {y^2} + 2xy = 144$ ......(iii)

Also, from equations (ii) and (iii), it can be obtained that,

\[2xy = 64\] ......(iv) 

Now, on subtracting equation (iv) from equation (ii), it can be obtained that,

\[{x^2} + {y^2} - 2xy = 16\] 

\[ \Rightarrow x - y =  \pm 4\] ......(v)

It can be calculated from equations (i) and (v) that, $x = 8$ and $y = 4$, when \[x - y = 4\] and $x = 4$ and $y = 8$, when \[x - y =  - 4\]. Henceforth, the remaining two observations are $4$ and $8$.

2. The mean and variance of ${\text{7}}$ observations are ${\text{8}}$ and ${\text{16}}$ respectively. If five of the observations are \[{\text{2,}}\,{\text{4,10,}}\,{\text{12}}\] and \[{\text{14}}\]. Find the remaining two observations.

Ans: Consider the remaining two observations to be $x$ and $y$. Thus the eight observations are \[2,\,4,10,\,12,\,14,\,x,\,y\].

Therefore, from the mean of the data it can be obtained that,

\[\overline x  = \dfrac{{2 + 4 + \,10 + 12 + \,14 + x + y}}{7} = 8\] 

\[ \Rightarrow 42 + x + y = 56\]

\[ \Rightarrow x + y = 14\] ......(i)

Again, from the variance of the data it can be obtained that,

\[{\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^7 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{N} = 16\]

\[\therefore \dfrac{1}{7}\left[ {{{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 6 \right)}^2} + {x^2} + {y^2} - (2)(8)\left( {x + y} \right) + \left( {2{{\left( 8 \right)}^2}} \right)} \right] = 16\]

\[ \Rightarrow \dfrac{1}{7}\left[ {36 + 16 + 4 + 16 + {x^2} + {y^2} - (16)\left( {14} \right) + 128} \right] = 16\]   [By, using (i)]

\[ \Rightarrow \dfrac{1}{7}\left[ {{x^2} + {y^2} + 12} \right] = 16\]

\[ \Rightarrow {x^2} + {y^2} = 100\] ......(ii)

Observe that form equation (i), it can be obtained that,

${x^2} + {y^2} + 2xy = 196$ ......(iii)

Also, from equations (ii) and (iii), it can be obtained that,

\[2xy = 96\] ......(iv) 

Now, on subtracting equation (iv) from equation (ii), it can be obtained that,

\[{x^2} + {y^2} - 2xy = 4\] 

\[ \Rightarrow x - y =  \pm 2\] ......(v)

It can be calculated from equations (i) and (v) that, $x = 8$ and $y = 6$, when \[x - y = 2\]and $x = 6$ and $y = 8$, when \[x - y =  - 2\]. Henceforth, the remaining two observations are $6$ and $8$.

3. The mean and standard deviation of six observations are ${\text{8}}$ and ${\text{4}}$ respectively. If each observation is multiplied by ${\text{3}}$, find the new mean and new standard deviation of the resulting observations.

Ans: Assume the observations to be \[{x_1},\,{x_2},\,{x_3},\,{x_4},\,{x_5},\,{x_6}\].

Therefore, the mean of the data is,

\[\overline x  = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = 8\] 

\[ \Rightarrow \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = 8\] ......(i)

Observe that when each of the observation is multiplied with $3$ and if we consider the resulting observations as ${y_i}$, then observe as shown below,

$\therefore {y_1} = 3{x_1}$ 

$ \Rightarrow {x_1} = \dfrac{1}{3}{y_1},\,\forall i = 1,\,2,\,3,\,....,\,6\& i \in {\mathbb{Z}^ + }$  

Now, the mean of the new data is,

\[\therefore \overline y  = \dfrac{{{y_1} + {y_2} + {y_3} + {y_4} + {y_5} + {y_6}}}{6}\] 

\[ \Rightarrow \overline y  = \dfrac{{3({x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6})}}{6}\]

\[ \Rightarrow \overline y  = 3 \times 8\]      [By using (i)]

\[ \Rightarrow \overline y  = 24\]

Therefore, the standard deviation of the data is,

\[\sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^6 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}}  = 4\]

\[ \Rightarrow {\left( 4 \right)^2} = \dfrac{{\sum\limits_{i = 1}^6 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{6}\]

\[ \Rightarrow \sum\limits_{i = 1}^6 {{{\left( {{x_i} - \overline x } \right)}^2}}  = 96\] ......(ii)

Again, it can be observed from equations (i) and (ii) that \[\overline y  = 3\overline x \] and \[\overline x  = \dfrac{{\overline y }}{3}\] and hence on substituting the values of ${x_i}$ and $\overline x $ in equation (ii) it can be clearly obtained as shown below,

 \[\therefore \sum\limits_{i = 1}^6 {{{\left( {\dfrac{{{y_i}}}{3} - \dfrac{{\overline y }}{3}} \right)}^2}}  = 96\]

\[ \Rightarrow \sum\limits_{i = 1}^6 {{{\left( {{y_i} - \overline y } \right)}^2}}  = 864\]

Henceforth, the standard deviation of the new data can be calculated as shown below,

\[\therefore \sigma  = \sqrt {\dfrac{{864}}{6}}  = \sqrt {144}  = 12\]

4. Given that $\overline {\text{x}} $ is the mean and $\sigma^{2}$ is the variance of ${\text{n}}$ observations ${{\text{x}}_{\text{1}}}{\text{,}}\,{{\text{x}}_{\text{2}}}{\text{,}}\,......{\text{,}}\,{{\text{x}}_{\text{n}}}$.Prove that the mean and variance of observations ${\text{a}}{{\text{x}}_{\text{1}}}{\text{,}}\,{\text{a}}{{\text{x}}_{\text{2}}}{\text{,}}\,{\text{a}}{{\text{x}}_{\text{3}}}{\text{, }}.....{\text{,}}\,{\text{a}}{{\text{x}}_{\text{n}}}$ are ${\text{a}}\overline {\text{x}} $ and ${{\text{a}}^{\text{2}}}{{\sigma}}^{\text{2}}$, respectively $\left( {{\text{a}} \ne {\text{0}}} \right)$.

Ans: Observe that the given observations are ${x_1},\,{x_2},\,......,\,{x_n}$ and the mean and variance of data is $\overline x $ and ${\sigma ^2}$ respectively.

\[\therefore {\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{y_i}{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}\]......(i)

Observe that when each of the observation is multiplied with $a$ and if we consider the resulting observations as ${y_i}$, then observe as shown below,

$\therefore {y_i} = a{x_i}$ 

$ \Rightarrow {x_i} = \dfrac{1}{a}{y_i},\,\forall i = 1,\,2,\,3,\,....,\,n\& i \in {\mathbb{Z}^ + }$  

Now, the mean of the new data is,

$\therefore \overline y  = \dfrac{1}{n}\sum\limits_{i = 1}^n {{y_i}}  = \dfrac{1}{n}\sum\limits_{i = 1}^n {a{x_i}}  = \dfrac{a}{n}\sum\limits_{i = 1}^n {{x_i}}  = a\overline x \,\,\,\,\,\,\,\left[ {\because \overline x  = \dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}} } \right]$ 

Again on substituting the values of ${x_i}$ and $\overline x $ in equation (i) it can be clearly obtained as shown below,

 \[\therefore {\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {\dfrac{{{y_i}}}{a} - \dfrac{{\overline y }}{a}} \right)}^2}} }}{n}\]

\[ \Rightarrow {a^2}{\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {{y_i} - \overline y } \right)}^2}} }}{n}\]

Henceforth, it can be clearly proved that the mean and variance of the new data is $a\overline x $ and ${a^2}{\sigma ^2}$, respectively.

5. The mean and standard deviation of ${\text{20}}$ observations are found to be ${\text{10}}$ and ${\text{2}}$, respectively. On rechecking it was found that an observation ${\text{8}}$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

(i) If wrong item is omitted 

Ans: Observe that the incorrect number of observations, incorrect mean and the incorrect standard deviation are $20$, $10$ and $2$, respectively. ${x_1},\,{x_2},\,......,\,{x_n}$ and the mean and variance of data is $\overline x $ and ${\sigma ^2}$ respectively.

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{n} = 10\] 

\[ \Rightarrow \dfrac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{{20}} = 10\] 

\[ \Rightarrow \sum\limits_{i = 1}^{20} {{x_i}}  = 200\]

Now, the correct mean is,

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{19} {{x_i}} }}{n}\] 

\[ \Rightarrow \overline x  = \dfrac{{192}}{{19}} = 10.1\,\,\,\,\,\left[ {\because \sum\limits_{i = 1}^{19} {{x_i}}  = 192} \right]\]

Again observe as shown below,

\[\therefore \sigma  = \sqrt {\dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} }  = 2\]

\[ \Rightarrow 4 = \dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} \]

\[ \Rightarrow 2080 = \sum\limits_{i = 1}^{20} {{x_i}^2} \]

Therefore, the correct standard deviation of the data is,

\[\sum\limits_{i = 1}^{20} {{x_i}^2}  - {\left( 8 \right)^2}\]

\[ \Rightarrow 2080 - 64\]

\[ \Rightarrow 2016\]

\[\therefore \sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^{19} {{x_i}^2} }}{n} - {{\left( {\overline x } \right)}^2}} \] 

\[ \Rightarrow \sigma  = \sqrt {\dfrac{{2016}}{{19}} - {{\left( {10.1} \right)}^2}} \]

\[ \Rightarrow \sigma  = \sqrt {1061.1 - 102.1}  = \sqrt {4.09}  = 2.02\]

(ii) If it is replaced by ${\text{12}}$.  

Ans: Observe that the incorrect sum of observations is $200$.

\[\therefore \sum\limits_{i = 1}^{20} {{x_i}}  = 200 - 8 + 12 = 204\]

Now, the correct mean is,

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{n}\] 

\[ \Rightarrow \overline x  = \dfrac{{204}}{{20}} = 10.2\,\,\,\,\,\left[ {\because \sum\limits_{i = 1}^{20} {{x_i}}  = 204} \right]\]

Again observe as shown below,

\[\therefore \sigma  = \sqrt {\dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} }  = 2\]

\[ \Rightarrow 4 = \dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} \]

\[ \Rightarrow 2080 = \sum\limits_{i = 1}^{20} {{x_i}^2} \]

Therefore, the correct standard deviation of the data is,

\[\sum\limits_{i = 1}^{20} {{x_i}^2}  - {\left( 8 \right)^2} + {\left( {12} \right)^2}\]

\[ \Rightarrow 2080 - 64 + 144\]

\[ \Rightarrow 2160\]

\[\therefore \sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^{20} {{x_i}^2} }}{n} - {{\left( {\overline x } \right)}^2}} \] 

\[ \Rightarrow \sigma  = \sqrt {\dfrac{{2160}}{{20}} - {{\left( {10.2} \right)}^2}} \]

\[ \Rightarrow \sigma  = \sqrt {108 - 104.04}  = \sqrt {3.96}  = 1.98\]

6. The mean and standard deviation of a group of ${\text{100}}$ observations were found to be ${\text{20}}$ and ${\text{3}}$, respectively. Later on it was found that ${\text{3}}$ observations are incorrect which were recorded as ${\text{21,}}\,{\text{21}}$ and ${\text{18}}$. Find the mean and standard deviation if the incorrect  observations are omitted.

Ans: Observe that the number of observations, incorrect mean and the incorrect standard deviation are $100$, $20$ and $3$, respectively. ${x_1},\,{x_2},\,......,\,{x_n}$ and the mean and variance of data is $\overline x $ and ${\sigma ^2}$ respectively.

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{100} {{x_i}} }}{n} = 20\] 

\[ \Rightarrow \dfrac{{\sum\limits_{i = 1}^{100} {{x_i}} }}{{100}} = 20\] 

\[ \Rightarrow \sum\limits_{i = 1}^{100} {{x_i}}  = 2000\]

Now, the correct mean is,

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{97} {{x_i}} }}{n}\] 

\[ \Rightarrow \overline x  = \dfrac{{2000 - 60}}{{97}} = 20\,\,\,\,\,\left[ {\because \sum\limits_{i = 1}^{97} {{x_i}}  = 1940} \right]\]

Again observe as shown below,

\[\therefore \sigma  = \sqrt {\dfrac{1}{{100}}\sum\limits_{i = 1}^{100} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} }  = 3\]

\[ \Rightarrow 9 = \dfrac{1}{{100}}\sum\limits_{i = 1}^{100} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} \]

\[ \Rightarrow 40900 = \sum\limits_{i = 1}^{100} {{x_i}^2} \]

Therefore, the correct standard deviation of the data is,

\[\sum\limits_{i = 1}^{100} {{x_i}^2}  - (2){\left( {21} \right)^2} - {(18)^2}\]

\[ \Rightarrow 40900 - 1206\]

\[ \Rightarrow 39694\]

\[\therefore \sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^{97} {{x_i}^2} }}{n} - {{\left( {\overline x } \right)}^2}} \] 

\[ \Rightarrow \sigma  = \sqrt {\dfrac{{39694}}{{97}} - {{\left( {20} \right)}^2}} \]

\[ \Rightarrow \sigma  = \sqrt {409.22 - 400}  = \sqrt {9.22}  = 3.04\].

Conclusion

NCERT Solutions for Statistics Class 11 Miscellaneous Exercise, are crucial for understanding how to work with data effectively. This chapter teaches methods like finding averages (mean, median, mode) and measures of spread (standard deviation) to analyze data accurately. It's important to focus on practicing these methods to become proficient in interpreting data and making informed decisions. By mastering these statistical techniques, students develop practical skills that are applicable across various fields, from business analytics to scientific research.

Class 11 Maths Chapter 13: Exercises Breakdown

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