Chapter 13: Statistics

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 $8$.

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 $8$.

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,

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

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.

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,

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

\[ \Rightarrow 2080 - 64\]

\[ \Rightarrow 2016\]

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

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

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

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,

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

\[ \Rightarrow 2080 - 64 + 144\]

\[ \Rightarrow 2160\]

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

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

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

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,

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

\[ \Rightarrow 40900 - 1206\]

\[ \Rightarrow 39694\]

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

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

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