Statistics Test - 10

Given $$\bar {x} = \dfrac {\sum x_{i}}{5} = 150$$
$$\Rightarrow \displaystyle \sum_{i = 1}^{5} x_{i} = 750 .... (i)$$

Mean of the date $$7, 8, 9, 7, 8, 7, \lambda, 8$$ is $$8$$

$$\displaystyle\sum_{i = 1}^{10} (x_i - 5) = 10$$

Let $$x$$ be the $$6^{th}$$ observation
$$\Rightarrow 45 + 54 + 41 + 57 + 43 + x = 48 \times 6 = 288$$
$$\Rightarrow x = 48$$
Variance $$= \left(\dfrac{\sum x_i^2}{6} - (\bar{X})^2\right)$$
$$\Rightarrow$$ variance $$=\dfrac{14024}{6} - (48)^2$$
$$=\dfrac{100}{3}$$
$$\Rightarrow$$ standard deviation $$=\dfrac{10}{\sqrt{3}}$$

$$\dfrac{\sum x_i}{20} = 10 \Rightarrow \sum x_i  = 200$$

$$x_1+...+x_4=44$$
$$x_5+.....+x_{10}=96$$
$$\vec{x}=14, \displaystyle\sum x_i=140$$
Variance $$=\dfrac{\displaystyle\sum x^2_i}{n}=\vec{x^2}=4$$
Standard deviation $$=2$$.

$$\sigma_{\mathrm{x}}^{2}=4$$
$$\sigma_{\mathrm{y}}^{2}=5$$
$$\overline{\mathrm{x}}=2$$
$$\overline{\mathrm{y}}=4$$
$$\displaystyle \frac{\Sigma \mathrm{x}_{\mathrm{i}}}{5}=2, \Sigma \mathrm{x}_{\mathrm{i}}=10;\Sigma \mathrm{y}_{\mathrm{i}}=20$$
$$
\sigma_{\mathrm{x}}^{2}=(\frac{1}{5}\Sigma \mathrm{x}_{\mathrm{i}}^{2})-(\overline{x})^{2}=\frac{1}{5}(\Sigma \mathrm{x}_{1}^{2})-4
$$

$${\textbf{Step -1: Let initial mean}}$$ $${\mathbf{\left( {\overline x } \right)}}$$ $${\textbf{and standard deviation}}$$ $${\mathbf{\left( {{\sigma _1}} \right)}}$$$${\textbf{of 10 observation are 20 and 2 respectively.}}$$  

               $${\text{Now, each of these observations is multiplied by p and reduced by q.}}$$

               $${\text{Thus, The new mean}}$$ $$ = \overline {{x_1}}  = p\overline x  - q \ldots \left( 1 \right)$$

               $${\text{Also, it is given that the new mean is half of the original mean}}{\text{.}}$$

               $$ \Rightarrow \overline {{x_1}}  = \dfrac{1}{2}\overline x  = \dfrac{1}{2} \times 20$$

               $$ \Rightarrow \overline {{x_1}}  = 10$$

               $${\text{Substitute this value of new mean in equation 1.}}$$

               $$ \Rightarrow 10 = p\left( {20} \right) - q$$

               $$ \Rightarrow 20p - q = 10 \ldots \left( 2 \right)$$

$${\textbf{Step -2: Find the value of p and q using the standard deviation.}}$$

               $${\text{New standard deviation is given by,}}$$

              $${\sigma _2} = \left| p \right|{\sigma _1} \ldots \left( 3 \right)$$

              $${\text{As it will not be affected by subtraction of q from each observation.}}$$

              $${\text{It is given that new standard deviation is half of the original.}}$$

              $$ \Rightarrow {\sigma _2} = \dfrac{1}{2}{\sigma _1} = \dfrac{1}{2} \times 2 = 1$$

              $${\text{Substitute this value in equation 3.}}$$

             $$ \Rightarrow \left| p \right| \times 2 = 1$$

             $$ \Rightarrow p =  \pm \dfrac{1}{2}$$

$${\textbf{Step -3: Find the value of q using p.}}$$

                $${\text{If }}$$ $$p = \dfrac{1}{2}$$

                $${\text{Then from equation 2 we have,}}$$

                $$ \Rightarrow 20 \times \dfrac{1}{2} - q = 10$$

                $$ \Rightarrow 10 - 10 = q$$

                $$\Rightarrow q = 0$$    $$[\textbf{Rejected, as q}\neq 0]$$

                $${\text{If }}$$ $$p =  - \dfrac{1}{2}$$

                $${\text{Using equation 2, we have,}}$$

                $$ \Rightarrow 20 \times \left( { - \dfrac{1}{2}} \right) - q = 10$$

                $$ \Rightarrow q =  - 10 - 10$$

                $$ \Rightarrow q =  - 20$$

$${\textbf{Hence, option C. i.e.}}{\mathbf{\left ( -20 \right )}} {\textbf{ is the correct answer.}}$$

 

Mean $$\quad \left( \mu  \right) =\cfrac { \sum { { x }_{ i } }  }{ 50 } =16$$
standard deviation $$\left( \sigma  \right) =\sqrt { \cfrac { \sum { { x }_{ i }^{ 2 } }  }{ 50 } -{ \left( \mu  \right)  }^{ 2 } } =16\Rightarrow \left( 256 \right) \times 2=\cfrac { \sum { { x }_{ i }^{ 2 } }  }{ 50 } $$
New mean
$$=\cfrac { \sum { ({ x }_{ i }-4)^{ 2 } }  }{ 50 } =\cfrac { \sum { { x }_{ i }^{ 2 } } +16\times 50-8\sum { { x }_{ i }^{  } }  }{ 50 } =\left( 256 \right) \times 2=16-8\times 16=400$$