Chapter 15: Statistics

Ans: The correct option is (A).

Given: Let’s consider the first 10 positive numbers

Multiply by \[ - 1\] and add \[{\text{1}}\] to each number 

For Example,

$1 \times ( - 1) =  - 1 + 1 = 0$

We know that, \[{\sigma ^2} = {\dfrac{\sum {{x_i}^2}}{n}}  - {\left( { {\dfrac{{\sum{x_i}}}{n}} } \right)^2}\]

Since the first 10 positive integers are \[1,2,3,4,5,6,7,8,9{\text{ and }}10\].

On multiplying each number by \[ - 1,\]

We get \[ - 1, - 2, - 3, - 4, - 5, - 6, - 7, - 8, - 9, - 10\]

On adding \[{\text{1}}\] in each number,

We get \[ - 1, - 2, - 3, - 4, - 5, - 6, - 7, - 8, - 9, - 10\]

${\sigma ^2} = \sum {\dfrac{{{x_i}^2}}{n}}  - {\left( {\sum {\dfrac{{{x_i}^2}}{n}} } \right)^2}$

$\sum {{x_i} = \dfrac{{ - 9 \times 10}}{2} =  - 45} $

$\sum {{x_i}^2 = {0^2} +  - {1^2} + {{( - 2)}^2} + {{( - 3)}^2} + ....... + {{( - 9)}^2} = \dfrac{{9 \times 10 \times 19}}{6} = 285} $

${\sigma ^2} = {\dfrac{{285}}{{10}}}  - {\left( {{\dfrac{{ - 45}}{{10}}} } \right)^2} = {\dfrac{{285}}{{10}}}  - {\left( {{\dfrac{{2025}}{{10}}} } \right)}$ 

\[ = \sqrt {\dfrac{{2850 - 2025}}{{100}}}  = {\left( {\sqrt {8.25} } \right)^2} = 8.25\]

Correct Answer: A

Ans: The correct option is (D).

Given: $\sum{x^2} = 18000,  \sum{x} =960$

We know that,

\[{\sigma ^2} = {\dfrac{\sum {{x_i}^2}}{N}}  - {{{\left( {\dfrac{{\sum {x_i}}}{n}} \right)}^2}} \]

Then,

\[{\sigma ^2} = {\dfrac{\sum {{x_i}^2}}{N}}  - {{{\left( {\dfrac{{\sum {x_i}}}{n}} \right)}^2}} \]

$N = 60$

${\sigma ^2} = {\dfrac{{18000}}{{60}}}  - {{{\left( {\dfrac{{960}}{{60}}} \right)}^2}} $

$= 300 - {(16)^2}$

$= 44$ 

Correct Answer: D

Ans: The correct option is (A).

Given: Coefficients of variations are $50$ and $60$

Arithmetic means are $30$ and $25$

We know that,\[{\text{C}}{{\text{V}}_1} = \dfrac{{{\sigma _1}}}{{\overline {{x_1}} }} \times 100\]

Then,

$C{V_1} = 50,   C{V_2} = 60$

$\overline {{x_1}}  = 30, \overline {{x_2}}  = 25$

$c{v_1} = \dfrac{{{\sigma _1}}}{{\overline {{x_1}} }} \times 100$

${\sigma _1} = \dfrac{{50 \times 30}}{{100}} = 15$

$C{V_2} = \dfrac{{{\sigma _2}}}{{\overline {{x_2}} }} \times 100$

${\sigma _2} = \dfrac{{60 \times 25}}{{100}} = 15$

${\text{now }}$

${\sigma _1} = {\sigma _2} = 15$ 

The difference in standard deviation is \[0.\]

Correct Answer: A

Ans: The correct option is (A).

Given: Some data are given in \[{}^{\circ}C  \text{ is }  5\]

We know that,\[F = \dfrac{9}{5}C + 32  \]

Then,

${\sigma _c} \text{ and } {\sigma _f} = ?$

$F = \dfrac{9}{5}c + 32$       

As We know that on adding and subtracting any number in data the standard deviation doesn’t change but on multiplying and dividing the data by the same number the standard deviation changes according to the number.

$x \to s$

$kx \to ks$

$x \to s$

$k + x \to s$ 

${\sigma _f} = \dfrac{9}{5}{\sigma _c}$

$\Rightarrow {\sigma _f} = 9$ ∵ Given ${\sigma _c}=5$

$\Rightarrow {\text{variance}}  = {(SD)^2} = {9^2} = 81$ 

Correct Answer: A

Ans: The formula for computing variation is 

Coefficient of variation C.V $ = \dfrac{\sigma }{\mu } \times 100$

Where, $\sigma  = {\text{standard deviation}}$

$\mu  = {\text{Mean}}$

Therefore,

${\text{C}}{\text{.V}}$$ = $$\dfrac{{{\text{Standard deviation}}}}{{{\text{Mean}}}} \times 100$

Ans: Mean of $n$ observations ${x_1},{x_2},.....,{x_n}$$ = \bar x$

Therefore, $\bar x = \dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}$

$ \Rightarrow \sum\limits_{i = 1}^n {{x_i} = n\bar x}$ then,

$\sum\limits_{i = 1}^n {({x_i} - \bar x)}$$ = $$\sum\limits_{i = 1}^n {{x_i} - \sum\limits_{i = 1}^n {\bar x} }$

$= n\bar x - \bar x\sum\limits_{i = 1}^n {1\left[ {{\text{and}}\sum\limits_{i = 1}^n 1  = n} \right]} $

$= n\bar x - \bar x.n$

$= 0$ 

Hence, the value of $\sum\limits_{i = 1}^n {({x_i} - \bar x)}$$ = 0$.

Then, $\sum\limits_{i = 1}^n{\left( {{x_i} - \bar x} \right)^2}$$ < $${\left( {{x_i} - a} \right)^2}$.

Ans: Square root of variance is standard deviation.

${\text{i}}{\text{.e, standard deviation }} = \sqrt {{\text{variance}}} $ 

Therefore, given

variance = 121

${\sigma ^2} = 121$

$\Rightarrow \sigma  = \sqrt {121} $

$\Rightarrow \sigma  = 11$ 

Hence, standard deviation is $11$

Ans: Standard deviation is not affected by the addition or subtraction of any value from the observation therefore independent of change in origin. But change in scale affects the standard deviation i.e., any value multiplied or divided to the observation changes standard deviation too.

Ans: Mean of the number of observations converts the series of observations into a minimum value and mean deviation is the arithmetic average of deviations of the observation from an observed value. Sum of squares of deviation is minimum when deviations are taken from mean.

Ans: Sum of all the deviation of the data about the data mean is $ = $ mean deviation and the median lies between mean and mode. Therefore, mean deviation of the data is least when measured from the median.