Chapter 15: Statistics

Ans: The correct option is (A).

Given: The lives in hours of $5$ bulbs are $1357$, $1090$, $1666$, $1494$, $1623$.

Use the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$ to calculate the mean of the given data. Here $\overline x $ is the mean.

Now, use the formula $M.D. = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - \overline x } \right|} }}{n}$ to calculate the mean deviation.

The total number of observations is $n = 5$. Using the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$, where $\overline x $ is the mean,

$\Rightarrow \overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^5 {{x_i}} } \right)}}{5}$

$\Rightarrow \overline x  = \dfrac{{1357 + 1090 + 1666 + 1494 + 1623}}{5}$

$\Rightarrow \overline x  = 1446{\text{ hours}}$ 

Now, the mean deviation is given as $M.D. = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - \overline x } \right|} }}{n}$. So for the given data,

$\Rightarrow M.D. = \dfrac{{\sum\limits_{i = 1}^5 {\left| {{x_i} - 1446} \right|} }}{5}$

$\Rightarrow M.D. = \dfrac{{\left| {1357 - 1446} \right| + \left| {1090 - 1446} \right| + \left| {1666 - 1446} \right| + \left| {1494 - 1446} \right| + \left| {1623 - 1446} \right|}}{5}$

$\Rightarrow M.D. = 178{\text{ hours}}$ 

Correct Answer: A

Ans: The correct option is (C).

Given: The marks obtained by $9$ students in the test are $50$, $69$, $20$, $33$, $53$, $39$, $40$, $65$, $59$.

First arrange the marks in ascending order as $20, 33, 39, 40, 50, 53, 59, 65, 69$ and find the median $\left( M \right)$ which will be ${\left( {\dfrac{{n + 1}}{2}} \right)^{th}}$ term after the arrangement.

Now, use the formula $M.D. = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - M} \right|} }}{n}$ to calculate the mean deviation from the median.

To find the median the data must be arranged in either ascending or descending order of their value. So the arrangement in ascending order is $20$, $33$, $39$, $40$, $50$, $53$, $59$, $65$, $69$.

The number of observations $\left( n \right)$ is odd that means the median $\left( M \right)$ is given as ${\left( {\dfrac{{n + 1}}{2}} \right)^{th}}$ term.

$\Rightarrow M = {\left( {\dfrac{{9 + 1}}{2}} \right)^{th}}{\text{ term}}$

$\Rightarrow M = {\left( 5 \right)^{th}}{\text{ term}}$

$\Rightarrow M = 50$ 

Now, the mean deviation from the median is given as $M.D. = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - M} \right|} }}{n}$. So for the given data,

$\Rightarrow M.D. = \dfrac{{\sum\limits_{i = 1}^9 {\left| {{x_i} - 50} \right|} }}{9}$

$\Rightarrow M.D.=\dfrac{{\left|{20-50}\right|+\left|{33-50}\right|+\left|{39-50}\right|+ \left|{40-50}\right|+\left|{50-50} \right|+\left|{53-50}\right|+\left|{59-50}\right|+\left| {65 - 50} \right|+\left|{69-50}\right|}}{9}$

$\Rightarrow M.D. = 12.67$ 

 Correct Answer: C

Ans: The correct option is (A).

Given: The data $6$, $5$, $9$, $13$, $12$, $8$, $10$.

Use the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$ to calculate the mean of the given data. Here $\overline x $ is the mean.

Now, use the formula $\sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}}$, where $\sigma $ is the standard deviation.

The total number of observations is $n = 7$. Using the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$, where $\overline x $ is the mean,

$\Rightarrow \overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^7 {{x_i}} } \right)}}{7}$

$\Rightarrow \overline x  = \dfrac{{6 + 5 + 9 + 13 + 12 + 8 + 10}}{7}$

$\Rightarrow \overline x  = 9$ 

Now, the standard deviation is given as $\sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}}$, where $\sigma $ is the standard deviation.

$\Rightarrow \sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^7 {{{\left( {{x_i} - 9} \right)}^2}} }}{7}} $

$\Rightarrow \sigma  = \sqrt {\dfrac{{{{\left( {6 - 9} \right)}^2} + {{\left( {5 - 9} \right)}^2} + {{\left( {9 - 9} \right)}^2} + {{\left( {13 - 9} \right)}^2} + {{\left( {12 - 9} \right)}^2} + {{\left( {8 - 9} \right)}^2} + {{\left( {10 - 9} \right)}^2}}}{7}} $

$\Rightarrow \sigma  = \sqrt {\dfrac{{52}}{7}} $ 

Correct Answer: A

Ans: The correct option is (C).

Given: $n$ observations ${x_1},{x_2},....,{x_n}$ and their arithmetic mean as \[\overline x \].

Use the definition of the variance which is ‘the ratio of the summation of the squares of the deviations of each observation from their mean and the total number of observations’.

Take the square root of this mathematical expression of the variance to get the standard deviation.

The arithmetic means is generally referred to as the mean $\overline x $ of the data.

According to the definition of variance it is the ratio of the summation of the squares of the deviations of each observation from their mean and the total number of observations. Mathematically it is given as,

${\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}$

Now, the standard deviation is the square root of the variance, so it is given by the relation,

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

Correct Answer: C

Ans: The correct option is (C).

Given: Total number of observations $ = 100$

Mean of the observations $ = 50$

Standard deviation $ = 5$

Use the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$ to calculate the sum of all the observations, i.e.  $\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)$. Here $\overline x $ is the mean.

Now, use the formula $\sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}}$, where $\sigma $ is the standard deviation, and calculate the variance. Break the summation into several parts and find the value of $\left( {\sum\limits_{i = 1}^n {{{\left( {{x_i}} \right)}^2}} } \right)$.

Since there are $100$ observations and the mean is $50$, so using the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$,

$ \Rightarrow 50 = \dfrac{{\left( {\sum\limits_{i = 1}^{100} {{x_i}} } \right)}}{{100}}$

$ \Rightarrow \left( {\sum\limits_{i = 1}^{100} {{x_i}} } \right) = 5000$        …… $\left( 1 \right)$

Now, the standard deviation is given as $\sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}}$, where $\sigma $ is the standard deviation. On squaring both the sides the variance is obtained. The provided value of $\sigma $ is $5$.

$\Rightarrow \sqrt {\dfrac{{\sum\limits_{i = 1}^{100} {{{\left( {{x_i} - \overline x } \right)}^2}} }}{{100}}}  = 5$

$\Rightarrow \dfrac{{\sum\limits_{i = 1}^{100} {{{\left( {{x_i} - \overline x } \right)}^2}} }}{{100}} = 25$

$\Rightarrow \sum\limits_{i = 1}^{100} {{{\left( {{x_i} - \overline x } \right)}^2}}  = 2500$ 

Breaking the summation into several parts,

\[ \Rightarrow \sum\limits_{i = 1}^{100} {{{\left( {{x_i}} \right)}^2}}  + \sum\limits_{i = 1}^{100} {{{\left( {\overline x } \right)}^2}}  - \sum\limits_{i = 1}^{100} {2\overline x {x_i}}  = 2500\]

Using the value from equation $\left( 1 \right)$,

$\Rightarrow \sum\limits_{i = 1}^{100} {{{\left( {{x_i}} \right)}^2}}  + \sum\limits_{i = 1}^{100} {{{\left( {50} \right)}^2}}  - \sum\limits_{i = 1}^{100} {2 \times 50 \times {x_i}}  = 2500$

$\Rightarrow \sum\limits_{i = 1}^{100} {{{\left( {{x_i}} \right)}^2}}  + 100 \times {\left( {50} \right)^2} - 100\sum\limits_{i = 1}^{100} {{x_i}}  = 2500$

$\Rightarrow \sum\limits_{i = 1}^{100} {{{\left( {{x_i}} \right)}^2}}  + 250000 - 500000 = 2500$

$\Rightarrow \sum\limits_{i = 1}^{100} {{{\left( {{x_i}} \right)}^2}}  = 252500$ 

Correct Answer: C

Ans: The correct option is (A).

Given: \[a,b,c,d,e \] be the observations of mean m and standard deviations.

We know that,

$Var(ax + b) = {a^{2}}\cdot Var(x)$

or

$S.D(ax + b) = a \cdot S.D(x)$ 

Then,

$Var(ax + b) = {a^{2}}.Var(x)$

$S.D(ax + b) = a \cdot S.D(x)$

$a = 1,b = k$

$\therefore S.D(x + k) = S.D(a + k,b + k,c + k,d + k,e + k)$

$= 1\cdot S.D(a,b,c,d,e) = s$ 

Correct Answer: A

Ans: The correct option is (C).

Given: \[{x_1},{x_2},{x_3},{x_{4}},{x_{5}}\] be the observations of mean \[{\text{m}}\] and standard deviation S. 

We know that,

 S.D = $\sqrt {\dfrac{{\sum {{x_i}^2} }}{n} - {{\left( {\dfrac{{\sum {{x_i}} }}{n}} \right)}^2} } $

 $\sum {{x_i}^2}  = {\left( {k{x_1}} \right)^2}{\text{ + }}.........{\text{ + }}{\left( {k{x_5}} \right)^2}$

$= {k^2}\sum {{x_i}^2} $

$\sum {{x_i}^{}}  = \left( {k{x_1}} \right){\text{ + }}.........{\text{ + }}\left( {k{x_5}} \right)$

$= k\sum {{x_i}} $ 

Then,

${\text{S}}{\text{.D}} = \sqrt {\dfrac{{\sum {{x_i}^2} }}{n} - {{\left( {\dfrac{{\sum {{x_i}}}}{n}} \right)}^2} } $

$\sum {{x_i}^2}  = {\left( {k{x_1}} \right)^2}{\text{  +  }}.........{\text{ + }}{\left( {k{x_5}} \right)^2}$

$= {k^2}\sum {{x_i}^2} $

$\sum {{x_i}^{}}  = \left( {k{x_1}} \right){\text{  +  }}.........{\text{  + }}\left( {k{x_5}} \right)$

$= k\sum {{x_i}} $

S.D $= \sqrt {\dfrac{{\sum {{x_i}^2} }}{5} - {{\left( {\dfrac{{\sum {{x_i}} }}{5}} \right)}^2} }$       Since, $\overline x  = {\dfrac{{\sum {x_i}}}{n}} ;\sigma  = S = {\text{ S}}{\text{.D}}$

$= \left| k \right|\sqrt {\dfrac{{\sum {{x_i}^2} }}{5} - {{\left( {\dfrac{{\sum {{x_i}^2} }}{5}} \right)}^2} } $

$= \left| k \right|s$ 

Correct Answer:  C

Ans: The correct option is (A).

Given: \[{x_1},{x_2},...{x_n}\] Be \[{\text{n}}\] observations

\[{w_i} = l{x_i} + k\]\[\text{ for } l = 1,2,...n,\]

\[landk\] are constants. The mean of $x_i$ is 48 and their Standard deviation is \[12\],

The mean of $w_i$’s is 55 and standard deviation of $w_i$’s is 15.

We know that,

$\sigma (ax + b) = \left| a \right|\sigma (x)$

$\sigma ({w_i}) = \sigma (l{x_i} + k)$ 

Then,

$\overline x  = 48$

$\sigma (x) = 12$

${w_i} = l{x_i} + k \ldots ..\left( 1 \right)$

${\text{for equation (1)}}$

$\sigma (ax + b) = \left| a \right|\sigma (x)$

$\Rightarrow \sigma ({w_i}) = \sigma (l{x_i} + k)$

$\Rightarrow \sigma (w) = (l)(12) \Rightarrow l = \dfrac{{15}}{{12}}$

$\Rightarrow l = 1.25$

Now, $\overline w  = l\overline x  + k$

$\Rightarrow 55 = (1.25 \times 48) + k$

$\Rightarrow k =  - 5$ 

Correct Answer: A

Ans: The correct option is (D).

Given: First 10 natural numbers, \[n = 10\]

To find: Standard deviations

We know that,

$\sigma  = \sqrt {\dfrac{{{n^2} - 1}}{{12}}}$

$\Rightarrow \sigma  = \sqrt {\dfrac{{{n^2} - 1}}{{12}}} $

$n = 10$

$\Rightarrow \sigma  = \sqrt {\dfrac{{{{10}^2} - 1}}{{12}}} $

$= \sqrt {\dfrac{{99}}{{12}}} $

$= 2.87$ 

Correct Answer: D

Ans: The correct option is (D).

Given: Let us consider the numbers  \[1,2,3,4,5,6,7,8,9,10.\]

\[{\text{1}}\] is added to each number.

We know that,

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

$\sum\limits_{i = 1}^{10} {X_i^2}  = {2^2} + {3^2} + ....... + {11^2}$

$= \dfrac{{11 \times 12 \times 23}}{6} - 1 = 505$

 ${X_1}, \ldots ,{X_n} = \sum\limits_{i = 1}^n {{X_i}}  = 2 + 3 + .... + 11 = 65$ 

Then,

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

$= \dfrac{{505}}{{10}} - {\left( {6.5} \right)^2}$

$= 8.25$ 

Correct Answer: D