Tag: range and mean deviation

Questions Related to range and mean deviation

The ________ is the difference between the greatest and the least value of the variate.

range and mean deviation statistics and probability maths coefficient of variance variance and standard deviation

  1. Range

  2. Data

  3. Average

  4. Variance

Show answer
Correct Option: A
Explanation:

$Range$ $as$ $the$ $name$ $indicates$ $gives$ $us$ $all$ $the$ $area$ $available$ $under$ $light$ $and$ $hence$ $statement$ $is$ $true.$

The mean deviation from the median is _________ that measured from any other value.

range and mean deviation statistics and probability maths coefficient of variance variance and standard deviation

  1. equal to 

  2. less than

  3. greater than

  4. None of these

Show answer
Correct Option: B
Explanation:

The value of the mean deviation is minimum if the deviations are taken from the median. So, it is less than that measured from any other value.

A series drawback of the mean deviation is that it cannot be used in statistical inference.

The difference between the maximum and the minimum obervations in data is called the ____________.

range and mean deviation statistics and probability maths coefficient of variance variance and standard deviation

  1. mean of the data

  2. range of the data

  3. mode of the data

  4. median of the data

Show answer
Correct Option: B
Explanation:

In arithmetic, the range of a set of data is the difference between the largest and smallest values.

So, difference between minimum and maximum values is called range.

For the measure of centre tendency, which the following is not true.

range and mean deviation statistics and probability maths coefficient of variance variance and standard deviation

  1. $Z=3M-2\bar{x}$

  2. $2\bar{x}+Z=3M$

  3. $2\bar{x}-3M=-Z$

  4. $2\bar{x}=Z-3M$

Show answer
Correct Option: D
Explanation:

We have, $Z=3M-2\bar{x}$
$\therefore 2\bar{x}+Z=3M$
or $2\bar{x}-3M=-Z$

$2\bar{x}=-Z+3M$
$\therefore 2\bar{x}=Z-3M $ is not true

Range of data $7, 8, 2, 1, 3, 13, 18$ is?

range and mean deviation statistics and probability maths coefficient of variance variance and standard deviation

  1. $10$

  2. $15$

  3. $17$

  4. None of the above

Show answer
Correct Option: C
Explanation:

$\begin{matrix} 7,8,2,1,3,13,18, \ range\, \, of\, \, data=\left( { \max  imum-\min  imum } \right)  \ =\left( { 18-1 } \right) =17\, \, \, \, \, Ans. \  \end{matrix}$

Coefficient of range $5, 2, 3, 4, 6, 8, 10$ is?

range and mean deviation statistics and probability maths coefficient of variance variance and standard deviation

  1. $\dfrac{2}{3}$

  2. $\dfrac{1}{3}$

  3. $\dfrac{3}{5}$

  4. $\dfrac{1}{2}$

Show answer
Correct Option: A
Explanation:
${ x _{ m } }=10{ x _{ 0 } }=2$
coefficient of range 
$\begin{array}{l} =\frac { { { x _{ m } }-{ x _{ 0 } } } }{ { { x _{ m } }t{ x _{ 0 } } } }  \\ =\frac { { 10-2 } }{ { 10+2 } } =\frac { 8 }{ { 12 } } =\frac { 2 }{ 3 }  \end{array}$

The highest score of a certain data exceeds in lowest score by $16$ and coefficient of range is $\cfrac{1}{3}$. The sum of the highest score and the lowest score is

range and mean deviation statistics and probability maths coefficient of variance variance and standard deviation

  1. $36$

  2. $48$

  3. $24$

  4. $18$

Show answer
Correct Option: B
Explanation:

Let the highest score be $x _{m}$ and 

the lowest score be $x _{0}$
Given that highest score exceeds lowest score by $16$
$\implies x _m=x _0+16\implies x _m-x _0=16$ ————(1)

Coefficient of range is given by $\dfrac{x _m-x _0}{x _m+x _0}$

Given that coefficient of range is $\dfrac 13$

$\implies \dfrac{x _m-x _0}{x _m+x _0}=\dfrac 13$ ———(2)

Substitute (1) in (2) we get

$\dfrac{16}{x _m+x _0}=\dfrac 13$

$\implies x _m+x _0=16\times3=48$

Therefore sum of the highest score and lowest score is $48$

For a frequency distribution $8^{th}$ decile is computed by the formula

range and mean deviation statistics and probability maths coefficient of variance variance and standard deviation

  1. $ \displaystyle D _{8}= l _{i}+\frac{\frac{N}{8}-C}{f}\times \left ( l _{2}-l _{1} \right )$

  2. $ \displaystyle l _{1}+\frac{\frac{8N}{10}-C}{f}\times \left ( l _{2}-l _{1} \right )$

  3. $ \displaystyle D _{8}= l _{1}+\frac{\frac{N}{10}-C}{f }\times \left ( l _{2}-l _{1} \right )$

  4. $ \displaystyle l _{1}+\frac{\frac{10N}{8}-C}{f }\left ( l _{2}-l _{1} \right )$

Show answer
Correct Option: B
Explanation:

A decile is any of the nine values that divide the sorted data into ten equal parts, so that each part represents 1/10 of the sample or population.
For a continuous distribution, the formula for $r^{th}$ decile is given by $D _r = l _1 + \frac{\frac{rN}{10} - C}{f} \times (l _2 - l _1)$
Substituting r = 8, we have $D _8 = l _1 + \frac{\frac{8N}{10} - C}{f} \times (l _2 - l _1)$ 

If $n> 1, x> -1, x\neq 0$, then the statement $\left ( 1+x \right )^{n}> 1+nx$ is true for

range and mean deviation statistics and probability maths coefficient of variance variance and standard deviation

  1. $ \;n\;\epsilon \;N$

  2. $\forall \;n\;> 1$

  3. $x> -1 \;and\; x\neq 0$

  4. None of these

Show answer
Correct Option: A
Explanation:

$P(1)$ is not true 


For $n=2,P\left( 2 \right) :{ \left( 1+x \right)  }^{ 2 }>1+2x$ is true if $x\neq 0$

Let $P(k):{ \left( 1+x \right)  }^{ k }>1+kx$ be two 

$\therefore{ \left( 1+x \right)  }^{ k+1 }=\left( 1+x \right) { \left( 1+x \right)  }^{ k}>\left( 1+x \right) \left( 1+kx \right)> 1+\left( k+1 \right) x+k{ x }^{ 2}>1+\left(k+1\right) x$

$\left( \because k{ x }^{ 2 }>0 \right) $
$\therefore$ By PMI
Given statement is true for every $n\in N$.

The coefficient of mean deviation from median of observations  $40, 62, 54, 90, 68, 76$  is

range and mean deviation statistics and probability maths coefficient of variance variance and standard deviation

  1. $2.16$

  2. $0.2$

  3. $5$

  4. None of these

Show answer
Correct Option: B
Explanation:

Arrange the given observations in ascending order
$40,54,62,68,76,90$
Here, number of terms $n=6 (even) $
$\displaystyle \therefore $ Median (M) $\displaystyle =\frac{\left ( \frac{n}{2} \right )th:term+\left ( \frac{n}{2}+1 \right )th:term}{2}=\frac{62+68}{2}=65$

$\Sigma \left | x _{i}-M \right |=25+11+3+3+11+25=78$
Mean deviation from median $\displaystyle =\frac{\Sigma \left | x _{i}-M \right |}{n}=\frac{78}{6}=13 $
$\therefore $ Coefficient of M.D.=$\displaystyle =\frac{M.D.}{median}=\frac{13}{65}=0.2$