Tag: variance and standard deviation

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Questions Related to variance and standard deviation

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

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

  1. $\dfrac{2}{3}$
  2. $\dfrac{1}{3}$
  3. $\dfrac{3}{5}$
  4. $\dfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
${ x _{ m } }=10{ x _{ 0 } }=2$
coefficient of range 
$\begin{array}{l} =\frac { { { x _{ m } }-{ x _{ 0 } } } }{ { { x _{ m } }t{ x _{ 0 } } } }  \\ =\frac { { 10-2 } }{ { 10+2 } } =\frac { 8 }{ { 12 } } =\frac { 2 }{ 3 }  \end{array}$
Multiple choice range and mean deviation statistics and probability maths coefficient of variance variance and standard deviation

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

  1. $36$
  2. $48$
  3. $24$
  4. $18$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $48$

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

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

  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 )$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $D _8 = l _1 + \frac{\frac{8N}{10} - C}{f} \times (l _2 - l _1)$ 

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

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

  1. $ \;n\;\epsilon \;N$
  2. $\forall \;n\;> 1$
  3. $x> -1 \;and\; x\neq 0$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $n\in N$.

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

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

  1. $2.16$
  2. $0.2$
  3. $5$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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.=$\displaystyle =\frac{M.D.}{median}=\frac{13}{65}=0.2$

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

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

  1. 2.16

  2. 1.2

  3. 5

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Arranging the given data in ascending order
40,54,62,68,76,90
Here, $n=6 (even)$
$M= \dfrac{\text{value of }3^{rd}\text{observation}+\text{value of }4^{th}\text{observation}}{2}$
Median $M=\dfrac{62+68}{2}=65$

Mean deviation about median $M.D=\dfrac{|40-65|+|54-65|+|62-65|+|68-65|+|76-65|+|90-65|}{65}$

$=\dfrac{25+11+3+3+11+25}{65}=1.2$

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

The difference between the maximum and the minimum observations in the data is

  1. class interval

  2. frequency

  3. cumulative frequency

  4. range
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The difference between maximum and the minimum observation in the data is range.

For example, suppose an experiment involves finding out the weight of lab rats and the values in grams are 320, 367, 423, 471 and 480. In this case, the range is simply computed as 480-320 = 160 grams.

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

The coefficient of range of a set of data is given to be $\dfrac18$. Then the ratio of the maximum value in the data to the minimum value is:

  1. $\dfrac81$
  2. $\dfrac98$
  3. $\dfrac97$
  4. $\dfrac87$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Coefficient of range of a set of data is given by $\dfrac{max-min}{max+min}$
$\dfrac{max-min}{max+min}=\dfrac{1}{8}$
$8max-8min=max+min$
$7max=9min$
$\dfrac{max}{min}=\dfrac{9}{7}$

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

The following are the wages of 8 workers in a factory. Find the range and coefficient of range. Wages are in dollars: 1400, 1450, 1520, 1380, 1485, 1495, 1575, 1440.

  1. $0.0231$
  2. $0.03112$
  3. $0.66$
  4. $0.02314$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The largest value of data is $x _m=1575$

The smallest value of data is $x _0=1380$
Range$=x _m-x _0=1575-1380=195$

Coefficient of data$=\dfrac{1575-1380}{1575+1380}=\dfrac{195}{2955}=0.0659\approx 0.66$

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

If the coefficient of range is $0.18$ and the largest value is $7.44$,then the smallest value is?

  1. $3.23$
  2. $4.15$
  3. $5.17$
  4. $5.14$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Coefficient of range$=\dfrac{x _m-x _0}{x _m+x _0}=\dfrac{7.44-x _0}{7.44+x _0}$

$0.18(7.44+x _0)=7.44-x _0$
$1.18x _0=7.44-7.44\times 0.18$
$1.18x _0=6.1008$
$x _0=5.17016\approx 5.17$