Tag: measures of dispersion

Questions Related to measures of dispersion

For the given data, SD = 10, AM = 20, the coefficient
of variation is____

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

  1. 47

  2. 24

  3. 44

  4. 50

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Correct Option: D
Explanation:

Coeffecient of variation $ = \frac {SD}{AM} \times 100 = \frac {10}{20} \times 100 = 50 $

For the given data, SD $= 10$, AM $= 20$ the coefficient of variation is ...........

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

  1. $47$

  2. $24$

  3. $44$

  4. $50$

Show answer
Correct Option: D
Explanation:

Coefficient of variation is the ratio of standard deviation to the mean.


Given that $SD=10$ and $AM=20$

Therefore of coefficient of variation is $\dfrac{SD}{AM}\times100=\dfrac{10}{20}\times100=50\%$

The mean of a distribution is $14$ and standard deviation is $5$. What is the value of the coefficient of variation?

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

  1. $57.7\%$

  2. $45.7\%$

  3. $35.7\%$

  4. None of these

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Correct Option: C
Explanation:

Coefficient of variation is given by $CV = \dfrac{SD}{Mean}\times 100 $
$\Rightarrow \dfrac{5}{14}\times 100 = 35.7\%$

If the standard deviation of a set of scores is $1.2$ and their mean is $10$, then the coefficient of variation of the scores is

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

  1. $12$

  2. $0.12$

  3. $20$

  4. $120$

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Correct Option: A
Explanation:

Given : standard deviation$(\sigma)=1.2,$ mean$(\overline {X})=10$.

Coefficient of variation(C.V.) $=\dfrac{\sigma}{\overline {X}}\times 100=\dfrac{1.2}{10}\times 100=12$
$\therefore$ C.V. $=12$
Hence, option $A$ is correct.

If $n=10, \bar{x}=12$ and $\sum x^2=1530$, then calculate the coefficient of variation.

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

  1. $20$

  2. $25$

  3. $30$

  4. $35$

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Correct Option: B
Explanation:

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

   
   $=\sqrt{\dfrac{1530}{10}-(12)^2}$

   $=\sqrt{153-144}$
   $=\sqrt{9}$
   $=3$

Coefficient of variation $=\dfrac{\sigma}{\overline{x}}\times 100$

                                       $=\dfrac{3}{12}\times 100$

                                       $=\dfrac{1}{4}\times 100$
                                       $=25$

If the standard deviation of $x _{1},x _{2},.....x _{n}$ is 3.5, then the standard deviatiuon of $-2x _{1}-3,-2x _{2}-3....,-2x _{n}-3$ is

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

  1. -7

  2. -4

  3. 7

  4. 1.75

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Correct Option: C

If $\sigma$ $f _i$ $x _i$  = 20 and $\sigma$ $f _i$ = 4, what is the mean of the data.

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

  1. $\dfrac{1}{5}$

  2. $80$

  3. $16$

  4. $5$

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Correct Option: A
Explanation:
$\sigma fixi=20$
and $\sigma fi=4$

Hence, Mean $=\dfrac{\sigma fi}{\sigma fixi}=\dfrac{4}{20}=\dfrac{1}{5}$

Hence Option $A$ is correct

The variance of the data $6,\ 8,\ 10,\ 12\,,14\,,\ 16,\ 18,\ 20,\ 22,\ 24$ is

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

  1. $15$

  2. $20$

  3. $30$

  4. $33$

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Correct Option: D
Explanation:
Mistake :$14$ is not given
Mean $\bar x=\dfrac{6+8+10+12+14+16+18+20+22+24}{10}=\dfrac{150}{10}=15$
Variance$=\dfrac{1}{n} \sum\limits _{i=1}^n(x _{i}-\bar x)^2$
$\implies \dfrac{1}{10}((6-15)^2+(8-15)^2+(10-15)^2+(12-15)^2+(14-15)^{2}+(16-15)^2+(18-15)^2+(20-15)^2$
$+(22-15)^2+(24-15)^2$

$\implies \dfrac{81+49+25+9+1+1+9+25+49+81}{10}$

$\implies \dfrac{330}{10}=33$

The variate x and u are related by $\displaystyle u= \frac{x-a}{h}$ then correct relation between $\displaystyle \sigma _{x}:and:\sigma _{u}$

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

  1. $\displaystyle \sigma _{x}= h\sigma _{u}$

  2. $\displaystyle \sigma _{x}= h+\sigma _{u}$

  3. $\displaystyle \sigma _{u}= h\sigma _{x}$

  4. $\displaystyle \sigma _{u}= h+\sigma _{x}$

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Correct Option: A
Explanation:

Given $\displaystyle u =\frac{x}{h}-\frac{a}{h}$
Since,S.D. is not depend on change of origin but it is depend on change of scale.
$\displaystyle \therefore \sigma _{u}=\frac{\sigma _{x}}{h}$
$\Rightarrow h\sigma _{u}=\sigma _{x}$

Standard deviation of a collection of data is $2\sqrt{2}$. If each value in a data set  is multipled by $3$, then the standard deviation of the new data is.

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

  1. $\sqrt{12}$

  2. $4\sqrt{2}$

  3. $6\sqrt{2}$

  4. $9\sqrt{2}$

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Correct Option: C
Explanation:
The standard deviation would also be multiplied by $3$.
Because the mean would also be $3x$ larger, the differences from the mean would be $3x$ larger too.
It is the same idea as if you were looking at your data set through an enlarging lens- everything would be $3x$ bigger, not only the data values, but also the mean, the differences from the mean, but just everything!
$\therefore$ the standard deviation becomes $2\sqrt{2}\times 3=6\sqrt{2}$