Tag: measures of dispersion

Questions Related to measures of dispersion

What are the advantages of squaring a difference for calculating variance and standard deviation?

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

  1. Squaring makes each term positive so that values above the mean do not cancel below the mean.

  2. Squaring adds more weight to the larger differences, and in many cases this extra weight is appropriate since points further from the mean may be more significant.

  3. It complicates the calculations

  4. All are incorrect

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Correct Option: A,B
Explanation:

Since,

$\sigma _x=\sqrt{\cfrac{\sum (x _i-\bar x)^2}{N}}$
So, we can say that opion $A$ and $B$ are correct.

Which of the following are positional measure of dispersion?

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

  1. Standard Deviation, Variance

  2. Percentile, Variance

  3. Quartile, Variance

  4. Percentile,Quartile

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Correct Option: D
Explanation:

Percentile,Quartile are positional measure of dispersion because it tells about the position of a particular data value has within a data set.
Standard deviation, Variance are computational measure of dispersion.

If the coefficient of variation and standard deviation of a distribution are 50% and 20 respectively, then its mean is

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

  1. 40

  2. 30

  3. 20

  4. none of these

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Correct Option: A
Explanation:

Given $\sigma = 20$, coefficient of variation $=50$ %
We know coefficient of variation $=\cfrac{\sigma }{\bar{x}}\times 100=50$
$\Rightarrow \bar{x} = 2\times \sigma = 40$

The sum of squares of deviations for $10$ observations taken from mean $50$ is $250 $. Then Co-efficient of variation is

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

  1. $10\%$

  2. $40\%$

  3. $50\%$

  4. None

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Correct Option: A
Explanation:
$\sum(x-\overline{x})^2=250$, $\overline{x}=50$
$\Rightarrow$  Standard deviation $(\sigma)=\sqrt{\dfrac{250}{10}}=\sqrt{25}=5$
$\Rightarrow$  Coefficient of variation $=\sqrt{\dfrac{\sum(x-\overline{x})^2}{n}}$
                                             $=\dfrac{\sigma}{Mean}\times 100$

                                             $=\dfrac{5}{50}\times 100$

                                             $=10\%$

The Coefficient of Variation is given by:

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

  1. $\dfrac{Mean}{\ Standard \ \ deviation } \times 100$

  2. $\dfrac{\ Standard \ \ deviation }{Mean}$

  3. $\dfrac{Standard \ \ deviation }{Mean }\times 100$

  4. $\dfrac{Mean}{Standard \ Deviation}$

Show answer
Correct Option: C
Explanation:

The coefficient of variation (CV) is a standardized measure of dispersion 

. It is defined as the ratio of the standard deviation to the mean.
$CV\quad =\quad \cfrac { \sigma  }{ Mean }\times100 $

If mean of a series is 40 and variance 1486, then coefficient of variation is 

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

  1. $0.9021$

  2. $0.9637$

  3. $0.8864$

  4. $0.9853$

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Correct Option: B
Explanation:

If mean of the given dist. be $\bar{x}$ and S.D be $\sigma $
then given $\bar{x} = 40, \sigma^2 = 1486$
$\therefore$ Coefficient of variation $=\cfrac{\sigma}{\bar{x}}=\cfrac{\sqrt{1486}}{40}=.9637$

If the coefficient of variation and standard deviation of a distribution are 50% and 20 respectively, the its mean is

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

  1. 40

  2. 30

  3. 20

  4. None of these

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Correct Option: A
Explanation:

We know if a distribution having mean $\bar{x}$ and standard deviation $\sigma$
then coefficient of variation $=\cfrac{\sigma}{\bar{x}}\times 100$
$\therefore \cfrac{20}{\bar{x}}\times 100=50\Rightarrow \bar{x} = 40$
Hence required mean is $=40$

The sum of the squares of deviation of 10 observations from their mean 50 is 250, then coefficient of varition is

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

  1. 10%

  2. 40%

  3. 50%

  4. None of these

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Correct Option: A
Explanation:

Given $\displaystyle \Sigma \left ( x _{i}-\overline{x} \right )^{2}=250$,$n=10,\overline{x}=50$

Now, $\sigma=\sqrt{\dfrac{1}{n}\Sigma \left ( x _{i}-\overline{x} \right )^{2}}$

$= \sqrt{\dfrac{1}{10}\times 250}=5$ 
Hence coefficient of variation $\displaystyle =\dfrac{\sigma }{\overline{x}}\times 100=\dfrac{5}{50}\times 100=10$%

The sum of the squares of deviation of 10 observations from their mean 50 is 250, then coefficient of variation is

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

  1. 10%

  2. 40%

  3. 50%

  4. none of these

Show answer
Correct Option: A
Explanation:

Given,   $\sum (x-\bar{x})^2 = 250, n = 10, \bar{x} =50$
Thus standard deviation $ = \sqrt{\cfrac{\sum (x-\bar{x})^2}{n}}=\sqrt{25}=5$
$\therefore$ Coefficient of variation $=\cfrac{\sigma}{\bar{x}}\times 100 =\cfrac{5}{50}\times 100$ % $= 10$%

The mean of a distribution is 4. If its coefficient of variation is 58%. Then the S.D. of the distribution is

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

  1. 2.23

  2. 3.23

  3. 2.32

  4. none of these

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Correct Option: C
Explanation:

Given,  mean $\bar{x} = 4,$ and coefficient of variation $=58$ %
If S.D of the given distribution is $\sigma$ then we know that,
Coefficient of variation $=\cfrac{\sigma}{\bar{x}}\times 100$ %
$\Rightarrow 58 = \cfrac{\sigma}{4}\times 100\Rightarrow \sigma = \cfrac{58\times 4}{100}=2.32$