Tag: statistics and probability

Questions Related to statistics and probability

The measure of dispersion is

  1. Mean deviation

  2. S.D.

  3. quartile deviation

  4. all of the above


Correct Option: D
Explanation:

Measures of dispersion include:
1)Sample standard deviation
2)Interquartile range (IQR) or Interdecile range
3)Range
4)Mean difference
5)Median absolute deviation (MAD)
6)Average absolute deviation (or simply called average deviation)
7)Distance standard deviation

The measure of dispersion is

  1. M.D.

  2. S.D.

  3. Q.D.

  4. All of these


Correct Option: D
Explanation:

Mean deviation, standard deviation as well as quartile deviation is the measure of dispersion.
Hence, all of these are measure of dispersion.
(It is well known fact)


Which one is correct?
Statement 1:Positional measure of dispersion describes about the position that a particular data value has within a data set.
Statement 2:Quartiles and percentiles are positional measure of dispersion.

  1. $1$ only

  2. $2$ only

  3. $1$ and $2$ both

  4. Neither $1$ nor $2$


Correct Option: C
Explanation:

Statement 1: is correct because positional measure of dispersion describes about the position that a particular data value has within a data set.
Statement 2:is correct because quartiles and percentiles are positional measure of dispersion.

If $\sum\limits _{i = 1}^9 {\left( {{x _i} - 5} \right) = 9}$ and $\sum\limits _{i = 1}^9 {{{\left( {{x _i} - 5} \right)}^2}}  = 45$, then the standard deviation of the $9$ items ${x _1},{x _2},.....,{x _9}$ is

  1. $2$

  2. $3$

  3. $9$

  4. $4$


Correct Option: A
Explanation:
S.D of $xi-5$ is

$\sigma =\sqrt{\dfrac{\sum _{i=1}^{9}(xi-5)^2}{9}-\left [ \dfrac{\sum _{i=1}^{9}(xi-5)^2}{9} \right ]^2}$

$\sigma =\sqrt{5-1}=2$

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


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?

  1. Standard Deviation, Variance

  2. Percentile, Variance

  3. Quartile, Variance

  4. Percentile,Quartile


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

  1. 40

  2. 30

  3. 20

  4. none of these


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

  1. $10\%$

  2. $40\%$

  3. $50\%$

  4. None


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:

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

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

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

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


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 

  1. $0.9021$

  2. $0.9637$

  3. $0.8864$

  4. $0.9853$


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 $=\cfrac{\sigma}{\bar{x}}=\cfrac{\sqrt{1486}}{40}=.9637$