Mathematics · Quantitative Aptitude

Statistics and Dispersion

559 Questions

Statistics and dispersion involve the calculation of mean, standard deviation, variance, and coefficient of variation for data sets. These questions also cover probability distributions and cumulative frequency analysis. Such quantitative aptitude topics are heavily featured in banking and SSC examinations.

Standard deviationNormal distributionMean calculationCumulative frequencyCoefficient of variation

Statistics and Dispersion Questions

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

For grouped data, Arithmetic mean by Assumed Mean Method =

  1. A + sd/N

  2. A + sfd/sf

  3. sfX/sf

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

In assumed mean method, any value can be taken as assumed mean whether it is there in the data or not but it should be centrally located in the data so that to simply the big figures in the data in order to ascertain mean of the given data through easy calculations. For grouped data, the formula for assumed mean is A+ sfd/sf where A is the assumed mean, sfd is the summation of frequency multiplied with X-A for all figures and sf is the summation of frequency or the number of element in the given data. 

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

The arithmetic mean of the first 100 natural numbers is _____.

  1. 50

  2. 52

  3. 51

  4. 50.5
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Arithmetic mean refers to the average amount in a given group of data. So arithmetic mean can be calculated by adding the first term and the last term of the series and then dividing the sum by 2. In the given series the first term is 1 and  the last term is 100, so the 

Arithmetic mean = ( 1+100 ) /2 
                             = 101 /2 
                             = 50.5 

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

The mean of a sample of size 10 is 15. If the value of each item is doubled, the mean of the sample will be _______.

  1. 15

  2. 30

  3. 11

  4. 22
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Mean refers to the average amount in a given group of data. So arithmetic mean can be calculated by adding the first term and the last term of the series and then dividing the sum by 2. In the given series the first term 'a' is doubled and  the last term 'b' is also doubled , so the 

Mean = {(a+a)+ (b+b)}  /2 

          = (2a+2b) /2 

          = 2 (a+b) /2 

          = 2 [ (a+b)/2} 

Therefore, the mean is also doubled. So,if the mean was 15 then now it will be 30.  

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

If the frequency of observations $X _{i}$ is $f _{i} (i = 1, 2, ..... n)$.

  1. $\overline {X} = \dfrac {X _{1}\times X _{2} \times X _{3}\times ......X _{n}}{N}$
  2. $\overline {X} = \dfrac {X _{1} + X _{2} + X _{3} + ......X _{n}}{N}$
  3. $\overline {X} = \dfrac {\sum X _{i}}{N}$
  4. $\overline {X} = \dfrac {\sum f _{i}X _{i}}{\sum f}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The arithmetic mean for grouped data is defined as the sum of the products of each value and its corresponding frequency, divided by the total frequency.

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

Consider following frequency distribution.

Class Intervals $0-10$ $10-20$ $20-30$ $30-40$
Frequency $8$ $10$ $12$ $15$

Arithmetic mean $=$?

  1. $39.65$
  2. $22.55$
  3. $32.55$
  4. $23.56$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

0−10 | 10−20 | 20−30 | 30−40 | | --- | --- | --- | --- | --- | | Frequency(f) | 8 | 10 | 12 | 15 | | Class mark ( mid points= x) | 5 | 15 | 25 | 35 | | Fx | 40 | 150 | 300 | 525 |

Mean = summation of fx / summation of f

          = 1015/ 45

          = 22.55

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

The mean of a sample of size $10$ is $15$. If the value of each item is reduced by $2$, the mean of the sample will be_____.

  1. $15$
  2. $13$
  3. $11$
  4. $22$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Mean refers to the average amount in a given group of data. So arithmetic mean can be calculated by adding the first term and the last term of the series and then dividing the sum by 2. In the given series the first term 'a' is decreased by 2 and  the last term 'b' is also decreased by 2 , so the 

Mean = {(a-2)+ (b-2)}  /2 

          = (a+b-4) /2 

          = {(a+b)/2} - 4/2

          = {(a+b)/2} - 2 

Therefore, the mean is also decreased by 2. So,if the mean was 15 then now it will be 13. 

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

Harmonic mean is a part of _______________.

  1. Positional average

  2. Mathematical average

  3. Both a & b

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Mathematical average refers to all such average where a figure is taken out through mathematical methods from the a given series that represents the whole series. Harmonic mean is a mathematical tool which is used to calculate average of a certain series. Therefore, it is a part of mathematical average. 

Multiple choice business economics and quantitative methods linear correlation spearman's coefficient of correlation spearman's rank correlation method correlation coefficients

The value of Spearman's rank coefficient lies between 

  1. $2$ and $3$
  2. $1$ and $2$
  3. $0$ and $1$
  4. $-1$ and $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Spearman's rank cofficient : $R=1-\dfrac { 6\sum { { d } _{ i }^{ 2 } }  }{ n({ n }^{ 2 }-1) } $

Its values lies between $-1$ and $1$
So option $D$ is correct.

Multiple choice business economics and quantitative methods linear correlation spearman's coefficient of correlation spearman's rank correlation method correlation coefficients

If $n=10, \sum x=4,\sum y=3, \sum x^2=8,\sum y^2=9$ and $\sum xy=3,$ then the coefficient of $r _{x,y}$ is

  1. $\frac{3}{4}$
  2. $\frac{1}{5}$
  3. $\frac{1}{6}$
  4. $\frac{1}{4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Correlation coefficient 
${ r } _{ x,y }=\dfrac { n\sum { xy } -\sum { x } \sum { y }  }{ \sqrt { \left[ n\sum { { x }^{ 2 }-{ \left( \sum { x }  \right)  }^{ 2 } }  \right] \left[ n\sum { { y }^{ 2 }-{ \left( \sum { y }  \right)  }^{ 2 } }  \right]  }  } $

$=\displaystyle\frac { 30-12 }{ \sqrt { 64\times 81 }  } $
$\Rightarrow r _{x,y}=\dfrac{1}{4}$

Multiple choice business economics and quantitative methods linear correlation spearman's coefficient of correlation spearman's rank correlation method correlation coefficients

FInd the rank correlation from the following data:

S. No. 1 2 3 4 5 6 7 8 9 10
Rank Differences -2 -4 -1 3 2 0 -2 3 3 -2

  1. 0.64

  2. 0.50

  3. 0.45

  4. 0.34
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Rank Difference $(d)$ | $d^2$ | | --- | --- | --- | | 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. | -2 -4 -1 3 2 0 -2 3 3 -2 | 4 16 1 9 4 0 4 9 9 4 |

 $\sum d^2=60,\quad n=10$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}$

$r=1-\cfrac{6(60)}{10(10^2-1)}$

$r=1-\cfrac{360}{990}$

$r=0.6363....\approx 0.64$

Multiple choice business economics and quantitative methods linear correlation spearman's coefficient of correlation spearman's rank correlation method correlation coefficients

Find the spearman's rank coefficient of correlation from the following data:

X 48 33 40 9 16 16 65 25 16 57
Y 13 13 24 6 15 4 20 9 6 19
  1. $0.76$
  2. $0.52$
  3. $0.61$
  4. $0.85$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Rank | $Y$ | Rank | $|d|$ | $d^2$ | | --- | --- | --- | --- | --- | --- | | 48 33 40 9 16 16 65 25 16 57 | 3 5 4 10 7 7 1 6 7 2 | 13 13 24 6 15 4 20 9 6 19 | 5 5 1 8 4 10 2 7 8 3 | 2 0 3 2 3 3 1 1 1 1 | 4 0 9 4 9 9 1 1 1 1 |

$n=10,\quad \sum d^2=39$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}=1-\cfrac{6\times 39}{10(10^2-1)}=1-\cfrac{234}{990}=0.76$

Multiple choice business economics and quantitative methods linear correlation spearman's coefficient of correlation spearman's rank correlation method correlation coefficients

The final position of twelve clubs in a football league and the average attendance at their home matches were as follows. Calculate a coefficient of correlation by ranks.

Club A B C D E F G H I J K L
Position 1 2 3 4 5 6 7 8 9 10 11 12
Attendance (thousands) 27 30 18 25 32 12 19 11 32 12 12 15

  1. 0.34

  2. 0.56

  3. 0.32

  4. 0.48
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Attendance | Rank | Position | $|d|$ | $d^2$ | | --- | --- | --- | --- | --- | --- | | A B C D E F G H I J K L | 27 30 18 25 32 12 19 11 32 12 12 15 | 4 3 7 5 1 9 6 12 1 9 9 8 | 1 2 3 4 5 6 7 8 9 10 11 12 | 3 1 4 1 4 3 1 4 8 1 2 4 | 9 1 16 1 16 9 1 16 64 1 4 16 |

$n=12,\quad \sum d^2=154$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}=1-\cfrac{6\times 154}{12(12^2-1)}=1-\cfrac{924}{1716}=0.48$

Multiple choice business economics and quantitative methods linear correlation spearman's coefficient of correlation spearman's rank correlation method correlation coefficients

Find the rank correlation coefficient between the heights of fathers and sons from the following data:

Heights of fathers in inches  65 66 67 67 68 69 70 72
Height of sons in inches 67 68 65 68 72 72 69 71
  1. $0.67$
  2. $0.58$
  3. $0.42$
  4. $0.92$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Rank | Height(Son) | Rank | $|d|$ | $d^2$ | | --- | --- | --- | --- | --- | --- | | 65 66 67 67 68 69 70 72 | 8 7 5 5 4 3 2 1 | 67 68 65 68 72 72 69 71 | 7 5 8 5 1 1 4 3 | 1 2 3 0 3 2 2 2   | 1 4 9 0 9 4 4 4 |

$n=08,\quad \sum d^2=35$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}=1-\cfrac{6\times 35}{8(8^2-1)}=1-\cfrac{210}{504}=0.58$

Multiple choice business economics and quantitative methods linear correlation spearman's coefficient of correlation spearman's rank correlation method correlation coefficients

What is correction factor(C.F) in the rank correlation coefficient.

  1. C.F $=\sum (m^{2}-1)$
  2. C.F $=\sum (m^{2}+1)$
  3. C.F $=\sum m^{2}(m^{2}-1)$
  4. C.F $=\sum m(m^{2}-1)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The correction factor in the rank correlation coefficient is given by $\sum m(m^2-1)$


where $m$ is the number of times the data repeats.
Hence, C.F $=\sum m(m^2-1)$.

Multiple choice business economics and quantitative methods linear correlation spearman's coefficient of correlation spearman's rank correlation method correlation coefficients

Based on the following data, find coefficient of rank correlation.

x 43 96 74 38 35 43 22 56 35 80
y 30 94 84 13 30 18 30 41 48 95
  1. $0.3456$
  2. $0.5621$
  3. $0.6303$
  4. $0.7326$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the ranks of students obtained in Physics be $x$ and the ranks of students obtained in Chemistry be $y$.

 $X$  $Y$  Rank $X$       $(x)$  Rank $Y$      $(y)$  $d=x-y$  $d^2$
 $43$ $30$   $5.5$  $7$ $-1.5$   $2.25$
 $96$  $94$  $1$ $2$   $-1$  $1$
 $74$  $84$  $3$  $3$  $0$  $0$
 $38$  $13$  $7$  $10$  $-3$  $9$
 $35$  $30$  $8.5$  $7$  $1.5$  $2.25$
 $43$  $18$  $5.5$  $9$ $-3.5$   $12.25$
 $22$  $30$  $10$ $7$   $3$  $9$
 $56$  $41$  $4$  $5$  $-1$  $1$
 $35$  $48$  $8.5$  $4$ $4.5$   $20.25$
 $80$  $95$  $2$  $1$  $1$  $1$
       $\sum$  $0$  $58$


In the $X$ series $43$ has repeated twice and given ranks $5.5$ instead of $5$ and $6$. 

For this the correction factor is $\dfrac{2(4-1)}{12}=\dfrac{1}{2}$.

Also $35$ has repeated twice and given ranks $8.5$ instead of $8$ and $9$. For this the correction factor is $\dfrac{2(4-1)}{12}=\dfrac{1}{2}$.

In the $Y$ series $30$ has repeated thrice and given ranks $7$ instead of $6,7,8$. 

For this the correction factor is $\dfrac{3(9-1)}{12}=2$.

So, the total correction factors $C.F=\dfrac{1}{2}+\dfrac{1}{2}+2=3$

The rank correlation coefficient is given by,

$r=1-\dfrac{6(\sum d^2-C.F)}{n(n^2-1)}$
$=1-\dfrac{6(58+3)}{10(100-1)}$
$=1-\dfrac{276}{10 \times 99}$
$=1-\dfrac{366}{990}$
$=1-0.3696$
$=0.6303$
Therefore the rank correlation coefficient is $0.6303$.