Mathematics · Quantitative Aptitude

Statistics and Dispersion

515 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 data handling analysis analysis of frequency distributions frequency table evs analysing data collection of data and presentation of data

In a truly normal frequency distribution _________.

  1. the mean always is the same as the standard deviation

  2. the mean is never the same as the mode

  3. the mode is never the same as the median

  4. the mean always is the same as the median
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a perfectly normal distribution, the curve is symmetrical, meaning the mean, median, and mode are all located at the same central point. Options A, B, and C are incorrect because they contradict the fundamental symmetry of the normal distribution.

Multiple choice median percentiles and quartiles range and mean deviation mode

Which of the following is a "Positional Average"?

  1. Arithmetic Mean

  2. Geometric Mean

  3. Harmonic Mean

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

Positional average refers to the average which are taken out through observation from the series where a particular value from the series is picked up which represents the whole series. In median, the middle most value of the series is taken as the representative value. Therefore, median is a positional average. 

Multiple choice median percentiles and quartiles range and mean deviation mode

_______ is that value of the variant which repeats maximum number of times in a distribution and around which other observations are densely distributed.

  1. Mean

  2. Median

  3. Harmonic mean

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

Mode is the highest occurring figure in a series. It is the value in a series of observation that repeats maximum number of times and which represents the whole series as most of the values in the series revolves around this value.

Multiple choice median percentiles and quartiles range and mean deviation mode

Calculation of Decile for individual data:
Find the $D _4$ and from the following data.
$20,22,24,26,28,30,32,34,36$

  1. 28

  2. 26

  3. 30

  4. 24
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For individual data, D_k = value at position k*(n+1)/10. For n=9, D_4 = value at 4*(10)/10 = 4th position. The 4th value in the sorted list 20, 22, 24, 26, 28, 30, 32, 34, 36 is 26.

Multiple choice median percentiles and quartiles range and mean deviation mode

Find $D _2$ for the following data:

Marks 10 20 30 40 50 60
No of Students 5 6 4 5 10 9

  1. 50 marks

  2. 20 marks

  3. 30 marks

  4. 40 marks
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total students = 5+6+4+5+10+9 = 39. D_2 position = 2*(39+1)/10 = 8th position. Cumulative frequencies: 5, 11, 15, 20, 30, 39. The 8th value falls in the 20 marks category.

Multiple choice median percentiles and quartiles range and mean deviation mode

Calculation of Decile for individual data:
Find the $D _8$ and from the following data.
$20,22,24,26,28,30,32,34,36$

  1. $D _8$=34
  2. $D _8$=32
  3. $D _8$=36
  4. $D _8$=28
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For n=9, D_8 = value at 8*(10)/10 = 8th position. The 8th value in the sorted list 20, 22, 24, 26, 28, 30, 32, 34, 36 is 34.

Multiple choice median percentiles and quartiles range and mean deviation mode

Find $D _4$ for the following data.

Marks 0-10 10-20 20-30 30-40 40-50 50-60
No of Students 5 6 5 5 10 9
  1. $D _4=30$
  2. $D _4=20$
  3. $D _4=40$
  4. $D _4=10$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total frequency = 44. D_4 position = 4*(44)/10 = 17.6. Cumulative frequencies: 5, 11, 16, 21, 31, 40. The 17.6th value falls in the 30-40 class interval.

Multiple choice economics frequency polygons collection, organisation and presentation of data frequency polygon and frequency curve diagrammatic presentation of data

Draw frequency polygon and frequency curve for the following data on land holding by the farmers.

Area in hectare 11 - 20 21 - 30 31 - 40 41 - 50 51 - 60 61 - 70 71 - 80
No. of farmers 58 103 208 392 112 34 12

  1. Area in hectaresClass markNo. of farmers

    1 -105.50

    11 -2015.558

    21 -3025.5103

    31 -4035.5208

    41 -5045.5392

    51 -6055.5112

    61 -7065.534

    71 -8075.512

    81 -9085.50

  2. Area in hectaresClass markNo. of farmers

    1 -105.50

    11 -2015.558

    21 -3025.5103

    31 -4035.5208

    41 -5045.5392

    51 -6055.5112

    61 -7065.535

    71 -8075.512

    81 -9085.50

  3. Area in hectaresClass markNo. of farmers

    1 -105.50

    11 -2015.558

    21 -3025.5103

    31 -4035.5208

    41 -5045.5392

    51 -6055.5112

    61 -7065.534

    71 -8075.515

    81 -9085.50

  4. Area in hectaresClass markNo. of farmers

    1 -105.50

    11 -2015.558

    21 -3025.5106

    31 -4035.5208

    41 -5045.5392

    51 -6055.5112

    61 -7065.534

    71 -8075.512

    81 -9085.50
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solution:

Area in hectares Class mark No. of farmers
1  -10 5.5 0
11 -20 15.5 58
21 -30 25.5 103
31 -40 35.5 208
41 -50 45.5 392
51 -60 55.5 112
61 -70 65.5 34
71 -80 75.5 12
81 -90 85.5 0
Multiple choice economics frequency polygons collection, organisation and presentation of data frequency polygon and frequency curve diagrammatic presentation of data
Class Interval Cumulative Frequency
10-19 8
20-29 19
30-39 23
40-49 30

Construct a frequency distribution table from the following cumulative frequency distribution $:$

  1. 8
    19
    23
    30
  2. 30
    23
    19
    8
  3. 8
    27
    50
    80
  4. 8
    11
    4
    7
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\Rightarrow$  The frequency distribution table are as follows :

$Class\,Interval$ $Cumulative\,Frequency$  $Frequency$ 
$10-19$  $8$  $8$ 
$20-29$  $19$  $19-8=11$ 
$30-39$  $23$  $23-19=4$ 
$40-49$  $30$  $30-23=7$ 
Multiple choice maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

The arithmetic mean in a measure of central tedency and is popularlyknown as mean. Arithmetic mean is obtained by dividing the sum of the values of all items of a series by the number of items of that series. Normally, arithmetic mean is denoted by $\bar X$ which is red as  '$X$ bar'. It can be computed for unclassified or ungrouped data or individual series as well as classified or grouped data or discrete or continuous series.

From the following data calculate arithmentic mean.

Marks 0-10 10-20 20-30 30-40 40-50 50-60
No. of students 10 20 30 50 40 30


  1. $25$
  2. $40$
  3. $15$
  4. $35$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

b'Let us take assumed mean $=$ 45
Calculation of deviations from assumed mean 

Marks No. of (X-45)/10
   X m students  f      d   fd
0-10 5 10 -4 -40
10-20 15 20 -3 -60
20-30 25 30 -2 -60
30-40 35 50 -2 -50
40-50 45 40 0 0
50-60 55 30 +1 30

                              N $=$180         +9              $\sum fd = $-180
Mean $A \displaystyle + \frac{\sum fd}{N} \times c = 45+ \frac{-180 \times 10}{180} = 35$'

Multiple choice maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

Following table gives frequency distribution of trees planted by different housing societies in a particular locality.
Find mean number of trees planted by housing society by using 'step deviation method'.

No. of tress 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35 - 40
No. of Societies 2 7 9 8 6 4
  1. $25.42$ trees
  2. $27.42$ trees
  3. $29.42$ trees
  4. $31.42$ trees
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider the following table, to calculate mean by "step deviation method":

$x _i$=mid value of class interval
Assumed mean $a=27.5$
class interval $c=5$

 $ci$  $f _i$  $x _i$  $d _i=\dfrac{x _i-a}{c}$ $f _id _i$
 10-15 2   12.5  $\dfrac{12.5-27.5}{5}=$-3  -6
 15-20 7  17.5  -2  -14
 20-25 9   22.5  -1  -9
 25-30 8  27.5  0  0
 30-35 6   32.5  1  6
 35-40 4   37.5  2  8
 $N=\Sigma f _i=36$          
 $\Sigma f _id _i=-15$

Mean $\overline x=a +\dfrac {\Sigma f _id _i}{N}\times c$

$\therefore \overline x=27.5 + \dfrac{-15}{36} \times 5=25.42$

Hence, option $A$ is corect.

Multiple choice maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

Following table gives age groupwise distribution of people suffering from 'Asthama' due to air pollution in certain city. Find mean age of person suffering from 'Asthama' .

Age (in years) 7 - 11 11 - 15 15 - 19 19 - 23 23 - 27 27 - 31 31 - 35 35 - 39
No. of People  5  9  13  21  16  15  12   9
  1. $23.88$years
  2. $24.88$ years
  3. $25.88 $years
  4. $26.88$ years
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider the following table, to calculate mean:

 $ci$ $ f _i$  $x _i$  $f _ix _i$
 7-11  5  9  45
 11-15  9  13  117
 15-19  13  17  221
 19-23  21  21  441
 23-27  16  25  400
 27-31  15  29  435
 31-35  12 33   396
 35-39  9  37  333
$N=\Sigma f _i=100$          
 $\Sigma f _ix _i=2388$

Mean $\overline x=\dfrac {\Sigma f _ix _i}{N}$
$\therefore \overline x=\dfrac{2388}{100}=23.88$
Mean age of people suffering from Asthama is $23.88$years
Hence, option $A$ is correct.
Multiple choice maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

Find the arithmetic mean for the following grouped frequency distribution:

Class-intervals$ $6-10$ $10-14$ $14-18$ $18-22$  $22-26$  $26-30$
Frequency  $ 4$   $ 6$   $9$   $ 12$    $ 7 $    $2$


  1. $15.8$
  2. $16.8$
  3. $17.8$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Consider the following table, to calculate mean:
|  $ci$ |  $f _i$ | $x _i$  |  $f _ix _i$ | | --- | --- | --- | --- | |  $6-10$ |  $4$ | $ 8$ | $ 32$ | | $ 10-14$ |  $6$ |  $12$ |  $72$ | |  $14-18$ |  $9$ |  $16$ | $ 144$ | |  $18-22$ | $12$ | $ 20$ |  $240$ | | $ 22-26$ |  $7$ | $ 24$ |  $168$ | |  $26-30$ | $ 2$ |  $28$ |  $56$ |
$N=\Sigma f _i=40$          
$\Sigma f _ix _i=712$

Mean $\overline x=\dfrac {\Sigma f _ix _i}{N}$
$\therefore \overline x=\dfrac{712}{40}=17.8$
Hence, option $C$ is correct.