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

Following table shows frequency distribution of advertisement on T.V. by shift of origin ans scale method.

Duration (in sec.) 25 - 30 30 - 35 35 - 40 40 - 45 45 - 50 50 - 55
No. of advertisements 10 32 15 9 7 2

Obtain mean duration of advertisement on T.V. by shift of origin and scale method.

  1. $31.97$ seconds
  2. $32.97$ seconds
  3. $34.97$ seconds
  4. $35.97$ seconds
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Consider the following table, to calculate mean by "shift of origin method":

$x _i$=mid value of class interval
Assumed mean $a=42.5$

 $ci$  $f _i$  $x _i$ $d _i=x _i-a$   $f _id _i$
 $25-30$  $10$  $27.5$  $-15$  $-150$
 $30-35$  $32$  $32.5$  $-10$  $-320$
 $35-40$  $15$  $37.5$  $-5$  $-75$
 $40-45$  $9$  $42.5$  $0$  $0$
 $45-50$  $7$  $47.5$  $5$  $35$
 $50-55$  $2$  $52.5$  $10$  $20$
 $N=\Sigma f _i=75$          
 $\Sigma f _id _i=-490$
Mean $\overline x=a +\cfrac {\Sigma f _id _i}{N}$
$\therefore \overline x=42.5 + \cfrac{-490}{75}=35.966667 \Rightarrow 35.97$

Mean duration of advertisement on T.V is $35.97$ seconds 
Hence, option $D$ is correct.

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

Consider the table given below

Marks 0-10 10-20 20-30 30-40 40-50 50-60
Number of Students 12 18 27 20 17 6

The arithmetic mean of the marks given above is

  1. $18$
  2. $28$
  3. $27$
  4. $6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$ x _i = \cfrac{\text{upper limit + lower limit}}{2} $

| Marks | No. of students $ \left( f _i \right) $ | $ x _i $  | $ x _i f _i $  | | --- | --- | --- | --- | | 0-10  | 12 | 5  | 60  | | 10-20  | 18  | 15  | 270  | | 20-30  | 27  | 25  | 675  | | 30-40  | 20  | 35  | 700  | | 40-50  | 17  | 45  | 765  | | 50-60  | 6  | 55  | 330  | |   | $ \Sigma{f _i} = 100 $  |   | $ \Sigma{x _i f _i} = 2800 $  | $ \therefore \; Mean=\cfrac { \Sigma { x _{ i }f _{ i } } }{ \Sigma { f _{ i } } } = \cfrac{2800}{100} = 28$
B) 28
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 using step deviation method for the following data shows distance covered by $40$ passengers to perform their work. (Round off your answer to the nearest whole number).

Distance (km) 1-5 5-9 9-13 13-17 17-21 21-25 25-29
Number of passengers 2 4 6 8 10 5 5


  1. $15$
  2. $16$
  3. $17$
  4. $19$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
 Distance (km) Mid point (X)  Number of passengers(F)  i=class interval width  A=11 Assumed mean  d'=$\dfrac{x-A}{i}$  fd' 
 1-5 3  2  4  11  -2  -4 
5-9  7  4  4  11  -1  -4 
9-13  11  6  4  11  0  0 
13-17  15  8  4  11  1  8 
17-21  19  10  4  11  2  20 
21-25  23  5  4  11  3  15 
25-29  27  5  4  11  4  20 
    $\Sigma f=40$        $\Sigma f d'=55$ 

The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \frac{\sum fd'}{\sum f} \times i$ 
A = Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \frac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =11 +\frac{55}{40}\times 4$
$= 11 + 5.5$
$= 16.5$ $\approx$ $17$

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

In a study on a certain population, the following data was given.

Population (X) 2000-2001 2001-2002 2002-2003 2003-2004 2004-2005 2005-2006 2006-2007
Number of people 10 20 30 40 50 60 70


Find the average number of population using step deviation method.

  1. $2002$
  2. $2003$
  3. $2004$
  4. $2005$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 X Midpoint (X)  Frequency (F)  i=class interval width  A=2003.5 Assumed mean  d'=$\dfrac{x-A}{i}$  fd' 
 2000-2001 2000.5  10  1  2003.5  -3  -30 
2001-2002  2001.5  20  1  2003.5  -2  -40 
2002-2003  2002.5  30  1  2003.5  -1  -30 
2003-2004  2003.5=A  40  1  2003.5  0   0
2004-2005  2004.5  50  1  2003.5  1  50 
2005-2006  2005.5  60  1  2003.5  2  120 
2006-2007  2006.5  70  1  2003.5  3  210 
    $\Sigma f=280$        $\Sigma fd'=280$ 


The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \dfrac{\sum fd'}{\sum f} \times i$ 
$A =$ Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \dfrac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =2003.5 +\dfrac{280}{280}\times 1$
$= 2003.5 + 1$
$= 2004.5$ $\approx$ $2005$

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

The frequency distribution of marks in English are given in the table:

Marks 50-60 60-70 70-80 80-90
Number of students 12 24 14 10

Find the mean by step deviation method.

  1. $58$
  2. $48$
  3. $69$
  4. $71$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 X Mid point(x)  Frequency (F)  i=class interval width  A=65 Assumed mean  d'=$\dfrac{x-A}{i} $  fd'
 50-60 55  12  10  65  -1  -12 
 60-70 65=A  24  10  65  0  0 
70 -80  75  14 10  65  1  14 
80-90  85  10  10  65  2  20 
    $\Sigma f=60$        $Sigma fd'=22$ 


The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \dfrac{\sum fd'}{\sum f} \times i$ 
$A =$ Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \dfrac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =65 +\dfrac{22}{60}\times 10$
$= 65 + 3.666$
$= 68.666$ $\approx$ $69$ marks

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

Using step deviation method find the mean.

X 20-40 40-60 60-80 80-100
frequency 4 8 12 16


  1. $40$
  2. $50$
  3. $60$
  4. $70$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 X Mid point (X)  Frequency (F)  i=class interval width  A=50 Assumed Means  d'=$\dfrac{x-A}{i}$  fd' 
 20-40 30  4 20  50  -1  -4 
40-60  50=A  8  20  50  0  0 
60-80  70  12  20  50  1  12 
80-100  90  16  20  50  2  32 
    $\Sigma f=40$        $\Sigma f d'=40$ 


The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \dfrac{\sum fd'}{\sum f} \times i$ 
$A =$ Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \frac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =50 +\dfrac{40}{40}\times 20$
$= 50 + 20$
$= 70$

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

 Marks 0-10  10-20  20-30  30-40  40-50  50-60  60-70  70- 80 80-90  90-100 
 Frequency 3  5  6  7  8   9  10  12  6 4 


Find the mean mark using step deviation method:

  1. $54$
  2. $55$
  3. $56$
  4. $57$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

 X Mid Point (X)  Frequency (F)  i=class interval width  A=45 Assumed mean  d'=$\dfrac{x-A}{i}$  fd' 
 0-10 5  3  10  45  -4  -12 
10-20  15  5  10  45  -3  -15
20-30  25  6  10  45  -2  -12 
30-40  35  7  10  45  -1  -7 
40-50  45=A 8  10  45  0  0 
50-60  55  9  10  45  1  9 
60-70  65  10  10  45  2  20 
70-80  75  12  10  45  3  36 
80-90  85  6  10  45  4  24 
90-100   95 4  10  45  5  20 
    $\Sigma f=70$        $\Sigma fd'=63$ 


The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \dfrac{\sum fd'}{\sum f} \times i$ 
$A =$ Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \dfrac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =45 +\dfrac{63}{70}\times 10$
$= 45 + 9$
$= 54$

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

 Marks 0-5  5-10  10-15  15-20  20-25  25-30  30-35  35-40  40-45  45-50 
Frequency  3  5  7  8  10  11  14  19  15  13 

For the following distribution, find the mean using step deviation method. (Round off your answer to the nearest whole number)

  1. $29$
  2. $31$
  3. $35$
  4. $37$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 X Mid point (x)  Frequency (F)  i=class  interval width  A=22.5 Assumed mean  d'=$\dfrac{x-A}{i}$  fd' 
0-5  2.5  3  5  22.5  -4  -12 
5-10  7.5  5  5  22.5  -3  -15 
10 -15 12.5  7  5  22.5  -2  -14 
15-20  17.5  8  5  22.5  -1  -8 
20-25  22.5=A  10  5  22.5  0  0 
25-30  27.5  11  5  22.5  1  11 
30-35  32.5  14  5  22.5  2  28 
35-40  37.5  19  5  22.5  3  57 
40-45  42.5  15  5  22.5  4  60 
45-50  47.5  13  5  22.5  5  65 
    $\Sigma f=105$        $\Sigma fd'=172$ 


The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \dfrac{\sum fd'}{\sum f} \times i$ 
$A =$ Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \dfrac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =22.5 +\dfrac{172}{105}\times 5$
$= 22.5 + 8.19$
$= 30.69$ $\approx$ $31$

Multiple choice statistics skewness of frequency distribution karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

Karl Pearson's coefficient of skewness of a distribution is 0.32.Its s.d.is 6.5 and mean is 29.6.The mode and median of the distribution are

  1. 27.52,28.91

  2. 26.92,27.23

  3. 25.67,26.34

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Karl Pearson's coefficient of skewness $\displaystyle =\frac{Mean-Mode}{S.D.}$ $\displaystyle \therefore 0.32=\frac{29.6-Mode}{6.5}\Rightarrow Mode=27.52$ Also Karl Pearson's coeff.of skewness $\displaystyle =\frac{3\left ( Mean-Median \right )}{S.D}$ $\displaystyle \because 0.32=\frac{3\left ( 29.6-Median \right )}{6.5}$ $\displaystyle \Rightarrow Median=28.91$

Multiple choice statistics skewness of frequency distribution karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

The sum of the deviations of the variates 6,8,10,16,20,24 

  1. -1

  2. 1

  3. 0

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

$\Rightarrow$   $Mean = \dfrac{6+8+10+16+20+24}{6}=14$

$\Rightarrow$   Sum of the deviations = $(6-14)+(8-14)+(10-14)+(16-14)+(20-14)+(24-14)$
$\Rightarrow$   Sum of the deviation = $-8-6-4+2+6+10$
$\therefore$    Sum of the deviation = $-18+18$
$\therefore$    Sum of the deviation = $0$

Multiple choice economics graphical representation revisiting histograms histogram histograms histograms for non-uniform class widths histograms with unequal class intervals

Histogram also gives value of _______ of the frequency distribution graphically

  1. mode

  2. mean

  3. median

  4. all of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The mode of a continuous frequency distribution can be determined graphically using a histogram by identifying the tallest bar and using specific geometric constructions within that interval.

Multiple choice economics graphical representation revisiting histograms histogram histograms histograms for non-uniform class widths histograms with unequal class intervals

The following frequency distribution is classified as 

Classes Frequency
0-50 25
0-30 18
0-10 5

  1. cumulative distribution in less than type

  2. Cumulative distribution in more than type

  3. discrete frequency distribution

  4. cumulative frequency distribution
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The data shows cumulative frequencies (5, 18, 25) for increasing upper boundaries (10, 30, 50), which is the definition of a 'less than' cumulative frequency distribution.

Multiple choice chemistry atomic theory, periodic classification and properties of elements introduction to periodic table necessity of classification need for classification

Predict block, group and period of $Sb$.

  1. Block-$p$, Period-$5$, Group-$15$
  2. Block-$p$, Period-$5$, Group-$14$
  3. Block-$s$, Period-$5$, Group-$13$
  4. Block-$s$, Period-$5$, Group-$11$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Antimony is a chemical element with symbol $Sb$ (from Latin: stibium) and atomic number $51$,group $15$ (pnictogens), $p$-block,period $5$.

Multiple choice statistics linear correlation coefficient of correlation correlation coefficient correlation coefficients

The regression coefficients of a bivariate distribution are -0.64 and -0.36. Then the correlation coefficient of the distribution is

  1. 0.48

  2. -0.48

  3. 0.50

  4. -0.50
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solution:

We have,
$b _{xy}=-0.64$ and $b _{yx}=-0.36$
$\therefore$ Correlation coefficient $=\sqrt{b _{xy}\times b _{yx}}$
$=\pm\sqrt{(-0.64)(-0.36)}=\pm0.48$
$\Longrightarrow \sigma=-0.48$
[$\because b _{xy}$ and $b _{yx}$ both are negative.]
Hence, B is the correct option.