Tag: measures of dispersion

Questions Related to measures of dispersion

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


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

  1. $15$

  2. $16$

  3. $17$

  4. $19$

Show answer
Correct Option: C
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
$\overline {X} =11 +\frac{55}{40}\times 4$
$= 11 + 5.5$
$= 16.5$ $\approx$ $17$

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.

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

  1. $2002$

  2. $2003$

  3. $2004$

  4. $2005$

Show answer
Correct Option: D
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
$\overline {X} =2003.5 +\dfrac{280}{280}\times 1$
$= 2003.5 + 1$
$= 2004.5$ $\approx$ $2005$

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.

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

  1. $58$

  2. $48$

  3. $69$

  4. $71$

Show answer
Correct Option: C
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

Using step deviation method find the mean.

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


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

  1. $40$

  2. $50$

  3. $60$

  4. $70$

Show answer
Correct Option: D
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
$\overline {X} =50 +\dfrac{40}{40}\times 20$
$= 50 + 20$
$= 70$

 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:

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

  1. $54$

  2. $55$

  3. $56$

  4. $57$

Show answer
Correct Option: A
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
$\overline {X} =45 +\dfrac{63}{70}\times 10$
$= 45 + 9$
$= 54$

 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)

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

  1. $29$

  2. $31$

  3. $35$

  4. $37$

Show answer
Correct Option: B
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
$\overline {X} =22.5 +\dfrac{172}{105}\times 5$
$= 22.5 + 8.19$
$= 30.69$ $\approx$ $31$

The measure of dispersion is

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

  1. Mean deviation

  2. S.D.

  3. quartile deviation

  4. all of the above

Show answer
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

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

  1. M.D.

  2. S.D.

  3. Q.D.

  4. All of these

Show answer
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.

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

  1. $1$ only

  2. $2$ only

  3. $1$ and $2$ both

  4. Neither $1$ nor $2$

Show answer
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

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

  1. $2$

  2. $3$

  3. $9$

  4. $4$

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

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

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