Tag: measures of dispersion

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Questions Related to measures of dispersion

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

The measure of dispersion is

  1. Mean deviation

  2. S.D.

  3. quartile deviation

  4. all of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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

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 $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
S.D of $xi-5$ is

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

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