Tag: mean of grouped data : step deviation method

Questions Related to mean of grouped data : 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


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
$\overline {X} =22.5 +\dfrac{172}{105}\times 5$
$= 22.5 + 8.19$
$= 30.69$ $\approx$ $31$