Direct Method

What is a Grouped Frequency Distribution Table

There are 3 methods for calculation of mean :

1. Direct Method
2. Assumed mean deviation method
3. Step deviation method.

1. Direct Method for Calculation of Mean

According to direct method

2. Assumed Mean Method
Arithmetic mean = \(a + \frac{{\sum\limits_{i = 1}^n {{f_i}{d_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}\)
Note : The assumed mean is chosen, in such a manner, that

1. It should be one of the central values.
2. The deviation are small.
3. One deviation is zero.

Working Rule :
Step 1 :       Choose a number ‘a’ from the central values of x of the first column, that will be our assumed mean.
Step 2 :      Obtain deviations d_i by subtracting ‘a’ from x_i. Write down hese deviations against the corresponding frequencies in the third column.
Step 3 :      Multiply the frequencies of second column with corresponding deviations d_i in the third column to prepare a fourth column of f_id_i.
Step 4 :      Find the sum of all the entries of fourth column to obtain ∑f_id_i and also, find the sum of all the frequencies in the second column to obtain ∑f_i.

Read More:

• Mean and its Advantages and Disadvantages
• Median of Grouped Frequency Distribution
• Mode in Statistics
• Pie Charts
• Frequency Polygon

3. Step Deviation Method
Deviation method can be further simplified on dividing the deviation by width of the class interval h. In such a case the arithmetic mean is reduced to a great extent.
Mean (\(\bar x\)) = a + \(\frac{{\Sigma {f_i}{u_i}}}{{\Sigma {f_i}}} \times h\)
Working Rule :
Step-1 :     Choose a number ‘a’ from the central values of x(mid-values)
Step-2 :    Obtain u_i = \(\frac{{{x_i} – a}}{h}\)
Step-3 :    Multiply the frequency f_i with the corresponding u_i to get f_iu_i.
Step-4 :    Find the sum of all f_iu_i i.e., ∑f_iu_i
Step-5 :     Use the formula  = a + \(\frac{{\Sigma {f_i}{u_i}}}{{\Sigma {f_i}}} \times h\) to get the required mean.

Grouped Frequency Distribution Table Example Problems with Solutions

Example 1:    

Find the mean by direct method.

Solution:

Mean = \(\frac{{\sum {f_i}{x_i}}}{{\sum {f_i}}}\) = \(\frac{{723}}{{200}}\) = 3.615

Example 2:    Find the mean of the following frequency distribution :

Solution:

Mean = \(\frac{{\sum {f}{x}}}{{\sum {f}}}\) = \(\frac{{10000}}{{200}}\) = 50

Example 3:    A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find  the mean number of plants per house.

Which method did you use for finding the mean and why ?
Solution:

Mean = \(\frac{{\sum {f}{x}}}{{\sum {f}}}\) = \(\frac{{162}}{{20}}\) = 8.1

Example 4:    Calculate the mean for the following distribution:

Solution:

∴ Mean = \(\frac{{\sum f\,x}}{{\sum f}} = \frac{{281}}{{40}}\)  = 7.025

Example 5:    Find the mean of the following frequency distribution :

Solution:

Mean = \(\frac{{\sum f\,x}}{{\sum f}} = \frac{{4930}}{{150}} = 32.8\overline 6\) or 32.87 (approx.)

Example 6:    Find the mean of the following distribution by direct method.

Solution:

Mean = \(\frac{{\sum f\,x}}{{\sum f}} = \frac{{578.5}}{{20}}\) = 28.9

Example 7:    For the following distribution, calculate mean using all the suitable methods.

Solution:

Mean = \(\frac{{\sum f\,x}}{{\sum f}} = \frac{{848}}{{64}}\)  = 13.25

Example 8:    The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city.

Solution:    Let assumed mean = 275

\(\bar x = a + \frac{{\Sigma {f_i}{d_i}}}{{\Sigma {f_i}}}\) = 275 + \(\frac{{ – 1750}}{{200}}\) = Rs 266.25

Example 9:    Calculate the arithmetic mean of the following distribution :

Solution:    Let assumed mean = 175 i.e. a = 175

Now , a = 175
\(\bar x = a + \frac{{\Sigma {f_i}{d_i}}}{{\Sigma {f_i}}}\) = 175 + \(\frac{{ – 4750}}{{180}}\)
= 175 – 26.39 = 148.61 approx.

Example 10:    Calculate the arithmetic mean of the following frequency distribution :

Solution:    Let assumed mean = 75 i.e., a = 75

a = 75, Σf_id_i= 150, Σf_i = 50
Mean \(\bar x = a + \frac{{\Sigma {f_i}{d_i}}}{{\Sigma {f_i}}}\) = 75 + \(\frac{{ 150}}{{50}}\) = 78

Example 11:    Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Solution:    Let assumed mean a  = 75.5

Mean = \(a + \frac{{\Sigma fd}}{{\Sigma f}} = 75.5 + \frac{{12}}{{30}}\) = 75.5 + 0.4 = 75.9

Example 12:    To find out the concentration of SO₂ in the air (in parts per million, i.e.ppm), the data was collected for 30 localities in a certain city and is presented below :

Find the mean concentration of SO₂ in the air.
Solution:    Let the assumed mean a = 0.10.

By step deviation method
Mean = a + \(\frac{{\Sigma {f_i}{u_i}}}{{\Sigma {f_i}}}\) × h
= 0.10 + \(\frac{{–1}}{{30}} \times 0.04\)
= 0.10 – 0.0013
= 0.0987
= 0.099 ppm

Example 13:    The weekly observation on cost of living index in a certain city for the year 2004–2005 are given below. Compute the mean weekly cost of living index.

Solution:    Let assumed mean be 1750 i.e., a = 1750

By step deviation method
Mean (\(\bar x\)) = a + \(\frac{{\Sigma {f_i}{u_i}}}{{\Sigma {f_i}}}\) × h
= 1750 + \(\frac{{ – 45}}{{52}} \times 100\)
= 1750 – 86.54
= 1663.46
Hence, the mean weekly cost of living index
= 1663.46

Example 14:    Find the mean marks from the following data by step deviation method

Solution:    Let assumed mean = 55 ⇒ a = 55

Here, a = 55, h = 10,
Σf_i = 85,  Σf_iu_i = –56
Mean (\(\bar x\)) = a + \(\frac{{\Sigma {f_i}{u_i}}}{{\Sigma {f_i}}}\) × h
h = 55 + \(\frac{{ – 56}}{{85}} \times 10\)
= 55 – 6.59 = 48.41
Hence, mean mark = 48.41.

Example 15:    Find the mean age of 100 residents of a colony from the follwing data :

Solution:    Let assumed mean a = 35

Here, a = 35,  h = 10
\(\bar x\) = a + \(\frac{{\Sigma {f_i}{u_i}}}{{\Sigma {f_i}}}\) × h
⇒  \(\bar x\)  = 35 + \(\frac{{ – 40}}{{100}} \times 10\) = 31
Hence, the mean age = 31 years

Example 16:    The following distribution show the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18.00. Find the missing frequency f.

Solution:    we have,

Mean  \(\bar x\) = \(\frac{{\Sigma fx}}{{\Sigma f}}\)  ⇒   18 = \(\frac{{752 + 20f}}{{44 + f}}\)
⇒   18 (44 + f) = 752 + 20f
⇒   752 + 20f = 792 + 18f
⇒   2f = 40
⇒      f = 20
Hence, the missing frequency is 20.

Example 17:    The arithmetic mean of the following frequency distribution is 50. Find the value of p.

Solution:    

Mean \(\bar x\) = \(\frac{{\Sigma fx}}{{\Sigma f}}\)  ⇒  50 = \(\frac{{5160 + 30P}}{{92 + P}}\)
⇒   50 (92 + P) = 5160 + 30 P
⇒   4600 + 50 P = 5160 + 30P
⇒   20 P = 560
⇒   P = 28

Example 18:    The mean of the following frequency distribution  is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies f1 and f2 :

Solution:    

30 + f₁ + f₂ = 50 ⇒ f₁ + f₂ = 20    ….(1)
Mean  = \(\frac{{\Sigma fx}}{{\Sigma f}}\) ⇒  62.8 = \(\frac{{2060 + 30{f_1} + 70{f_2}}}{{50}}\)
⇒  62.8 = \(\frac{{206 + 3{f_1} + 7{f_2}}}{5}\)
⇒  206 + 3f₁ + 7f₂ = 314
⇒   3f₁ + 7f₂ = 108                     ….(2)
3f₁ + 3f₂ = 60                            ….(3)
[Multiplying (1) by 3]
On Subtracting (3) from (2), we get
4f₂ = 48  ⇒  f₂ = 12
Putting f₂ = 12 in (1), we get
f₁ = 8