What is a Grouped Frequency Distribution Table

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 }} }})

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}} times h)

Working Rule :

Step-1 :     Choose a number ‘a’ from the central values of x(mid-values)

Step-2 :    Obtain u_i = (frac{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}} times h) to get the required mean.

Grouped Frequency Distribution Table Example Problems with Solutions

Example 1:    

Mid-values	2	3	4	5	6
Frequencies	49	43	57	38	13

Find the mean by direct method.

Solution:

Mid Values	frequencies (fi)	fixi
2	49	98
3	43	129
4	57	228
5	38	190
6	13	78
Total	N = Σfi = 50	Σfixi = 2750

Mean = (frac}}) = (frac) = 3.615

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

Class Interval	Frequency
10-30	90
30-50	20
50-70	30
70-90	20
90-110	40

Solution:

Class Interval	f	Mid value (x)	f × x
10-30	90	20	1800
30-50	20	40	800
50-70	30	60	1800
70-90	20	80	1600
90-110	40	100	4000
	Σf = 200		Σfx = 10000

Mean = (frac}}) = (frac) = 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 locali
ty. Find  the mean number of plants per house.

Number of plants	0 – 2	2 – 4	4 – 6	6 – 8	8 – 10	10 – 12	12 – 14
No. of houses	1	2	1	5	6	2	3

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

Solution:

Number of plants	Number of houses (f)	Mid value (x)	f × x
0-2	1	1	1
2-4	2	3	6
4-6	1	5	5
6-8	5	7	35
8-10	6	9	54
10-12	2	11	22
12-14	3	13	39
	Σf = 20		Σfx = 162

Mean = (frac}}) = (frac) = 8.1

Example 4:    Calculate the mean for the following distribution:

Variable	5	6	7	8	9
Frequency	4	8	14	11	3

Solution:



∴ Mean = (frac = frac)  = 7.025

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



Solution:



Mean = (frac = frac = 32.8overline 6) or 32.87 (approx.)

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

Class interval	0 – 10	11 – 20	21 – 30	31 – 40	41 – 50
Frequency	3	4	2	5	6

Solution:



Mean = (frac = frac) = 28.9

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

Size of Item	1 – 4	4 – 9	9 – 16	16 – 27
Frequency	6	12	26	20

Solution:



Mean = (frac = frac)  = 13.25

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

Expenditure (in rupees)	100-150	150-200	200-250	250-300	300-350	350-400	400-450	450-500
Frequency	24	40	33	28	30	22	16	7

Solution:    Let assumed mean = 275



(bar x = a + frac}}) = 275 + (frac) = Rs 266.25

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

Class Interval	Frequency
0 – 50	17
50 –100	35
100 –150	43
150–200	40
200– 250	21
250– 300	24

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



Now , a = 175

(bar x = a + frac}}) = 175 + (frac)

= 175 – 26.39 = 148.61 approx.

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

Class interval	50– 60	60–70	70–80	80–90	90– 100
Frequency	8	6	12	11	13

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



a = 75, Σf_id_i= 150, Σf_i = 50

Mean (bar x = a + frac}}) = 75 + (frac) = 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.

Number of heart beats per minute	Frequency
65– 68	2
68–71	4
71–74	3
74–77	8
77– 80	7
80– 83	4
83– 86	2

Solution:    Let assumed mean a  = 75.5



Mean = (a + frac = 75.5 + frac) = 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
}}) × h

= 0.10 + (frac 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}}) × h

= 1750 + (frac 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}}) × h

h = 55 + (frac 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}}) × h

⇒  (bar x)  = 35 + (frac 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)  ⇒   18 = (frac)

⇒   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)  ⇒  50 = (frac)

⇒   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) ⇒  62.8 = (frac})

⇒  62.8 = (frac}{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