Mean

Mean or average or arithmetic mean is one of the representative values of data. We can find the mean of observations by dividing the sum of all the observations by the total number of observations.

Mean of raw data:

If x₁, x₂, x₃, ……. x_n are n observations, then

Arithmetic Mean = (x₁, x₂, x₃, ……. x_n)/n

= (∑x_i)/n

∑ (Sigma) is a Greek letter showing summation

1. Weights of 6 boys in a group are 63, 57, 39, 41, 45, 45. Find the mean weight.

Solution:

Number of observations = 6

Sum of all the observations = 63 + 57 + 39 + 41 + 45 + 45 = 290

Therefore, arithmetic mean = 290/6 = 48.3

Mean of tabulated data:

If x₁, x₂, x₃, x₄, ……. x_n are n observations, and f₁, f₂, f₃, f₄, ……. f_n represent frequency of n observations.

Then mean of the tabulated data is given by

= (f₁ x₁ + f₂ x₂ + f₃ x₃ + ……. f_n x_n)/(f₁ + f₂ + f₃ + …… f_n) = ∑(f_ix_i)/∑f_i

2. A die is thrown 20 times and the following scores were recorded 6, 3, 2, 4, 5, 5, 6, 1, 3, 3, 5, 6, 6, 1, 3, 3, 5, 6, 6, 2.

Prepare the frequency table of scores on the upper face of the die and find the mean score.

Solution:

 

Number on the upper face of die	Number of times it occurs (frequency)	fixi
1	2	1 × 2 = 2
2	2	2 × 2 = 4
3	5	3 × 5 = 15
4	1	4 × 1 = 4
5	4	5 × 4 = 20
6	6	6 × 6 = 36

Therefore, mean of the data   = ∑(f_ix_i)/∑f_i

                                        = (2 + 4 + 15 + 4 + 20 + 36)/20

                                        = 81/20

                                        = 4.05



3. If the mean of the following distribution is 9, find the value of p.

X	4	6	p + 7	10	15
f	5	10	10	7	8

Solution:

Calculation of mean

xi	fi	xifi
4	5	20
6	10	60
p + 7	10	10(p + 7)
10	7	70
15	8	120

∑f_i = 5 + 10 + 10 + 7 + 8 = 40

∑ f_ix_i = 270 + 10(p + 7)

Mean = ∑(f_ix_i)/∑f_i

9 = {270 + 10(p + 7)}/40

⇒ 270 + 10p + 70 = 9 × 40

⇒ 340 +10p = 360

⇒ 10p = 360 - 340

⇒ 10p = 20

⇒ p = 20/10

⇒ p = 2



Mean of grouped data:

While calculating the mean of the grouped data, the values x₁, x₂, x₃, ……. x_n are taken as the mid-values or the class marks of various class intervals. If the frequency distribution is inclusive, then it should be first converted to exclusive distribution.


4. The following table shows the number of plants in 20 houses in a group

Number of Plants	0 - 2	2 - 4	4 - 6	6 - 8	8 - 10	10 - 12	12 - 14
Number of Houses	1	2	2	4	6	2	3

Find the mean number of plans per house

Solution:

We have
   

Number of Plant	Number of Houses (fi)	Class Mark (xi)	fixi
0 - 2	1	1	1 × 1 = 1
2 - 4	2	3	2 × 3 = 6
4 - 6	2	5	2 × 5 = 10
6 - 8	4	7	4 × 7 = 28
8 - 10	6	9	6 × 9 = 54
10 - 12	2	11	2 × 11 = 22
12 -14	3	13	3 × 13 = 39

∑f_i = 1 + 2 + 2 + 4 + 6 + 2 + 3 = 20

∑f_i x_i =1 + 6 + 10 + 28 + 54 + 22 + 39 = 160

Therefore, mean = ∑(f_ix_i)/ ∑f_i = 160/20 = 8 plants

● Statistics

• Real Life Statistics
  
• Terms Related to Statistics
  
• Frequency Distribution of Ungrouped and Grouped Data
  
• Use of Tally Marks
  
• Class Limits in Exclusive and Inclusive Form
  
• Construction of Bar Graphs
  
• Mean
  
• Mean of the Tabulated Data
  
• Mode
  
• Median
  
• Construction of Pie Chart
  
• How to Construct a Line Graph?
  

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