In this part of lecture notes we will learn about MEAN, MODE and MEDIAN.

 Mean

Mean in statistics is the same as average in Arithmetic. This is also called Arithmetic Mean.

 

Mean of Raw (or ungrouped) data:

The mean of n observations x_1, \ x_2, \ x_3, \ ..., x_n   is given by the formula:

Mean=\frac{Sum \ of\ observations}{Number of observations}=\frac{x_1, \ x_2, \ x_3, \ ..., x_n}{n} = \frac{\Sigma{x_i}}{n}

 

Here \Sigma \ (sigma)\ shows summation.

Similarly, we have:

Mean\ of\ n\ numbers=\frac{Sum \ of \ the \ numbers}{Numbers \ of \ addend}

 

m5xExample:

The heights of 5 plants in a garden are: 150 cm, 170 cm, 158 cm, 160 cm \ and\  162 cm .

Find the mean height.

Solution:

Sum of the given observations = (150 + 170 + 158 + 160 + 162) cm = 800 cm  .

Number of observations = 5.

\therefore Mean\ of\ n\ numbers=\frac{Sum \ of \ the \ numbers}{Numbers \ of \ observations} =\frac{800}{5}\ cm=160\ cm

Hence, the mean height is 160 cm.

 

Mean Of Tabulated Data

Let the frequencies of n observations x_1, \ x_2, \ x_3, \ ..., x_n  be f_1, \ f_2, \ f_3, \ ..., f_n  respectively.

We define mean as:

\therefore Mean = \frac{f_1.x_1+f_2.x_2+f_3.x_3+...+f_n.x_n}{f_1, \ f_2, \ f_3, \ ..., f_n }=\frac{\Sigma(f_i.x_i)}{\Sigma{f_i}}  

Example:

 The height of 72 plants in a garden is given below:

Height (in cm) 58 60 62 64 66 68
Number of plants 10 12 16 11 9 6

Find the mean height of plant.

Solution:

We may prepare the table as shown below:

Height ( in cm )

x_i

No. of Plants 

f_i

f_i.x_i

58

10 580

60

12 720

62

16 992

64

11

704

66

9

594

68 6

408

(\Sigma f_i)=64

(\Sigma f_i.x_i)=3998

Therefore,

Mean height = (3998/64) cm = 62.47 cm.

 

Mean of grouped Data

In a class interval, we take x as the value of the value of the mark (average of the class range) and use the formula,

Mean=\frac{\Sigma(f_i.x_i)}{\Sigma f}  

m6x

Example:

The following table shows the weight of 50 persons in a group.

Weight (in kg) 40-44 44-48 48-52 52-56 56-60
Number of persons 8 12 9 16 5

 Solution:

We have

Weight ( in cm )

Frequency f_i

class mark   x_i

f_i .x_i

40-44

8 42 42×8=336

44-48

12 46 46×12=552

48-52

9

50

50×9=450

52-56

16

64

54×16=854

56-60 5 58

58×5=290

(\Sigma f_i)=50

(\Sigma f_i.x_i)=2492

∴ Mean weight = 2482/50 kg = 49.84 kg

Hence, the mean weight is 49.84 kg.

 

Mode

The observation which occur maximum number of times is called the mode of the given data. In a tabulated data the observation with maximum frequency is the mode.

 Example

Given below is the data showing the size number of shoes sold by a shoe company in a day are 11, 12, 7, 5, 8, 8, 9, 7, 8, 8, 6, 5, 5, 6, 10, 8, 9, 8 . Find the mode of the data.

 Solution

In the given data, 8 occurs maximum number of time

So, the mode of the given data = 8.

 

Median

The median is the value separating the higher half of a data sample, from the lower half.

The basic advantage of the median in describing data compared to the mean (often simply described as the “average”) is that it is not skewed so much by extremely large or small values, and so it may give a better idea of a ‘typical’ value.m7x

The first step is to arrange data in ascending or descending order of magnitudes, the value of the middle term is called the median of the data.

  • When the number of observations is odd, then there will be only one midterm and this term is the median. This is \frac{n+1}{2}
  • When the number of observations is even, then there will be two middle terms. The average of these two middle terms will be the median of the data.

This is the mean of (\frac{n}{2}) \ \&\ (\frac{n}{2}+1) terms.

 

Example:

Find the median of the following data: 55, 23, 41, 36, 41, 45, 50, 55  

Solution

Arranging in ascending order: 23, 36, 41, 41, 45, 50, 55, 55  

Number of terms = 8.

So, 4th and 5th terms are the two middle terms.

Therefore,

Median=Mean \ of \ the \ 4^{th} \ and \ 5^{th} \ terms \\ =\frac{1}{2}(41+45) = 43  

Hence, Median = 43 m8x

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