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:

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

Example:

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 $= \frac{3998}{64} = 62.47 \ cm$

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,

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

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.

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, $Median = 43$