Median for Discrete and Continuous Frequency Type Data (grouped data)

In our article , we explained how to calculate the . In this article we shall discuss how to as well as . For both cases, the method for calculating the median is the same. We shall also discuss a few .

Median for Discrete and Continuous Frequency Type Data (grouped data): -

For the or a continuous variable the calculation of the median involves identifying the , i.e. the class containing the median. This can be done by calculating the less than type . It should be recalled that less than type cumulative frequencies correspond to the of the respective classes. we calculate $\dfrac{N}{2}$ and find out the less than type cumulative frequency just greater or equal to $\dfrac{N}{2}$ . Let that less than type cumulative frequency be denoted by $F_{k}$ . Now, we find out the class boundary corresponding to which the less than type cumulative frequency is equal to $F_{k}$ . Let us denote that class boundary by $x_{k}$ . The median class will be that class for which the upper-class boundary is $x_{k}$ . Now, that we know the median class, the median can be calculated using the following formula:

where l =

N =

cf = cumulative frequency of class preceding the median class

f =

h =

Example: The following distribution represents the number of minutes spent per week by a group of teenagers in going to the movies. Find the median number of minutes spent per by the teenagers in going to the movies.

Solution:

Let us convert the class intervals given to class boundaries and construct the less than type cumulative frequency distribution.

here $\dfrac{N}{2}=\dfrac{300}{2}=150$

and the cumulative frequency just greater than or equal to 150 is 198.

$\therefore F_{k}=198$ and

$\because F_{k}$ is the less than type cumulative frequency corresponding to the class boundary 399.5

$\therefore x_{k}=399.5$

$\therefore$ the median class is the class for which upper class boundary is $x_{k}=399.5$

In other words, 299.5-399.5 is the median class, i.e. the class containing the median value.

$\therefore$ using the formula for median we have,

where l = lower limit of median class = 299.5

N = total no. of observations = 300

cf = cumulative frequency of class preceding the median class = 123

f = frequency of median class = 75

h = class size or width = 100

therefore,

So, the median number of minutes spent per week by this group of 300 teenagers in going to the movies is 335.5, i.e. there are 150 teenagers for whom the number of minutes spent per week in going to the movies is less than 335.5 and there are another 150 teenagers for whom the number of minutes spent per week in going to the movies is greater than 335.5.

Note:

It is quite clear that in calculating the median of any grouped frequency distribution using this method, the nature of the variable (i.e. discrete or continuous) is of little consequence. Whatever be the nature of the variable, for grouped frequency distributions, this method is exhaustive and will ensure correct calculation of the median.

Properties of Median:

1. If we have two sets of va;ues having medians $M_{1}$ and $M_{2}$ respectively, then the combine set median, say, $M$ , lies between $M_{1}$ and $M_{2}$

2. If $y=g(x)$ be a real-valued, monotonic function of $x$ , then median of $y$ is given by $\tilde{y}=g(\tilde{x})$ , $\tilde{x}$ being the median of $x$ .

Example: Values of two variables, $x$ and $y$ , are related as $y= 12x-311.5$ .If the median of $x$ be $\tilde{x}=130$ , find the median of $y$ , i.e. $\tilde{y}$ .

Solution:

Here, the median of $y$ is $\tilde{y} =12 \times 130 -311.5=124.5$

Exercise:

1. Obtain the median for the following frequency distribution of house rent for a sample of 30 families in a certain locality:

2. Frequency distribution of I.Q. of 309 6-year old children is given below:

Find the median I.Q. of these children.

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