9th Class Mathematics Statistics Median

Median

Category : 9th Class

*       Median

 

Median of a data is the value of the variable which divides it into two equal parts. It means it is the value of variable so that the number of observation above it is equal to the number of observation below it. Suppose\[{{x}_{1}},{{x}_{2}}.....{{x}_{n}}\]are n observation in ascending or descending order. The median of the above observation is:

(i) n is odd then median is the value of\[{{\left( \frac{n+1}{2} \right)}^{th}}\] observation.

(ii) n is even then median is the value of arithmetic mean of \[{{\left( \frac{n}{2} \right)}^{th}}\]and \[{{\left( \frac{n}{2}+1 \right)}^{th}}\] observation i.e  

\[\text{Mean}=\frac{{{\left( \frac{n}{2} \right)}^{th}}\text{observation}+{{\left( \frac{n}{2}+1 \right)}^{th}}\text{observation}}{2}\]

Method for finding the median for grouped data. Step for finding the median

Step 1: Forgiven frequency distribution, prepare the commutative frequency table and obtain\[N=\sum {{f}_{i}}\].

Step 2: Find (N/2).

Step 3: Look at the cumulative frequency Just greater than (N/2) and find the corresponding class, known as median class.

Step 4: Then by using median formula, calculate median, which is given below:

Median \[=l+\left[ h\times \frac{\frac{N}{2}-C}{f} \right]\]

where I = lower limit of median class, h = width of median class,

f = frequency of median class

c =cumulative frequency of the class for preceding the median class

\[N=\sum fi\]  

 

Find the median class of daily wages from the following frequency distribution  

Daily Wages (In Rs.) Frequency Daily Wages (In Rs.) Frequency
100 - 150 6 250 - 300 20
150 - 200 3 300 - 350 10
200 - 250 5    

 

(a) 250                                 

(b) 260                                 

(c) 270                                  

(d) 280                     

(e) None of these  

 

Answer: (c)

Explanation

Class Interval Frequency \[({{f}_{i}})\] Cf
100 - 150 6 6
150 - 200 3 9
200 - 250 5 14
250 - 300 20 34
300 - 350 10 44
  \[\sum {{f}_{i}}=44=N\]  

 

(In the above table c f represents cumulative frequency)

median\[=250+\left\{ 50+\left( \frac{22-14}{20} \right) \right\}=270\]

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