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Example – 2 . Find the median for the given distribution :

Solution :- At first we prepare cumulative frequency distribution table as under :

Now, n = 120. So, n/2 = 60. This observation lies in the class 40 – 50. Then,

 l = lower limit of the class = 40,

cf = cumulative frequency of the class preceding (40 – 50) = 52,

f = frequency of median class = 26 , h = class size = 10.

Using formula, median = l + [(n/2 – cf)/f] × h

= 40 + [(60 – 52)/26] × 10

= 40 + (8/26) × 10

= 40 + 3.07

= 43.07.

Median by  Ogive :-

Let us consider the frequency distribution given below :

 Solution :- To draw Ogive, we first construct cumulative frequency table as under :          

Fig. – 6.

From the graph, Median = n/2th observation = 60th observation = 43.

Quartiles and Inter quartile Range

When data is arranged in ascending order and if the distribution is divided into four quarter inserting three points in between them, then these points are called quartiles. If these three points be denoted by Q₁,Q₂ and Q₃, then Q₂ is the median. Q₁ is called lower quartile and Q₂ the upper quartile.  

Calculation of Lower Quartile

Q₁ = L₁ + (i₁/f₁)(n/4 – cf )

Where, L₁ = Actual lower limit of the class containing the lower quartile, n/4; 

Lower quartile class is the class in which n/4 belongs.
              i₁ = class size of this class,
              f₁ = frequency of this class,
              c_f = cumulative frequency of the class preceding lower quartile class,

Calculation of Upper Quartile
                   Q₃ = L₃ + (i₃/f₃)(3n/4 – cf )

Where, L₃, i₃, f3are lower limit, class-size, and frequency of the class containing the upper quartile and cf denotes the cumulative frequency of the class preceding upper quartile class. 

Inter Quartile Range = Q₃ – Q₁.

Semi Inter Quartile Range = (Q₃ –Q₁)/2.