PERCENTILES

Percentiles are the values which divide a given set of values into 100 equal parts. 

The points of sub-divisions being 

P₁, P₂, P₃, P₄, .......................P₉₉

P₁ is the value for which one-hundredth of the observations are less than are equal to P₁ and the remaining ninety-nine hundredths observations are greater than or equal to P₁, once the values are arranged in ascending order of magnitude.  

Example 1 : 

Following are the wages of the laborers : 

$82, $56, $90, $50, $120, $75, $75, $80, $130, $65

Find P₈₂.

Solution :

Number of values given (n)  =  10

Arrange the wages in ascending order : 

$50, $56, $65, $75, $75, $80, $82, $90, $120, $130

Formula to find percentile is   

(n + 1)p^th value

To find P₈₂, substitute 10 for n and 82/100 for p in the above formula.

(n + 1)p^th value  =  (10 + 1) ⋅ (82/100)^th value

Simplify. 

(n + 1)p^th value  =  (11) ⋅ (82/100)^th value

(n + 1)p^th value  =  (902/100)^th value

(n + 1)p^th value  =  9.02^th value

Find 9.02^th value in the ascending order of wages :

9.02^th value  =  9th value + 0.02(10th value - 9th value)

9.02^th value  =  120 + 0.02(130 - 120)

9.02^th value  =  120 + 0.02(10)

9.02^th value  =  120 + 0.2

9.02^th value  =  120.20

So, P₈₂ is $120.20.

Example 1 : 

Following distribution relates to the distribution of monthly wages of 100 workers. 

Compute P₂₃.

Solution :

Computation of P₂₃

The formula to find percentile for grouped frequency distribution is 

Number of values given (N)  =  100

For P₂₃, we have

p  =  23/100

Then, we have

Np  =  100 ⋅ (23/100)  =  23

In the cumulative frequency table above, 23 comes between 5 and 28.

Then, we have 

N_l  =  5

N_u  =  28

l₁  =  499.50

C  =  699.50 - 499.50  =  200

To find P₂₃, substitute the above values in the formula.

P₂₃  =  499.50 + [(23 - 5)/(28 - 5)] ⋅ 200

P₂₃  =  499.50 + (18/23) ⋅ 200

P₂₃  =  499.50 + (18/23) ⋅ 200

P₂₃  =  499.50 + 156.52

P₂₃  =  656.02

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