(III) Percentiles
Centiles is another term for percentiles. Any given observation is essentially divided into a total of 100 equal parts by a centile or percentile. These percentiles or centiles are represented as P1, P2, P3, P4,……….P99. P1 is a typical value of peaks for which 1/100 of any given data is either less than P1 or equal to P1.
Usually, the percentile formula is used to compare a person’s performance to that of others. Remembering that a candidate’s percentile in tests and scores indicates how they rank among other candidates is important. The ratio of values below the value ‘x’ to all matters is used to determine the percentile of the value ‘x’. The percentiles can be calculated for weight, wealth, and several other factors. The formulas for calculating percentiles are:
[Tex]P_{1}=[\frac{N+1}{100}]^{th}~item [/Tex]
[Tex]P_{2}=[\frac{2(N+1)}{100}]^{th}~item [/Tex]
[Tex]P_{3}=[\frac{3(N+1)}{100}]^{th}~item [/Tex]
…………….[Tex]P_{99}=[\frac{99(N+1)}{100}]^{th}~item [/Tex]
Where, n is the total number of observations, P1 is First Percentile, P2 is Second Percentile, P3 is Third Percentile, ……….P99 is Ninety Ninth Percentile.
Example 1:
Calculate the P20, P90 from the following weights in the family: 25, 17, 32, 11, 40, 35, 13, 5, and 46.
Solution:
First of all, organise the numbers in ascending order.
5, 11, 13, 17, 25, 32, 35, 40, 46
[Tex]P_{20}=[\frac{20(N+1)}{100}]^{th}~item [/Tex]
[Tex]P_{20}=[\frac{20(9+1)}{100}]^{th}~item [/Tex]
P20 = 2nd item
P20 = 11
[Tex]P_{90}=[\frac{90(N+1)}{100}]^{th}~item [/Tex]
[Tex]P_{90}=[\frac{90(9+1)}{100}]^{th}~item [/Tex]
P90 = 9th item
P90 = 40
Example 2:
Calculate P10 and P75 for the data related to the age (in years) of 99 members in a housing society.
Solution:
[Tex]P_{10}=[\frac{10(N+1)}{100}]^{th}~item [/Tex]
[Tex]P_{10}=[\frac{10(99+1)}{100}]^{th}~item [/Tex]
P10 = 10th item
Now, the 10th item falls under the cumulative frequency of 20 and the age against this cf value is 10.
P10 = 10 years
[Tex]P_{75}=[\frac{75(N+1)}{100}]^{th}~item [/Tex]
[Tex]P_{75}=[\frac{75(99+1)}{100}]^{th}~item [/Tex]
P75 = 75th item
Now, the 75th item falls under the cumulative frequency of 85 and the age against this cf value is 40.
P75 = 40 years
Example 3:
Determine the value of P50 for the company’s salary listed below.
Solution:
In case N is an even number, the following formula is used:
[Tex]P_{50}=[\frac{50(N)}{100}]^{th}~item [/Tex]
[Tex]P_{50}=[\frac{50(60)}{100}]^{th}~item [/Tex]
P50 = 30th item
Now, the 30th item falls under the cumulative frequency 38 and the salary against this cf value lies between 700-800.
[Tex]P_{50}=l+\frac{\frac{50(N)}{100}-m}{f}\times{c} [/Tex]
[Tex]P_{50}=700+\frac{\frac{50(60)}{100}-22}{16}\times{100} [/Tex]
[Tex]P_{50}=700+\frac{30-22}{16}\times{100} [/Tex]
P50 = ₹750
Also Read:
Range | Meaning, Coefficient of Range, Merits and Demerits, Calculation of Range
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