Percentiles - Calculation of Percentiles

Example 1
1) Find \[\mathbf{P}_{40}\] for the following data.
10, 15, 8, 16, 19, 11, 12, 14, 9

Solution: Arrange the data in ascending order i.e. 8, 9, 10, 11, 12, 14, 15, 16, 19
n = 9
\[\mathrm{P}_{40}=\text{size of }40\left(\frac{n+1}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{40}=\text{size of }40\left(\frac{9+1}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{40}=\text{size of }40\left(\frac{10}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{40}=\mathrm{size~of}\left(\frac{40\times10}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{40}=\mathrm{size~of}\left(\frac{400}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{40}=\mathrm{size~of}4^{\text{th Observation}}\]
\[\mathrm{P}_{40}=\mathrm{size~of}4^{\text{th Observation}}\] is 11
\[\therefore\mathbf{P}_{40}=11\]
\[\boxed{\quad\mathbf{Ans}:\mathbf{P}_{40}=11}\]

Example 2
2) Calculate \[\mathbf{P}_{85}\] from the following data.
79, 82, 36, 38, 51, 72, 68, 70, 64, 63

Solution: Arrange the data in ascending order i.e. 36, 38, 51, 63, 64, 68, 70 72, 79, 82
n = 10
\[\mathrm{P}_{_{85}}=\text{size of 85}\left(\frac{n+1}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{_{85}}=\text{size of 85}\left(\frac{10+1}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{_{85}}=\text{size of 85}\left(\frac{11}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{85}=\mathrm{size~of}(85\times0.11)^{\text{th Observation}}\]
\[\mathrm{P}_{85}=\mathrm{size~of}(9.35)^{\text{th Observation}}\]
\[\mathrm{P}_{85}=\mathrm{size~of~}9^{\text{th Observation}}\]+ 0.35\[(10^\text{th observation}{-9^{\text{th observation}}})\]
\[\mathrm{P}_{85}=79+0.35(82-79)\]
\[\mathrm{P}_{85}=79+0.35\times3\]
\[\mathbf{P}_{85}=79+1.05\]
\[\therefore\mathbf{P}_{85}=80.05\]
\[\boxed{\quad\mathrm{Ans:P}_{85}=80.05}\]

1) Find out \[\mathbf{P}_{20}\] and \[\mathbf{P}_{60}\] for the following data:

Solution: Arrange the data in ascending order. 

\[\mathrm{P}_{_{20}}=\text{size of 20}\left(\frac{n+1}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{_{20}}=\text{size of 20}\left(\frac{40+1}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{_{20}}=\text{size of 20}\left(\frac{41}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{20}=\mathrm{size~of}\left(\frac{20\times41}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{_{20}}=\mathrm{size~of}\left(\frac{820}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{_{60}}=\mathrm{size~of~8.2^{th~Observation}}\]
\[\mathrm{P}_{_{60}}=\mathrm{size~of~8.2^{th~Observation}}\] lies in cf 9 
Hence \[\mathbf{P}_{20}=59\]
\[\therefore\mathbf{P}_{20}=59\]
Calculation of \[\mathbf{P}_{60}\]

\[\mathrm{P}_{_{60}}=\text{size of 60}\left(\frac{n+1}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{_{60}}=\text{size of 60}\left(\frac{40+1}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{_{60}}=\text{size of 60}\left(\frac{41}{100}\right)^{\text{th Observation}}\] 
\[\mathrm{P}_{60}=\mathrm{size~of}\left(\frac{60\times41}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{60}=\mathrm{size~of}\left(\frac{2460}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{_{60}}=\mathrm{size~of}\left(24.6\right)^{\text{th Observation}}\]
\[\mathrm{P}_{60}=\mathrm{size~of~24.6^{th~Observation}}\]
lies in cf 25
Hence \[\mathbf{P}_{60}=61\]
\[\therefore\mathbf{P}_{60}=61\]
\[\boxed{\quad\mathrm{Ans:P}_{20}=59,\mathrm{P}_{60}=61}\]

Apply the steps as mentioned in Quartiles of continuous data.

1) Find \[\mathbf{P}_{65}\] from the following data

Solution:

Step I
\[\mathrm{P}_{_{65}}=\mathrm{size~of}\left(\frac{65n}{100}\right)^{\text{th Observation}}\]

\[\mathrm{P}_{_{65}}=\mathrm{size~of}\left(\frac{65n}{100}\right)^{\text{th Observation}}\]
\[\mathrm{P}_{65}=\mathrm{size}\mathrm{of}\left(\frac{3250}{100}\right)^{\text{th Observation}}\]
\[P_{65}=size~of~32.5^{thObservation}\]
\[P_{65}=size~of~32.5^{thObservation}\] lies in cf 42
Hence, the percentile class is 15-20.
l =15 f = 12 cf = 30 n = 50 h = 5
Step II
\[\mathbf{P}_{65}=l+\left(\frac{\frac{65n}{100}-cf}{f}\right)\times h\]
\[\mathbf{P}_{65}=15+\left(\frac{\frac{65\times50}{100}-30}{12}\right)\times5\]
\[\mathrm{P}_{65}=15+\left(\frac{\frac{3250}{100}-30}{12}\right)\times5\]
\[\mathbf{P}_{65}=15+\left(\frac{32.5-30}{12}\right)\times5\]
\[\mathbf{P}_{65}=15+\left(\frac{2.5}{12}\right)\times5\]
\[\mathrm{P}_{65}=15+\left(\frac{2.5\times5}{12}\right)\]
\[\mathrm{P}_{65}=15+\left(\frac{12.5}{12}\right)\]
\[\mathrm{P}_{65}=15+1.04\]
\[\therefore\mathbf{P}_{65}=16.04\]
\[\boxed{\quad\mathrm{Ans:P}_{65}=16.04}\]