Deciles - Calculation of Deciles

Example 1
1) Calculate \[\mathrm{D}_4\] and \[\mathrm{D}_8\] for the following data. 10, 15, 7, 8, 12, 13, 14, 11, 9

Solution: Arrange the data in ascending order. 7, 8, 9, 10, 11, 12, 13, 14, 15
\[\mathrm{D}_4=\mathrm{size~of}4\left(\frac{n+1}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_4=\mathrm{size~of}4\left(\frac{9+1}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_4=\mathrm{size~of}4\left(\frac{10}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_4=\mathrm{size~of}(4\times1)^{\text{th Observation}}\]
\[\mathrm{D}_4=\mathrm{size~of}4^{\text{th Observation}}\]
\[\therefore\mathbf{D}_{4}=10\]
Calculation of D8
\[\mathrm{D}_8=\mathrm{size~of}8\left(\frac{n+1}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_8=\mathrm{size~of}8\left(\frac{9+1}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_8=\mathrm{size~of}8\left(\frac{10}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_8=\mathrm{size~of}(8\times1)^{\text{th Observation}}\]
\[\mathrm{D}_8=\mathrm{size~of}8^{\text{th Observation}}\]
\[\therefore\mathbf{D}_{8}=14\]
\[\boxed{\quad\mathbf{Ans}:\mathbf{D}_4=\mathbf{10},\mathbf{D}_8=\mathbf{14}}\]

Example 2
2) Calculate \[\mathrm{D}_8\] from the given data: 14, 13, 12, 11, 15, 16, 18, 17, 19, 20

Solution: First arrange the data in ascending order. 

11, 12, 13, 14, 15, 16, 17, 18, 19, 20
n = 10
\[\mathrm{D}_8=\mathrm{size~of}8\left(\frac{n+1}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_8=\mathrm{size~of}8\left(\frac{10+1}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_8=\mathrm{size~of}8\left(\frac{11}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_8=\mathrm{size~of}(8\times1.1)^{\text{th Observation}}\]
\[\mathrm{D}_8=\mathrm{size~of}(8.8)^{\text{th Observation}}\]
\[\mathrm{D_s=size~of~8^{th~observation}+0.8~(9^{th~observation}-8^{th~observation})}\]

\[\mathrm{D}_{_8}=18+0.8(19-18)\]
\[\mathrm{D}_8=18+(0.8\times1)\]
\[\mathrm{D}_8=18+0.8\]
\[\therefore\mathbf{D}_{8}=18.8\]
\[\boxed{\begin{array}{c}\mathbf{Ans:D_8=18.8}\end{array}}\]

1. Find out \[\mathrm{D}_2\] and \[\mathrm{D}_4\] for the following data.

Solution:

\[\mathrm{D}_2=\mathrm{size~of}2\left(\frac{n+1}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_2=\mathrm{size~of}2\left(\frac{39+1}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_2=\mathrm{size~of}2\left(\frac{40}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_2=\mathrm{size~of}(2\times4)^{\text{th Observation}}\]
\[\mathrm{D_2=size~of~(8)^{th~Observation}}\]
\[\mathrm{Size~of}8^{\text{th Observation}}\] lies in cf 11
Hence \[\mathrm{D}_2\] = 20 marks
\[\therefore\mathbf{D}_{2}=20\]
Calculation of \[\mathrm{D}_4\]
\[\mathrm{D}_4=\mathrm{size~of}4\left(\frac{n+1}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_4=\mathrm{size~of}4\left(\frac{39+1}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_4=\mathrm{size~of}4\left(\frac{40}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_4=\mathrm{size~of}(4\times4)^{\text{th Observation}}\]
\[\mathrm{D}_4=\mathrm{size~of}16^{\text{th Observation}}\]
\[\mathrm{Size~of}16^{\text{th Observation}}\] lies in cf 11
Hence \[\mathrm{D}_4\] = 40 marks
\[\therefore\mathbf{D}_{4}=40\]
\[\boxed{\quad\mathbf{Ans}:\mathbf{D}_{2}=20,\mathbf{D}_{4}=40}\]

Apply the steps as mentioned in Qualities of continuous data.

1) Find out \[\mathrm{D}_5\] and \[\mathrm{D}_7\] for the following data of marks of 100 students in a class test.

Solution:

Calculation of \[\mathrm{D}_5\]
Step I
\[\mathrm{D}_{s}=\mathrm{size~of}\left(\frac{5n}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_{s}=\mathrm{size~of}\left(\frac{5\times100}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_{s}=\mathrm{size~of}\left(\frac{500}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_8=\mathrm{size~of}50^{\text{th Observation}}\]
\[\mathrm{Size~of}50^{\text{th Observation}}\] lies in cf 60
Hence Decile class = 20-30
∴ l = 20 f = 40 cf = 20 n = 100 h = 10
Step II
\[\mathbf{D}_5=l+\left(\frac{\frac{5n}{10}-cf}{f}\right)\times h\]
\[\mathrm{D}_{5}=20+\left(\frac{\frac{5\times100}{10}-20}{40}\right)\times10\]
\[\mathrm{D}_{5}=20+\left(\frac{\frac{500}{10}-20}{40}\right)\times10\]
\[\mathbf{D}_{5}=20+\left(\frac{50-20}{40}\right)\times10\]
\[\mathrm{D}_{5}=20+\left(\frac{30}{40}\right)\times10\]
\[\mathrm{D}_{5}=20+\frac{300}{40}\]
\[\mathrm{D}_{5}=20+7.5\]
\[D_5=27.5 marks\]
\[\therefore\mathbf{D}_{5}=27.5\]
Calculation of \[\mathrm{D}_7\]
Step I
\[\mathrm{D}_7=\mathrm{size~of}\left(\frac{7n}{10}\right)^\text{th Observation}\]
\[\mathrm{D}_{\gamma}=\mathrm{size~of}\left(\frac{7\times100}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_{\gamma}=\mathrm{size~of}\left(\frac{700}{10}\right)^{\text{th Observation}}\]
\[\mathrm{D}_{\gamma}=\text{size of }70^{\text{th Observation}}\]
\[\mathrm{D}_{\gamma}=\text{size of }70^{\text{th Observation}}\] lies in cf 80
Hence Decile class = 30-40
∴ l = 30 f = 20 cf = 60 n = 100 h = 10
Step II

\[\mathbf{D}_{7}=l+\left(\frac{\frac{7n}{10}-cf}{f}\right)\times h\]
\[\mathbf{D}_7=30+\left(\frac{\frac{7\times100}{10}-60}{20}\right)\times10\]
\[\mathbf{D}_7=30+\left(\frac{\frac{700}{10}-60}{20}\right)\times10\]
\[\mathbf{D}_7=30+\left(\frac{70-60}{20}\right)\times10\]
\[\mathbf{D}_7=30+\left(\frac{10}{20}\right)\times10\]
\[\mathrm{D}_7=30+\left(\frac{100}{20}\right)\]
\[\mathrm{D}_{7}=30+5\]
\[\mathrm{D}_7\] = 35 marks

∴ \[\mathrm{D}_7\] = 35
\[\boxed{\quad\mathbf{Ans}:\mathbf{D}_{5}=27.5,\mathbf{D}_{7}=35}\]