# Compute percentile coefficient of kurtosis from the following data and interpret the result. Hourly wages (Rs) 23-27 28-32 33-37 38-42 43-47 48-52 Number of workers 22 16 9 4 3 1

Solution:

 x f cf 23-27 22 22 28-32 16 38 33-37 9 47 38-42 4 51 43-47 3 54 48-52 1 55 N = 55

Now,

Calculating P75:

P75 = $$75 \left ( \frac{N}{100} \right )^{th} item$$ = $$75 \left ( \frac{55}{100} \right )^{th}$$ = 41.25th

In cf, the value just greater than 41.25 is 4. so, class  = 33 – 37

Now,

P75 = $$L + \frac{\frac{75 N}{100} – cf}{f} \times i$$

= $$33 + \frac{41.25 – 38}{9} \times 4$$

= 34.45

Calculating P25:

P25 = $$25 \left ( \frac{N}{100} \right )^{th} item$$ = $$25 \left ( \frac{55}{100} \right )^{th}$$ = 13.75th

In cf, the value just greater than 13.755 is 22. so, class  = 23 – 27

Now,

P25 = $$L + \frac{\frac{25 N}{100} – cf}{f} \times i$$

= $$23 + \frac{13.75 – 0}{22} \times 4$$

= 23.625

Calculating P90:

P90 = $$90 \left ( \frac{N}{100} \right )^{th} item$$ = $$90 \left ( \frac{55}{100} \right )^{th}$$ = 49.5th

In cf, the value just greater than 49.5 is 51. so, class  = 38 – 42

Now,

P90 = $$L + \frac{\frac{90 N}{100} – cf}{f} \times i$$

= $$23 + \frac{49.5 – 47}{4} \times 4$$

= 40.5

Calculating P10:

P10 = $$10 \left ( \frac{N}{100} \right )^{th} item$$ = $$10 \left ( \frac{55}{100} \right )^{th}$$ = 5.5th

In cf, the value just greater than 5.5 is 21. so, class  = 23-27

Now,

P10 = $$L + \frac{\frac{10 N}{100} – cf}{f} \times i$$

= $$23 + \frac{5.5 – 0}{22} \times 4$$

= 24

Now,

Percentile cofficient of Kurtosis

= $$\frac{P_{75} – P_{25}}{2(P_{90} – P_{10})}$$

= $$\frac{34.45 – 23.625}{2(40.5 – 24)}$$

= $$\frac{10.825}{33}$$

= 0.33