Question

A shooter missed his target in the last 10 trials by the following distance (mm) during the practice session:

20, 32, 24, 41, 18, 27, 15, 36, 35, 25

Find the quartile deviation and coefficient of quartile deviation of such distance missed by the shooter.

Answer 1

Writing the measures of mistargets in ascending order :

15, 18, 20, 24, 25, 27, 32, 35, 36, 41

Here, n = 10

First Quartile :

Q₁ = Value of \(\left(\frac{n+1}{4}\right)\)th observation

= Value of \(\left(\frac{10+1}{4}\right)\) = 2.75 th observation

= Value of 2nd observation +0.75 (Value of 3rd observation – Value of 2nd observation)

= 18 + 0.75 (20 – 18)

= 18 + 0.75 (2)

= 18 + 1.50 = 19.50 mm

Third Quartile :

Q₃ = Value of 3\(\left(\frac{n+1}{4}\right)\) th observation

= Value of 3 (2.75) = 8.25th observation

= Value of 8th observation + 0.25 (Value of 9th observation – Value of 8th observation)

= 35 + 0.25 (36-35)

= 35 + 0.25 = 35.25 mm

Quartile deviation of measures of mistargets :

Q_d = \(\frac{\mathrm{Q}_{3}-\mathrm{Q}_{1}}{2}\)

Putting Q₃ = 35.25; Q1 = 19.50, we get

Q_d = \(\frac{35.25-19.50}{2}=\frac{15.75}{2}\) = 7.875 ≈ 7.88

Coefficient of quartile deviation = \(\frac{\mathrm{Q}_{3}-\mathrm{Q}_{1}}{\mathrm{Q}_{3}+\mathrm{Q}_{1}}\)

= \(\frac{35.25-19.50}{35.25+19.50}\)

= \(\frac{15.75}{54.75}\)

= 0.29