Find range and coefficient of range of the following:
(a) Per day earning of seven agricultural labourers in ₹:

60, 72, 36, 85, 35, 52, 72

(b) Mean temperature deviation from normal (2002).

Jan. Feb. Mar. Apr. May June

+1.5 +2.4 +3.1 −1.5 −0.4 +3.3

July Aug. Sep. Oct. Nov. Dec.

−0.1 −0.6 −1.5 −0.6 −1.9 −6.1

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Solution

(a) Range = Largest item (L) − Smallest item (S)
= 85 − 35 = 50

$Coefficient of Range = \frac{L - S}{L + S} = \frac{85 - 35}{85 + 35} = \frac{50}{120} = 0.42$

(b) Range = Largest item (L) − Smallest item (S)
= +3.3 − (−6.1)

= +3.3 + 6.1 = +9.4

$Coefficient of Range = \frac{L - S}{L + S} = \frac{+ 3.3 - (- 6.1)}{+ 3.3 + (- 6.1)} = \frac{+ 3.3 + 6.1}{+ 3.3 - 6.1} = \frac{+ 9.4}{- 2.8} = - 3.36$

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Jan.	Feb.	Mar.	Apr.	May	June
+1.5	+2.4	+3.1	−1.5	−0.4	+3.3
July	Aug.	Sep.	Oct.	Nov.	Dec.
−0.1	−0.6	−1.5	−0.6	−1.9	−6.1

(a) Range = Largest item (L) − Smallest item (S) = 85 − 35 = 50 Coefficient of Range=L-SL+S=85-3585+35 =50120=0.42 (b) Range = Largest item (L) − Smallest item (S) = +3.3 − (−6.1) = +3.3 + 6.1 = +9.4 Coefficient of Range=L-SL+S =+3.3--6.1+3.3+-6.1 =+3.3+6.1+3.3-6.1=+9.4-2.8 =-3.36