Paper 4: Fundamentals of Business Mathematics & Statistic

5.20 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Measures of Central Tendency and Measures of Dispersion

=^15 (0.3010 + 0.6021 + 0.7782 + 0.9031 + 1.0000) = 51 × 3.5844 = 0.7169

∴ G. M. = antilog 0.7169 = 5.211

H.M. ( )

5 5

1 1 1 1 (^11) 60 30 20 25 12
2 4 6 8 10 120

= =

+ + + + + + + +

= 5 × 120 600137 137= =4.379

Again A.M. = 51 (2+4+6+8+10) =^15 × 30 = 6

We get A.M. =6, G. M. = 5.211, H.M. = 4.379 i.e. A.M. ≥ G. M. ≥ H. M

Note : In only one case the above relation is not true. When all the variates are equal, we will find that

AM = GM = HM

Example 24 : A.M. and G.M. of two observations are respectively 30 and 18. Find the observations. Also find

H.M.

Now x y 2 + =^30 or, x + y = 60 .......(1) again xy 18=

Or, xy = 324 or, (60 – y). y = 324, from (1)

Or, y^2 – 60 y + 324 = 0 or, (y –54) (y –6) = 0, y = 54, 6

∴y = 54 , x = 6 or, y = 6, x = 54.

∴ Required observations are 6, 54.

H. M.^22254 10.80.

16 154 9 1 10

54

∴ = = + = × =

+

SELF EXAMINATION QESTIONS :

1. Find H.M. of the numbers :
   (i) 3, 6, 24, 48 [Ans. 7.1]
   (ii) 2, 4, 6, 8 [Ans. 3.84]


2. Calculate H.M. of the following numbers
   (i) 1,^12 ,^13 , ......, 101 [Ans. 0.18]

(ii) 1, 21 ,^13 ,.....^1 n [Ans. n 1^2 + ]

(iii) 1, 31 ,^15 ,.....,2n 1^1 − [Ans.^1 n]