Paper 4: Fundamentals of Business Mathematics & Statistic

5.18 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Measures of Central Tendency and Measures of Dispersion

Example 17 : Find the H. M. of 3, 6, 12 and 15.

H.M.^44240 6.15

1 1 1 1 20 10 5 4 39

3 6 12 15 60

= = + + + = =

+ + +

Example 18 : Find the H.M. of 1, 21 , 31 ,.......n^1

H.M.

( )

n n

1 2 3 ... n n 2 n 1

2

= + + + + =

+ −

2n

=n(n 1)+

[Note. The denominator is in A..P. use S = n 2 {2a (n 1)d+ − }

Example 19 : A motor car covered distance of 50 miles four times. The first time at 50 m. p. h, the second at

20 m. p. h., the third at 40 m. p. h, and the fourth at 25 m.p.h Calculate the average speed and explain the

choice of the average.

Average Speed (H.M) = 501 201 401 251

=^4

+ + + = 20 50 25 40 1000

4

= + + + = 4 ×^1000135 = 29.63

= 30 (app.) m. p. h.

For the statement x units per hour, when the different values of x (i.e. distances) are given, to find average,

use H.M. If again hours (i.e., time of journey) are given, to find average, we are to use A.M. In the above

example, miles (distances) are given, so we have used H.M.

Weighted H.M. The formula to be used is as follows :

1 2 n

1 2 n

H.M. N

f f ..... f

x x x

=

• 
  
  
  
  • 
    
    
    
    • , ∑f = N
    
    
    
    
  
  
  
  

Example 20 :

(a) A person travelled 20 k.m. at 5 k.m.p.h. and again 24 k.m. at 4 k.m.p.; to find average speed.

(b) A person travelled 20 hours at 5 k.m.p.h. and again 24 hours at 4.m.p.h.; to find average speed.

(a) We are to apply H.M. (weight) in this case, since, distances are given.

Average speed (H.M.) 20 2420 24 4 6 10^4444

5 4

= + = =

+ + = 4.4 k.m.p.h.

(b) We are to apply A.M. (weighted), since times of journey are given.

Average speed (A.M.) =20 5 24 4 100 96 196× +20 24+ × = 44 + = 44 = 4.45 k.m.p.h (app.)