Paper 4: Fundamentals of Business Mathematics & Statistic

FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 10.15

p( 1 )=

11 11 11 1 11

1! 1! 2.7188

eee

e

−−−

====

p(0) + p(1) = 2.7183 2.7183^11 +=.736
1 – [p(0)+p(1)]
= 1 – 0.736 = 0.264
Example 14 :
If the mean of a Poisson distribution’s 4, find (1) S.D. (4) m 3 (5) m 4
Solution :
m = 4

1. S.D. = m== 42


2. μ 3 ==m 4


3. μ 4 =+ =+=mm 3448522
   Example 15 :
   Find the probability that at most 5 defective bolts will be found in a box of 200 bolts, if it is known that 2%
   of such bolts are expected to be defective. (e-4=0.0183)

Solution : m = n × p
= 200 × .02
= 4
P(o) = P(0)+P(1)+P(2)+P(3)+P(4)+P(5)

=

4 44442345

e− (^14) 2! 3! 4! 5!

⎛⎞⎜ ⎟⎟

⎜⎜⎜+++++⎟⎟

⎝⎠

= e–4(1+4+8+10.67+8.53)

=

11111 1 1

ee 2.718 2.178

−++=− +⎡⎤ ⎡ ⎤⎢⎥ ⎢ ⎥

⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦

= 1 – 0.736 = 0.264

Example 16 :
One fifth per cent of the blades produced by a blade manufacturing factory turn out to be defective. The
blades are supplied in packets of 10. Use poisson distribution to calculate the approximate number of
packets containing no defective, one defective and two defective blades respectively in a consignment
of 100,000 packets.
(Given e^02 = .9802)
Solution :

Here p = 5001 ,10n=

m= np = 5001 ×=10 0.02