10.16 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Theoretical Distribution

Using the formula for Poisson distribution, the probability of x defective blades is

() 0.02()0.02

!

e x

px= x

The frequencies of 0, 1, 2, 3 defective blades given by

( ) 1,00,000 0.02()0.20

!

e x

fx x X

−

=

Number of packets with no defective blade

= 1,00,000 × e–0.02 = 1,00,000 × 0.9802

= 98,020

Number of packets with one defective blades

P(1) =

0.02().02

!

e− x

= 1,00,000 × e–0.02 × 0.02

= 98,020 × .02

= 98,020 × 1002

= 1960.4

= 1960

Number of packets with two defective blades is

= 1,00,000 × e–0.02×(0.02) 2

= 98020 × .0002

= 19.6040

= 20

Example 17 :

It is known from past experience that in a certain plant there are on the average 4 industrial accidents per

month. Find the probability that in a given year there will be less than 4 aecident. Assume Poisson distribution.

Solution :

m= 4

p(x = r) =

(^44)
!!
em emr r
r r
−−

The required probability that there will be less than 4 accidents is given as :
p(x<4) = p(x = 0) + p(x = 1)+ p(x = 2) + p(x = 3)