1 The number of industrial injuries per working week in a pa

1. The number of industrial injuries per working week in a particular factory is known to follow a Poisson distribution with mean 0.5. Find the probability that a) in a particular week there will be: i. no accident, ii. less than 2 accidents, iii. more than 2 accidents; b) in a 3 weeks period there will be no accidents.

Solution

Possion Distribution
PMF of P.D is = f ( k ) = e- x / x!
Where   
= parameter of the distribution.
x = is the number of independent trials
a)
P( X = 0 ) = e ^-0.5 * 0.5^0 / 0! = 0.6065

b)
P( X < 2) = P(X=1) + P(X=0) +
= e^-0.5 * 0 ^ 1 / 1! + e^-0.5 * ^ 0 / 0! +   
= 0.9098

c)
P( X < = 2) = P(X=2) + P(X=1) + P(X=0)   
= e^-0.5 * 0.5 ^ 2 / 2! + e^-0.5 * 0 ^ 1 / 1! + e^-0.5 * ^ 0 / 0!
= 0.9856
P( X > 2) = 1 -P ( X <= 2) = 1 - 0.9856 = 0.0144

d)
Mean rate for 3 week period is = 3*0.5 = 1.5

P( X = 0 ) = e ^-1.5 * 1.5^0 / 0! = 0.2231