Poisson Distribution Real Life Examples

In this class, We discuss Poisson Distribution Real Life Examples.

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Example 1:

The average number of accidents on a national highway daily is 1.8.

Determine the probability that the number of accidents

1) At atleast one

2) atmost one

Solution:

Given average number of accidents = 1.8 = lambda value.

If we apply binomial distribution to this example, we need n and p values.

n is the number of cars going on the highway. Identifying n is not possible.

p is the probability of a car doing an accident, and p is also not possible.

P(atleast 1) = P(X >=1) = 1- P(X = 0)

P(X = 0) = (e^-1.8 (1.8)^0)/0!

= 1- 0.1653

= 0.834

2) P(X <=1) = P(X=0) + P(X = 1) = P(atmost 1)

= 0.4628

Example 2:

A distributor of bean seeds determines from the extensive test that 5 percent of a large batch of seeds will not germinate.

He sells the seeds in a package of 200 and guarantees 90 percent germination.

Determine the probability of particular packet violet the guarantee.

Solution:

Probability of seeds not germinating = 0.05 = 5 percent.

λ = mean of seeds not germinating in a sample of 200.

λ = np = 200*0.05 = 10 seeds.

given 90 percent assurance.

In a packet, if more than 20 seeds fail, we lose the assurance.

P(X > 20) = 1-P(X<= 20)

= 1 – x= 0 to 20 Σ (e^-λ λ^x)/x!

= 1 – 0.9984

= 0.0016 very fewer chances