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