Introduction:
The allocation was first introduced by Simeon-Denis poisson (1781–1840) and in print, together with his probability theory, in 1838 in his work (Research on the Probability of Judgments in Criminal and Civil Matters). The job focused on sure random variables N that tally, among other things, the number of separate occurrences that take place during a time-interval of given length. In probability hypothesis and statistics, Poisson division is a discrete probability division expresses probability of a number of actions.
Poisson Probability Table Conditions:
Poisson probability distribution states two conditions
1.The amount of successes in two disjoint time intervals is autonomous.
2.The probability of a success during a small time interval is comparative to the entire length of the time interval.
Poisson probability formula:
The Probability distribution of a Poisson random variable, X instead of the number of successes happening in a given time interval or a specified area of space is given by the formula:
P(X)=e-`nu` `nu`X / X!
where
X=0,1,2,3....
e = expected value
µ = mean figure of successes in the given time interval or area of space.
Mean and Variance for poisson probability:
If µ is the standard number of successes happening in a given time interval or area in the Poisson distribution, then the mean and the variance of the Poisson distribution are both equal to µ.
E(X) = µ
and
V(X) = s2 = µ
Note: In a Poisson distribution, only one factor, µ is needed to determine the probability of an occurrence.
Poission Probability Table:
The below shows the value of poisson probability for the mean and X values of
`lambda` =0.1 to 5.5
X =0 to 17
poisson probability table
Examples of Poisson Probability Distribution:
Example 1:
A life assurance salesman sells on the usual 3 life insurance policies per week. Use Poisson's law to work out the possibility that in a given week he will sell
(a) a few policies
(b) 2 or supplementary policies but less than 5 policies.
(c) assume that there are 5 working days per week, what is the probability that in a given day he will sell single policy?
Solution:
Here, µ = 3
(a) "Some policies" means "1 or more policies". We can work this out by finding 1 minus the "zero policies" probability:
P(X > 0) = 1 - P(x0)
Now P(X)=e-`nu` `nu`X / X! So P(X`@` )= e-3 30 / 3! = 4.9787 `xx` 10-2
So
Probability= P(X > 0)
=1 - P(x0)
=1-4.9787`xx`10-2
=0.95021
(b) P(2`<=` X `>=` 5) = P(X2)+P(X3)+P(X4)
= (e-3 32 / 2!) + (e-3 33 / 3!) + (e-3 34 / 4!)
=0.61611
(c) Average number of policies sold per day: `3/5 ` = 0.6
So on a given day P(X)=e-0.6 (0.6)1 / 1!
= 0.32929
Example 2:
A business makes electric motors. The probability an electric motor is imperfect is 0.01. What is the probability that a model of 300 electric motors will contain precisely 5 defective motors?
Solution:
The average number of defectives in 300 motors is µ = 0.01 × 300 = 3
The probability of getting 5 defectives is:
P(X ) = e-3 35 / 5!
= 0.10082
The allocation was first introduced by Simeon-Denis poisson (1781–1840) and in print, together with his probability theory, in 1838 in his work (Research on the Probability of Judgments in Criminal and Civil Matters). The job focused on sure random variables N that tally, among other things, the number of separate occurrences that take place during a time-interval of given length. In probability hypothesis and statistics, Poisson division is a discrete probability division expresses probability of a number of actions.
Poisson Probability Table Conditions:
Poisson probability distribution states two conditions
1.The amount of successes in two disjoint time intervals is autonomous.
2.The probability of a success during a small time interval is comparative to the entire length of the time interval.
Poisson probability formula:
The Probability distribution of a Poisson random variable, X instead of the number of successes happening in a given time interval or a specified area of space is given by the formula:
P(X)=e-`nu` `nu`X / X!
where
X=0,1,2,3....
e = expected value
µ = mean figure of successes in the given time interval or area of space.
Mean and Variance for poisson probability:
If µ is the standard number of successes happening in a given time interval or area in the Poisson distribution, then the mean and the variance of the Poisson distribution are both equal to µ.
E(X) = µ
and
V(X) = s2 = µ
Note: In a Poisson distribution, only one factor, µ is needed to determine the probability of an occurrence.
Poission Probability Table:
The below shows the value of poisson probability for the mean and X values of
`lambda` =0.1 to 5.5
X =0 to 17
poisson probability table
Examples of Poisson Probability Distribution:
Example 1:
A life assurance salesman sells on the usual 3 life insurance policies per week. Use Poisson's law to work out the possibility that in a given week he will sell
(a) a few policies
(b) 2 or supplementary policies but less than 5 policies.
(c) assume that there are 5 working days per week, what is the probability that in a given day he will sell single policy?
Solution:
Here, µ = 3
(a) "Some policies" means "1 or more policies". We can work this out by finding 1 minus the "zero policies" probability:
P(X > 0) = 1 - P(x0)
Now P(X)=e-`nu` `nu`X / X! So P(X`@` )= e-3 30 / 3! = 4.9787 `xx` 10-2
So
Probability= P(X > 0)
=1 - P(x0)
=1-4.9787`xx`10-2
=0.95021
(b) P(2`<=` X `>=` 5) = P(X2)+P(X3)+P(X4)
= (e-3 32 / 2!) + (e-3 33 / 3!) + (e-3 34 / 4!)
=0.61611
(c) Average number of policies sold per day: `3/5 ` = 0.6
So on a given day P(X)=e-0.6 (0.6)1 / 1!
= 0.32929
Example 2:
A business makes electric motors. The probability an electric motor is imperfect is 0.01. What is the probability that a model of 300 electric motors will contain precisely 5 defective motors?
Solution:
The average number of defectives in 300 motors is µ = 0.01 × 300 = 3
The probability of getting 5 defectives is:
P(X ) = e-3 35 / 5!
= 0.10082
No comments:
Post a Comment