subject: Poisson Distribution Applet [print this page] Introduction to Poisson distribution applet:
The Poisson distribution is the limited form for binomial distribution. The examples for the Poisson distribution are the number of cars passing through a certain street in the time period t and the number of printing mistakes at each page of the book. The Poisson distribution includes the probability distribution function, mean and variance for the Poisson variables. This article has the details about Poisson distribution applet.
Formula Used for the Poisson Distribution Applet:
Random variable x taking non- negative integral values is said to follow Poisson distribution if its probability mass function is given by
P[X = x] =' (e^ - lambda lambda^x)/ (x!)'
Poisson distribution has only one parameter 'lambda' > 0.
The formula for the mean value is' lambda' = n p
The variance is also 'lambda' .
Where n is the number of trails and p is the possibilities for the event.
Examples for Poisson Distribution Applet:
Example 1 for Poisson distribution applet:
A manufacturer of cotton pins knows that 4% of his product is defective. If he sells pins in boxes of 100 and guarantees that not more than 1 pin will be defective. Calculate the probability for the given box which will fail to meet the guaranteed quality.
Solution:
The value of p is p = '4/100' , n = 100
The mean value is 'lambda' = n p = ('4/100' ) (100) = 4
By the Poisson distribution
P[X = x] = '(e^ - lambda lambda^x)/ (x!)'
Probability for the given box will to meet the guaranteed quality = P[X > 1]
P[X > 1] = 1- P[X = 1]
P[X > 1] = 1- (P (0) +P (1) + P (2) +P (3))
P[X > 1] = 1- 'e^-4' (1 + 4+ 8 +10.67)
P[X > 1] = 1- 'e^-4' (23.67)
P[X > 1] = 1- 'e^-4' (23.67)
P[X > 1] = 1- 0.0183(23.67)
P[X > 1] = 1- 0.4331
P[X > 1] = 0.5669
The probability for the given box will fail to meet the guaranteed quality is 0.5669.
Example 2 for Poisson distribution applet:
A car- hire firm has three cars. The number of demands for a car with mean of 5.8. Compute the proportion of days on which neither the car is used nor the proportion of days on which some demand is refused.
Solution:
Let X defines the number of demands for the car.
The mean value is 5.8.
By the Poisson distribution
P[X = x] = '(e^ - lambda lambda^x)/ (x!)'
Proportion of the days on which neither car is used = P[X = 0] = 'e^-5.8' = 0.003
Proportion of days on which some demand is refused = P[X > 3]
P[X > 3] = 1- P[X = 3]
P[X > 3] = 1- [P (0) +P (1) + P (2) +P (3)]
P[X > 3] = 1- 'e^-5.8' (1+ 5.8+ 16.82+ 32.52)
P[X > 3] = 1 - (0.003) (56.14)
P[X > 3] = 1 - 0.1684
P[X > 3] = 0.8316
The proportion of days on which neither the car used is 0.003. The proportion of days on which some demand refused is 0.8316.
