Poisson Distribution is a discrete probability function which takes average rate of success and Poisson random variable as inputs and gives the output values of poisson distribution. It can also be used for the number of events in other specified intervals such as distance, area or volume.

**Formula:**

where **e** is the base of the natural logarithm equal to 2.71828..
**k** is the number of occurrences of an event; the probability of which is given by the function
**x!** is the factorial of k

λ is a positive real number.

__Example:__

1. Calculate the Poisson Distribution whose λ (Average Rate of Success)) is 3 & X (Poisson Random Variable) is 6.

Substitute these values in the above formula,

f = 3^{6} x e^{-3}/6!

= 729 x 0.0498 / 720

= 0.0504

2. A manufacturer of television set known that on an average 5% of their product is defective. They sells television sets in consignment of 100 and guarantees that not more than 2 set will be defective. What is the probability that the TV set will fail to meet the guaranteed quality? e^{-5} = 0.0067

__Solution:__

Success = the TV is defective

X = number of successes

p = probability of success = 5% = 0.05

n = 100 , λ = np = 100 x 0.05 = 5

Poisson Distribution is

P(X=x) = e^{-λ} λ^{x} /x! ; x=0,1,2,3,4

Guarantee: X not less than 2 => X= 0,1,2

P(X > 2) = 1 - [P(0)+ P(1) + P(2)]

= 1 - e^{-5} [1 + 5 + 25/2 ]

= 1 - e^{-5} (37/2)

= 1 - (0.0067) x 37/2

= 1 - 0.12395

= 0.87605

**Practice Problems:**

1. If 3% of the electric bulbs manufactured by a company are defective find the probability that in a sample of 100 bulbs exactly 5 bulbs are defective. Given that e^{-3} = 0.0498.

2. It is known from the past experience that in a certain plant there are on the average of 4 industrial accidents per month. Find the probability that in a given year will be less that 3 accidents. Given e^{-4} = 0.0183

3. Suppose that the change of an individual coal miner being killed in a mining accident during a year is 1.1499. Use the Poisson Distribution to calculate the probability that in the mine employing 350 miners- there will be atleast one accident in a year. Given e^{-0.25}= 0.7788

4. A taxi firm has two cars which it hires out day by day. The number of demands for a car on each day is distributed as Poisson distribution with mean 1.5. Calculate the proportion of days on which neither car is used and the proportion of days on which some demands is refused

When you try such calculations on your own, this *poisson distribution calculator* can be used to verify your results of calculations.

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