1. Definition of Poisson’s Distribution
2. Properties of Poisson’s Distribution
3. Example of Poisson’s Distribution
4. Fitting of Poisson’s Distribution
5. Applications of Poisson’s Distribution
2. CONTENT
1. Definition of Poisson’s Distribution
2. Properties of Poisson’s Distribution
3. Example of Poisson’s Distribution
4. Fitting of Poisson’s Distribution
5. Applications of Poisson’s Distribution
3. 1. POISSON’S DISTRIBUTION
Poisson’s distribution is used in situations where,
1. 𝑝, probability of success i.e. chance of occurrence of an event is indefinitely
very small (𝑝 → 0. )
2. 𝑛,number of trials indefinitely large, i.e. 𝑛 → ∞
3. 𝑚, is a positive real number. say parameter of Poisson distribution 𝑚 = 𝑛. 𝑝
Definition: A Discrete random variable 𝑋 taking values 0,1,2…. Is said to
follow poisson's distribution with parameters 𝑚, and its given by,
𝑃 𝑋 = 𝑟 =
𝑒−𝑚
𝑚 𝑟
𝑟!
; 𝑟 = 0,1,2 & 𝑚 > 0
= 0 ; 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Where, 𝑚 = parameter of Poisson distribution
𝑟 = Last sentence of the problem!
4. 2. PROPERTIES OF POISSON’S DISTRIBUTION
Mean 𝜇 = 𝑚 = 𝑛. 𝑝
Variance 𝜎2 = 𝑚
Thus the mean and variance of P. D. are equal and each
is equal to parameter of the distribution
Standard deviation 𝜎 = 𝑚
Coefficient of skewness = 1
𝑚
5. 3. EXAMPLE OF POISSON’S DISTRIBUTION
If 2% of the electric bulbs manufactured by a company are defective. Find the
probability that in a sample of 100 bulbs a) 3 are defective b) at least two are
defective.
Ans: Given, 2% 𝑏𝑢𝑙𝑏𝑠 𝑎𝑟𝑒 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒 = 𝑝 = 2% = 0.02
ℎ𝑒𝑟𝑒, 𝑛 = 100
Here p is very small and n is large so we use poisson’s distribution, so P. D.
formula
𝑃 𝑋 = 𝑟 =
𝑒−𝑚
𝑚 𝑟
𝑟!
Here, 𝑚 = 𝑚𝑒𝑎𝑛 = 𝑛. 𝑝 = 100 × 0.02 = 2
a) Three bulbs are defective means 𝑟 = 3
𝑃 𝑟 = 3 =
𝑒−2
23
3!
= 0.1804
Probability that 3 bulb are defective is 0.1804
6. CONT….
b) At least two bulbs are defective:
Probability that at least two bulb are defective means 𝑟 = 2,3,4,5 … .100
As, Total Probability = 1.
Also, 𝑃 𝑥 ≥ 2 = [𝑃 𝑥 = 2 + 𝑃 𝑥 = 3 + ⋯ + 𝑃(𝑥 = 100]
𝑃 𝑥 ≥ 2 = 1 − [𝑃 𝑥 = 0 + 𝑃(𝑥 = 1)]
𝑃 𝑥 ≥ 2 = 1 − [
𝑒−220
0!
+
𝑒−221
1!
] = 0.541
Probability that at least two bulbs are defective is 0.541
7. 4. FITTING OF POISSON’S DISTRIBUTION
Fitting of a distribution to a data means,
1. Estimation the parameters of distribution on the basis of data
2. Computing probabilities
3. Computing expected frequencies.
When a Poisson distribution is to be fitted to an observed data the following
procedure is adopted
1. Find 𝑚𝑒𝑎𝑛 𝑥 =
𝑓𝑥
𝑓
2. Poisson’s Parameter 𝑚 = 𝑥
3. Poisson’s Distribution function 𝑃 𝑋 = 𝑟 =
𝑒−𝑚 𝑚 𝑟
𝑟!
; 𝑟 = 0,1,2 …
4. Put 𝑟 = 0; 𝑃 0 = 𝑃 𝑟 = 0 =
𝑒−𝑚 𝑚0
0!
= 𝑒−𝑚
5. Find the expected frequency of r= 0 𝑖. 𝑒. 𝐹 0 = 𝑁 × 𝑃 0 , 𝑤ℎ𝑒𝑟𝑒 𝑁 = 𝑓
6. The other expected frequencies can be found using
𝐹 𝑟 + 1 =
𝑚
𝑟 + 1
× 𝐹 𝑟 𝑓𝑜𝑟 𝑟 = 0,1,2 … .
8. EXAMPLE OF FITTING OF POISSON’S DISTRIBUTION
The following mistakes per page were observed in a book,
Fit a Poisson distribution and estimate the expected frequencies.
Ans:
1. 𝑚𝑒𝑎𝑛 𝑥 =
𝑓𝑥
𝑓
=
143
325
= 0.44
2. Poisson’s Parameter 𝑚 = 𝑥 = 0.44
3. 𝑃 𝑋 = 𝑟 =
𝑒−𝑚 𝑚 𝑟
𝑟!
=
𝑒−0.440.44 𝑟
𝑟!
4. Put 𝑟 = 0; 𝑃 0 =
𝑒−𝑚 𝑚0
0!
= 𝑒−0.44
= 0.6440 (From the poisson Table)
5. 𝐹 0 = 𝑁 × 𝑃 0 , = 325 × 0.6440 = 209.43~210
6. 𝐹 𝑟 + 1 =
𝑚
𝑟+1
× 𝐹 𝑟 𝑓𝑜𝑟 𝑟 = 0,1,2,3,4
F 1 = 𝐹 0 + 1 =
0.44
0+1
× 209.43 = 92.15~92
F 2 = 𝐹 1 + 1 =
0.44
1+1
× 92.15 = 20.27~20
F 3 = 𝐹 2 + 1 =
0.44
2+1
× 20.27 = 2.972~3
F 4 = 𝐹 3 + 1 =
0.44
3+1
× 2.972 = 0.33~0
The fitted Poisson’s distribution is,
No of mistakes 0 1 2 3 4
No of pages 211 90 19 5 0
𝒙 𝒇 𝒇. 𝒙
0 211 0
1 90 90
2 19 38
3 5 15
4 0 0
Total 325 143
𝒙 0 1 2 3 4 Total
Observed
Frequency
211 90 19 5 0 325
Expected
Frequency
210 92 20 3 0 325
9. 5. APPLICATIONS OF POISSON’S DISTRIBUTION
Poisson’s distribution is used in,
1. To find number of defective items found in a large
size lot.
2. To find number of deaths due to snake-bite in a
village
3. To know number of persons affected due to
radiation
4. To know arrival pattern of defective vehicles,
patients in a hospital or telephone calls.
5. In biology, to count number of bacteria.