- 1. Presentation on topic : What is Poisson distribution ? Give examples where it can be applied. Presented by : Sangeeta Saini M.Com (p) 641
- 2. Poisson Distribution. Definition of Poisson Distribution. Properties of Poisson distribution. Examples of Poisson distribution.
- 3. Poisson Distribution Poisson Distribution is a discrete probability distribution and it is widely used in statistical work . This distribution was developed by French mathematician Dr. Simon Denis Poisson in 1837 and the distribution is named after him. The Poisson Distribution is used in those situations where the probability of happening of an event is small ,i.e. the event rarely occurs.
- 4. DEFINITION OF POISSON DISTRIBUTION Poisson Distribution is defined and given by the following probability function: Where P(X=x ) = probability of obtaining x number of success. m=np = parameter of distribution . e = 2.7183 base of natural logarithms P(X=x) = e-m .mx X!
- 5. Number of success (X) Probability P(X) 0 e-m.m0=e-m 1 e-1.m1 =me-m 2 e-2.m2 =m2e-m 2! 2 . . . . . . . . . . x e-m.mx x! 0! 1!
- 6. Properties of Poisson Distribution Discrete probability distribution •The Poisson distribution is a discrete probability distribution in which the number of success are given in whole number such as , 0,1,2,……… etc. Value of p and q •Poisson distribution is used in those situations where the probability of occurrence of an event is very small (p-0) and probability of non occurrence of event is very large (q-1). Main parameter •It has only one parameter m and its value is equal to np •m= np Constant of Poisson distribution •The constant of Poisson distribution can be obtained by from following formula: •Mean =m= np •Variance =m •S.D.= root of m •Moment coeff. Of skewness = 1/ root of m •Moment coeff. Of kurtosis = 3+ 1/m Equality of mean and variance •Its mean and variance are equal. •Mean = variance
- 7. Examples of Poisson Distribution It is used in statistical quality control to count the number of defects of an item. In Biology, to count the number of bacteria. In insurance problems to count the number of casualities. To count the number of errors per page in a typed material . To count the number of incoming telephone calls in a town. To count the number of defective blades in a lot of manufactured blades in a factory. To count the number of deaths at a particular crossing in a town as a result of road accident. To count the number of suicides committed by lover point in a year.
- 8. Applications of Poisson Distribution Application of Poisson distribution formula Fitting of Poisson distribution
- 9. Application of Poisson distribution formula When the value of p is given. Example : It is given that 2% of the screw manufactured by a company are defective. Poisson distribution to find the probability that a packet of 100 screws contains : (a) no defective screws (b) one defective screw (c) two or more defectives.{ Given : e-2= 0.135 } Solution : let p = probability of defective screw = 2 % = 2/100 n=100 m=n.p 100×2/100 =2 , m=2 (a) No defective screw P(X=x) = e-m .mx ÷ x! P(X=0) = e-2. 20 ÷ 0! = e-2 =0.135
- 10. (b) P(one defective ) P(X=1) = e-2. 21÷ 1! = e-2×2 = 0.135 × 2 = 0.270 (c)P ( two or more defectives) =1- [P(0)+P(1)] = 1-[0.135+0.270] = 1-0.405 = 0.595
- 11. When the value of m is given Example : Between the hours of 2 P.M. to 4 P.M., the average number of phone calls per minute coming into a switch board of company is 2.5. find the probability that during one particular minute there will be : (a) no phone call at all . (b)exactly 3 calls. (Given e-2 = 0.1353 , e-0.5 = 0.6065 ) Solution : average number of phone calls =m=2.5 P(X=x) = e-m .mx ÷ x! (a) P(no call) =P(X=0) = e-2.5. 2.50÷ 0! = e-2.5 e-2.5= e-2.e-0.5 = 0.1353 ×0.6065 = 0.0821 (b) P( exactly 3 calls) =P(X=3) = e-2.5. 2.53÷ 3! = (0.0821).(15.625)/ 3×2×1 = 0.2138
- 12. Fitting of a Poisson Distribution Step 1. firstly , we compute mean from the observed frequency data . Mean = /N Step 2. the value of e-m is obtained . If the value of e-m is not given in the question . e-m = Reciprocal [ Antilog (m × 0.4343)] Step 3. then ,we compute the probability of 0,1 , 2, 3 or x success by using Poisson distribution. P(X=x) = e-m .mx ÷ x!
- 13. Step 4. the expected frequency are then obtained by multiply each probability with n , total frequencies. Number of successes (X) Probability P(X) Expected frequency Fe (X) 0 e-m.m 0=e-m 0! N.P(0) =Ne-m 1 e-1.m1 =me-m 1! N.P(1)= N.me-m 2 e-2.m2 =m2e-m 2! 2 N.P(2)= N. m2e-m 2 . . . . . . . . . . . x e-m.mx x! N.P(X)= N. e-m.mx X!
- 14. Deaths 0 1 2 3 4 Frequency 109 65 22 3 1 Also find mean and variance of above distribution. Solution : Fitting of Poisson distribution Deaths Frequency (f) fX 0 109 0 1 65 65 2 22 44 3 3 9 4 1 4 200 122
- 15. Mean = 122 /200 = 0.61 m= 0.61 Now obtain the value of e-0.61 e-m = Reciprocal [ Antilog (m× 0.4343)] Putting m=0.61 e-m = Reciprocal [ Antilog (0.61× 0.4343)] = Rec. [antilog(0.26492)] = Rec. [1.841] =0.5432
- 16. No. of deaths probability Expected frequency 0 P(0)=e-0.61(0.61)0/0! =e- 0.61= 0.5432 N.P(0)=200×0.5432 = 108.64 = 109 1 P(1)=e-0.61(0.61)1/1! =e- 0.61×0.61= 0.5432× 0.61 = 0.33 N.P(1)=200×0.33=66 2 P(2)=e-0.61(0.61)2/2! = 0.10 N.P(2)=200×0.10=20 3 P(3)=e-0.61(0.61)3/3! = 0.021 N.P(3)=200×0.021=4.2=4 4 P(4)=e-0.61(0.61)4/4! = 0.0031 N.P(4)=200×0.0031=0.62= 1 Thus X 0 1 2 3 4 Fe 109 66 20 4 1 Mean= 0.61 Mean= variance =0.61
- 17. Conclusion In conclusion ,we can say that , the Poisson Distribution is useful in rare events where the probability of success (p) is very small and probability of failure (q) is very large and value of n is very large.