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### poisson distribution

• 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.
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