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+//Caption: Poisson distribution
+//Example6.4
+//Page179
+clear;
+clc;
+//(a): Exactly 0 customer will arrive in 10 minutes interval
+X1= 0; //no. of customer
+Mean = 4;//mean arrival rate of 4 per 10 minutes
+[P1,Q1]=cdfpoi("PQ",X1,Mean)
+disp(P1,'Exactly 0 customer will arrive P(X=0,4) is =')
+//(b): Exactly 2 customers will arrive in 10 minutes interval
+X2 =2; //no. of customer
+P2 = exp(-Mean)*(Mean^X2)/(factorial(2))
+disp(P2,'Exactly 2 customer will arrive P(X=2,4) is =')
+//(c): at most 2 customers will arrive in 10 minutes interval
+[P3,Q3]=cdfpoi("PQ",X2,Mean)
+disp(P3,'Atmost 2 customer will arrive P(X<=2,4) is =')
+//(d): at least 3 customers will arrive in 10 minutes interval
+X3 =3;
+P4 = 1-P3
+disp(P4,'At least 3 customer will arrive P(X>=3,4) is=')
+//Result
+//Exactly 0 customer will arrive P(X=0,4) is =   
+// 
+//    0.0183156  
+// 
+// Exactly 2 customer will arrive P(X=2,4) is =   
+// 
+//    0.1465251  
+// 
+// Atmost 2 customer will arrive P(X<=2,4) is =   
+// 
+//    0.2381033  
+// 
+// At least 3 customer will arrive P(X>=3,4) is=   
+// 
+//    0.7618967  
+\ No newline at end of file