10 SCREEN 12
20 PRINT "The probability distribution of a Poisson random variable X representing"
30 PRINT "the number of successes occurring in a given time interval or a specified"
40 PRINT "region of space is given by the formula:"
50 PRINT "P(x)=((e^-m)*(m^x))/x! where x= 0,1,2,3... e=2.71828 (but use your calculator's e button)"
70 PRINT "m = mean number of successes in the given time interval or region of space"
80 PRINT "Mean and Variance of Poisson Distribution"
90 PRINT "If m is the average number of successes occurring in a given time interval or region in the"
100 PRINT "Poisson distribution, then the mean and the variance of the Poisson distribution are both equal to m. e(x)=m AND V(x)=s^2=m"
120 PRINT "Note: In a Poisson distribution, only one parameter, m is needed to determine the probability of an event."
130 PRINT "EXAMPLE: A life insurance salesman sells on the average 3 life insurance policies per week."
140 PRINT "Use Poisson's law to calculate the probability that in a given week he will sell"
150 PRINT "(a) some policies (b) 2 or more policies but less than 5 policies."
160 PRINT "(c) Assuming that there are 5 working days per week, what is the probability that in a given day he will sell one policy?"
170 PRINT "Answer: Here, m = 3 (a) Some policies means 1 or more policies. We can work this out by finding 1 minus the zero policies probability"
180 PRINT "P(x>0)=1-P(x0) Now P(x)=((e^-m)*(m ^ x))/x! So P(x0)=((e^-3)*(3^0))/0!=4.9787x10^-2! So Probability=P(X>0)=1-P(x0)=1-4.9787x10^-2=.95021"
200 PRINT "(b) (P(2<=X<5)=P(x2)+P(x3)+P(x4)=(((e^-3)*(3^2))/2!)+(((e^-3)*(3^3))/3!)+(((e^-3)*(3^4))/4!)=0.61611"
210 PRINT "(c) Average number of policies sold per day: 3/5=0.6 So on a given day, P(X)=((e^-0.6)*(0.6^1))/1!=0.32929"