<p>in year 1,2,3,4,5 : </p>
<p>year 1 : pulled over 3 times
year 2 : pulled over 1 time
year 3 : pulled over 2 times
year 4 : pulled over 1 time
year 5 : pulled over 3 times </p>
<p>what is the probability that year 6, i will be pulled over at least once?</p>
<p>Recommendation: Drive slower.</p>
<p>(Where does it say "get your homework solved HERE" on this forum...? They should take that down, wherever it says it.)</p>
<p>Given the data, 1.</p>
<p>So this problem does not involve the poissant distribution?</p>
<p>Poisson. I was working nonparametrically. You could use a Poisson model if you wanted to. Then it would be 1-exp(-2). But it requires extra assumptions.</p>
<p>whats the answer?</p>
<p>1 - e^(-rt(2)) and I think that equals about 83.1%. lambda = sqrt(mean) = sqrt(2) and then x in this case is 0.</p>
<p>Square root of the mean???? What funny stats book have you been using?</p>
<p>P(X=x) = (exp(-lambda) lambda^x)/x! </p>
<p>is the usual parametrization of the Poisson distribution, with E(X)=Var(X)=lambda. If the sample mean and variance of a large sample are not approximately equal, you have to deal with over/underdispersion in your model...</p>
<p>/I would go for the nonparametric estimate, but your instructor probably won't appreciate. I also don't see why you'd have to express 1-exp(-2) in decimal notation.</p>
<p>Mean and standard deviation of a poisson random variable with parameter lamda are mean = lambda and standard deviation = sqrt lambda respectively. Thats all the book I have says and in the problems it just gave lambda so I went by that. Thanks for clarifying.</p>