Let X be a poisson random varable with a mean of 4 use the f

Let X be a poisson random varable with a mean of 4. use the formula and calculate

a) P(X=3)

b) P(X < 2)

c) P(X less than or equal to 2)

d) P(X greater than or equal to 2)

e) P(X > 2)

Solution

A)

Note that the probability of x successes out of n trials is          
          
P(x) = u^x e^(-u) / x!          
          
where          
          
u = the mean number of successes =    4      
          
x = the number of successes =    3      
          
Thus, the probability is          
          
P (    3   ) =    0.195366815 [answer]

***********

b)

Note that P(fewer than x) = P(at most x - 1).          
          
Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    4      
p = the probability of a success =    0      
x = our critical value of successes =    2      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   1   ) =    0.091578194
          
Which is also          
          
P(fewer than   2   ) =    0.091578194 [answer]

*************

c)

Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    4      
          
x = the maximum number of successes =    2      
          
Then the cumulative probability is          
          
P(at most   2   ) =    0.238103306 [answer]

***********

d)

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    4      
          
x = our critical value of successes =    2      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   1   ) =    0.091578194
          
Thus, the probability of at least   2   successes is  
          
P(at least   2   ) =    0.908421806 [answer]

**************

e)

Note that P(more than x) = 1 - P(at most x).          
          
Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    4      
          
x = our critical value of successes =    2      
          
Then the cumulative probability of P(at most x) from a table/technology is          
          
P(at most   2   ) =    0.238103306
          
Thus, the probability of at least   3   successes is  
          
P(more than   2   ) =    0.761896694 [answer]