The number of frogs found by a frog finding system in any 30

The number of frogs found by a frog finding system in any 30 second interval under specified conditions average 1.81. Assume the frogs appear randomly and independently.

1. what is the probability that no frogs are found in a one-minute interval?

2. what is the probability of observing at least five frogs but not more than eight frogs in two minutes of observation?

Solution

a)

If there are an average of 1.81 frogs in 30s, then there are 1.81*2 = 3.62 frogs in 1 min.

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 =    3.62      
          
x = the number of successes =    0      
          
Thus, the probability is          
          
P (    0   ) =    0.026782676 [ANSWER]

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

b)

In 2 minutes, there are an average of 1.81*4 = 7.24 frogs.

Note that P(between x1 and x2) = P(at most x2) - P(at most x1 - 1)          
          
Here,          
          
x1 =    5      
x2 =    8      
          
Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    7.24      
          
          
Then          
          
P(at most    4   ) =    0.152201292
P(at most    8   ) =    0.697307497
          
Thus,          
          
P(between x1 and x2) =    0.545106205   [ANSWER]