The number N of calls arriving at a switchboard during a per
Solution
Here, we see a Poisson distribution of mean = 10.
a)
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 = 10
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 ) = 0.000499399
Thus, the probability of at least 2 successes is
P(at least 2 ) = 0.999500601 [answer]
****************
b)
Using a cumulative poisson distribution table or technology, matching
u = the mean number of successes = 10
x = the maximum number of successes = 3
Then the cumulative probability is
P(at most 3 ) = 0.010336051 [answer]
*****************
c)
Note that P(between x1 and x2) = P(at most x2) - P(at most x1 - 1)
Here,
x1 = 4
x2 = 6
Using a cumulative poisson distribution table or technology, matching
u = the mean number of successes = 10
Then
P(at most 3 ) = 0.010336051
P(at most 6 ) = 0.130141421
Thus,
P(between x1 and x2) = 0.11980537 [answer]

