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Somebody please assist me solve

(2) The number of incoming calls reaching the switchboard of a computer per minute is thought to be a Poisson random variable with rate 6.5. Compute
i)The probability that during any given minute, the switchboard will
receive between 4 and 6 calls inclusive.
ii)The probability that the operator will have a breathing spell of at least
30 seconds between successive calls.
iii)The mean time lag between two calls.

Sagot :

P(n) = (e^-λ . λ^n)/n! where λ = 6.5
P(4) = (e^-6.5 x 6.5^4) / 4! = 0.111822…
P(5) = (e^-6.5 x 6.5^5) / 5! = 0.145368…
P(6) = (e^-6.5 x 6.5^6) / 6! = 0.157829…
P(between 4 and 6 inclusive) = 0.111822… + 0.145368… + 0.157829… = 0.4150 (4sf)

In a 30 second interval λ = 6.5 / 2 = 3.25
P(0) = (e^-3.25 x 3.25^0) / 0! = 0.03877
The probability of getting no calls in a 30 second interval is 0.03877 (ie less than 4%)

If the average number of calls in a minute is 6.5 then the average interval between calls will be 60/6.5 = 9.23 seconds.
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