Answer:
(a) P (X = 0) = 0.0498.
(b) P (X > 5) = 0.084.
(c) P (X = 3) = 0.09.
(d) P (X ≤ 1) = 0.5578
Step-by-step explanation:
Let <em>X</em> = number of telephone calls.
The average number of calls per minute is, <em>λ</em> = 3.0.
The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 3.0.
The probability mass function of a Poisson distribution is:
(a)
Compute the probability of <em>X</em> = 0 as follows:
Thus, the probability that there will be no calls during a one-minute interval is 0.0498.
(b)
If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.
Compute the probability of <em>X</em> > 5 as follows:
P (X > 5) = 1 - P (X ≤ 5)
Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.
(c)
The average number of calls in two minutes is, 2 × 3 = 6.
Compute the value of <em>X</em> = 3 as follows:
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Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.
(d)
The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.
Compute the probability of <em>X</em> ≤ 1 as follows:
P (X ≤ 1 ) = P (X = 0) + P (X = 1)
Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.
