X more than 5 (a) x > 5 (b) 5x (c) 5 + x (d) x - 5
1 answer:
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Victoria's method is linear because the number of minutes increase by an equal number (15) every month.
Workings in the attachments below. The green line is the function that has been set up for Victoria. The lines that form a curved looking graph belong to the function that was set up for Zach.
Workings in the attachments below. The green line is the function that has been set up for Victoria. The lines that form a curved looking graph belong to the function that was set up for Zach.
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.