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Question: A telephone switchboard receiving number of phone calls follows Poisson distribution with parameter ...

A telephone switchboard receiving number of phone calls follows Poisson distribution with parameter 33per hour. The probability that it receives 55 calls in the 22 hours{\text{hours}} duration is:
A. e−3355!\dfrac{{{e^{ - 3}}{3^5}}}{{5!}}
B. 1−e−3355!1 - \dfrac{{{e^{ - 3}}{3^5}}}{{5!}}
C. e−6355!\dfrac{{{e^{ - 6}}{3^5}}}{{5!}}
D. 1−e−6355!1 - \dfrac{{{e^{ - 6}}{3^5}}}{{5!}}

Explanation

Solution

Here we must be aware of what the Poisson distribution is used to calculate. By the use of it we are able to calculate the probability of the particular number of the successes in the time interval when we know the mean rate of the success.
By the Poisson distribution it is given as:
P(x;μ)=μxe−μx!P(x;\mu ) = \dfrac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}
Here μ\mu is the mean number of the successes that occur in the given interval or parameter.
And xx is the actual number of successes that occur in the specified region.
And P(x;μ)=μxe−μx!P(x;\mu ) = \dfrac{{{\mu ^x}{e^{ - \mu }}}}{{x!}} is the probability that xx successes occur in the Poisson experiment when the mean number of successes is μ\mu .

Complete step by step solution:
Here we are given that a telephone switchboard receiving a number of phone calls follows Poisson distribution with parameter 33per hour.
This statement tells us that the average number of the calls in an hour is 33
And we are given to find the probability that it receives 55 calls in the 22 hours{\text{hours}} duration.
This statement tells us that we need to find the rate of the success in the 22hours.
So as we know that the he receives 33 calls in one hour and according to this we come to know that the average in the 22hours is
μ=3(2)=6hours\mu = 3(2) = 6{\text{hours}}
Number of successes that are needed is 55 as we are told to find the probability of receiving 55 calls in two hours.
So we get x=5x = 5
Now applying the formula we get
P(x;μ)=μxe−μx!P(x;\mu ) = \dfrac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}
Required probabilityP(x;μ)=μxe−μx!=35e−65!P(x;\mu ) = \dfrac{{{\mu ^x}{e^{ - \mu }}}}{{x!}} = \dfrac{{{3^5}{e^{ - 6}}}}{{5!}}

Hence C is the correct result.

Note:
Here the student must be aware of the concept of the Poisson distribution and how we can find the probability using it. One must be aware of the formula and the meaning of the symbols or the variables that are used in the formula.
P(x;μ)=μxe−μx!P(x;\mu ) = \dfrac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}
Here μ\mu is the mean number of the successes that occur in the given interval or parameter.
And xx is the actual number of successes that occur in the specified region.
And P(x;μ)=μxe−μx!P(x;\mu ) = \dfrac{{{\mu ^x}{e^{ - \mu }}}}{{x!}} is the probability that xx successes occur in the Poisson experiment when the mean number of successes is μ\mu .