Question

The number of requests for assistance received by a towing service is a Poisson process with...

The number of requests for assistance received by a towing service is a Poisson process with rate θ = 4 per hour.

a. Compute the probability that exactly ten requests are received during a particular 2-hour period.

b. If the operators of the towing service take a 30-min break for lunch, what is the probability that they do not miss any calls for assistance?

c. How many calls would you expect during their break?

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Concepts and reason

The concept used to solve this problem is the concept of the using Poisson distribution for the events which are very less likely to happen.

The Poisson distribution is a discrete probability distribution which can be used to determine the probability of a given number of events occurring in a fixed interval of time. It can be used to model situations where the chances of the happening of events is very low.

Fundamentals

Poisson distribution has following conditions,

The first is9{
U
, that is, the number of trials is indefinite.

The second is06d
, which represents that the probability of success is small for each trial.

The third isnp = 2
, which is finite, where is a positive real number.

Consider a random variable having mean rate of occurrences following a Poisson distribution. The probability distribution for the variable can be written as follows,

ix
2=(x=x)

Here is the parameter of the distribution which is equal to the mean of the distribution.

The probabilities of X = x
instances of the event can be calculated by using the density function.

(a)

It is provided that the number of requests received by a towing service with per hour and. Let X be the probability that exactly 10 requests have been received during 2-hour period is calculated as,

X~P(2)
2=nxp
4x2

101
„(8)*, 0 =(01=x)d
1 m 2 =(x= x)

It can be calculated using Excel. The screenshot of the same is shown as,

=POISSON.DIST(10,8,FALSE)
D
E
0.09926153383

(b)

Let Y be the probability that they do not miss any calls during lunch break of half an hour. It is provided that n=4, p=0.5

Y - P(1)
a=nxp
= 4x0.5
= 2

(7)x_3 =(0=)d
vry a =((= 1)d

Using excel the probability is calculated and screenshot of the same

fx
=POISSON.DIST(0,2,FALSE)
0.13533528324!

(c)

The total number of calls they would expect during the break is calculated as,

2=nxp
= 4x0.5
= 2
Y - P(2)

Ans: Part a

The probability of 10 requests received during the 2-hour periodP(X = 10)
is 0.099.

Part b

The probability that they do not miss calls during lunch break of half an hour is0.1353
.

Part c

The total number of calls they would expect during break is 2.

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