Answer:
a. 0.5413
b. 20
c. 0.3724
d. 4.4721
Step-by-step explanation:
Solution:-
- We will start by defining a random variable X.
X : The number of support requests arrived
- The event defined by the random variable ( X ) is assumed to follow Poisson distribution. This means the number of request in two distinct time intervals are independent from one another. Also the probability of success is linear within a time interval.
- The time interval is basically the time required for a poisson event to occur. Consequently, each distributions is defined by its parameter(s).
- Poisson distribution is defined by " Rate at which the event occurs " - ( λ ). So in our case the rate at which a support request arrives in a defined time interval. We define our distributions as follows:
X ~ Po ( λ )
Where, λ = 1 / 30 mins
Hence,
X ~ Po ( 1/30 )
a)
- We see that the time interval for events has been expanded from 30 minutes to 1 hour. However, the rate ( λ ) is given per 30 mins. In such cases we utilize the second property of Poisson distribution i.e the probability of occurrence is proportional within a time interval. Then we scale the given rate to a larger time interval as follows:
λ* = 
- We redefine our distribution as follows:
X ~ Po ( 2/1 hr )
- Next we utilize the probability density function for poisson process and accumulate the probability for 2 to 4 request in an hour.

- The required probability is:
Answer
b)
We will repeat the process we did in the previous part and scale the poisson parameter ( λ ) to a 10 hour work interval as follows:
λ* = 
- The expected value of the poisson distribution is given as:
E ( X ) = λ
Hence,
E ( X ) = 20 (10 hour work day) .... Answer
c)
- We redefine our distribution as follows:
X ~ Po ( 20/10 hr )
- Next we utilize the probability density function for poisson process and accumulate the probability for 20 to 24 request in an 10 hour work day.

- The required probability is:
Answer
c)
The standard deviation of the poisson process is determined from the application of Poisson Limit theorem. I.e Normal approximation of Poisson distribution. The results are:
σ = √λ
σ = √20
σ = 4.4721 ... Answer