Support requests arrive at a software company at the rate of 1 every 30 minutes. Assume that the requests arrive as events in a Poisson process. a) What is the probability that the number of requests in an hour is between 2 and 4 inclusive? Give your answer to four decimal places.

b) What is the expected number of requests in a 10 hour work day? Give an exact answer.

c) What is the probability that the number of requests in a 10 hour work day is between 20 and 24 inclusive? Give your answer to four decimal places.

d) What is the standard deviation of the number of requests in a 10 hour work day? Give your answer to four decimal places.

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

the answer for this question is a= 13,622.84

step-by-step explanation:

school a:

minimum value is 2

maximum value is 22

the lower quartile is 2.5

the median is 10

the upper quartile is 15.5

school b:

minimum value is 9

maximum value is 20

the lower quartile is 12

the median is 16

the upper quartile is 18

box plot for school a isn't symmetrical. the data tails on the right

box plot for school b isn't symmetrical. the data tails on the left

hope i

-lavira: )