Let X denote the data transfer time (ms) in a grid computing system (the time required for data transfer) between a "worker" computer and a "master" computer. Suppose that X has a gamma distribution with mean value 37.5 ms and standard deviation 21.6 (suggested by an article)a. what are the values of alpha and beta?b. what is the probability that data transfer time exceeds 50ms?c. what is the probability that data transfer time is between 50 and 75 ms?please include worked out solutions so that I can follow them easier to understand better! Thank you!

Respuesta :

Answer:

a. The value of alpha is 3.014 and the value of beta is 12.442

b. The probability that data transfer time exceeds 50ms is 0.238

c. The probability that data transfer time is between 50 and 75 ms is 0.176

Step-by-step explanation:

a. According to the given data we have that the mean and standard deviation of the random variable X are 37.5 ms and 21.6.

Therefore, E(X)=37.5 and V(X)=(21.6)∧2  

To calculate alpha we would have to use the following formula:

alpha=E(X)∧2/V(X)

alpha=(37.5)∧2/21.6∧2

alpha=1,406.25 /466.56

​alpha=3.014

To calculate beta we would have to use the following formula:

β=  V(X) ∧2/E(X)

β=(21.6)  ∧2/37.5

β=466.56 /37.5

β=12.442

b. E(X)=37.5 and V(X)=(21.6)∧2  

Therefore, P(X>50)=1−P(X≤50)

Hence, To calculate the probability that data transfer time exceeds 50ms we use the following formula:

P(X>50)=1−P(X≤50)

=1−0.762

=0.238

The probability that data transfer time exceeds 50ms is 0.238

c. E(X)=37.5 and V(X)=(21.6)∧2  

​Therefore, P(50<X<75)=P(X<75)−P(X<50)  

Hence, To calculate the probability that data transfer time is between 50 and 75 ms we use the following formula:

P(50<X<75)=P(X<75)−P(X<50)

=0.938−0.762

=0.176

The probability that data transfer time is between 50 and 75 ms is 0.176