Answer:
a. Alpha equals 3.014 while beta equals 12.442
b. The likelihood that the data transfer duration surpasses 50ms is 0.238
c. The chance that data transfer time falls between 50 and 75 ms is 0.176
Step-by-step explanation:
a. Given the data, the mean and standard deviation for the random variable X are 37.5 ms and 21.6, respectively.
Thus, E(X)=37.5 and V(X)=(21.6)∧2
To find alpha, we need to apply the formula:
alpha=E(X)∧2/V(X)
alpha=(37.5)∧2/21.6∧2
alpha=1,406.25
/466.56
alpha=3.014
To determine beta, the following formula is employed:
β= V(X)
∧2/E(X)
β=(21.6) ∧2/37.5
β=466.56
/37.5
β=12.442
b. With E(X)=37.5 and V(X)=(21.6)∧2,
Hence, P(X>50)=1−P(X≤50)
To find the probability of data transfer time exceeding 50ms, we use the formula:
P(X>50)=1−P(X≤50)
=1−0.762
=0.238
The chance of data transfer time exceeding 50ms is 0.238
c. With E(X)=37.5 and V(X)=(21.6)∧2,
Thus, P(50<X<75)=P(X<75)−P(X<50)
To find the probability that data transfer time is between 50 and 75 ms, we apply the formula:
P(50<X<75)=P(X<75)−P(X<50)
=0.938−0.762
=0.176
The probability that data transfer time falls between 50 and 75 ms is 0.176
