The weekly downtime Y (in hours) for a certain instrument of a company has approximately
Posted: Tue Aug 03, 2021 11:22 am
The weekly downtime Y (in hours) for a certain instrument of a company has approximately a gamma distribution with a = 3 and ß = 2. The loss in OMR to the company as a result of downtime is given by L = 30Y + 2Y2. What is the expected value of L. (a) 276 (b) 456 (0) 385 (d) 367
Correct option:
(a) 276
Explanation:
Mean of Gamma Distribution = E(Y) = аз = 3 X 2 = 6
Variance of Gamma Distribution = \sigma ^{2} = a 32 = 3 x 22 = 12
We have:
o2 = E(Y) – (E(Y))
Substituting values, we get:
12 = E(Y?) - 64
So, we get:
E(Y? = 36 + 12 = 48
Given:
L = 30Y + 2Y
So, we get:
E(L) = (30 x E(Y)] + [2 x E(Y?)
Substituting values, we get:
E(L) = (30 x 6] + 2 x 48
= 180 + 96
= 276
Correct option:
(a) 276
Explanation:
Mean of Gamma Distribution = E(Y) = аз = 3 X 2 = 6
Variance of Gamma Distribution = \sigma ^{2} = a 32 = 3 x 22 = 12
We have:
o2 = E(Y) – (E(Y))
Substituting values, we get:
12 = E(Y?) - 64
So, we get:
E(Y? = 36 + 12 = 48
Given:
L = 30Y + 2Y
So, we get:
E(L) = (30 x E(Y)] + [2 x E(Y?)
Substituting values, we get:
E(L) = (30 x 6] + 2 x 48
= 180 + 96
= 276