Question

Let the Poisson random variable U (see p. 227) be the number of calls for technical assistance received by a computer company during the firm’s nine normal work- day hours. Suppose the average number of calls per hour is 7.0 and that each call costs the company $50. Let V be a Poisson random variable representing the number of calls for technical assistance received during a day’s remaining 3.10 Order Statistics 193 fifteen hours. Suppose the average number of calls per hour is 4.0 for that time period and that each such call costs the company $60. Find the expected cost and the vari- ance of the cost associated with the calls received during a twenty-four-hour day.

Answer 1

Answer:

Expected cost = $6,750

Variance of the cost = $373,500

Step-by-step explanation:

During normal 9 work hours: average number of calls = 7.0

Cost of each call = $50

During 15 off hours:

average number of calls = 4.0

Cost of each call = $60

Let's take U as the number of calls during the normal 9 hours.

I.e, $U = U_1+U_2+U_3+U_4....+ U_9$

Therefore,

$E(U) = E(U_1)+E(U_2)+E(U_3)+E(U_4)....+E(U_9)$

= 7+7+7+7+7+7+7+7+7

= 63

In Poisson random variable, Variance= mean, thus:

$Var(U) =Var(U_1)+Var(U_2).....+Var(U_9)$

= 7+7+7+7+7+7+7+7+7

=63

Let's take V as the number of calls during the day's remaining 15 hours.

E(V) = Var(V)

= 15(4)

=60

The expected cost and the variance of cost associated with the calls received during a 24 hour day:

The expected cost =

$50U + $60V

= $50(63) + $60(60)

= $3150 + $3600

= $6750

The variance of the cost :

= Var(50U + 60V)

= 50²Var(U) + 60²Var(V)

= 2500*63 + 3600*60

= $373,500

Therefore, the expected cost is $6,750 and the variance of the cost is $373,500