Answer:
a. $90,000
b. 15 advertisement slots
c. 0.53
Step-by-step explanation:
Let the purchase price of the slots, c = 4,500
The salvage value of the slots, s = 2,000
The selling price of the slots, p = 8,000
a. If the store manager sells all the advertising slots in advance, the revenue received, 'R', is given as follows;
R = The number of slots, n × The purchase price for selling immediately, c
∴ R = 20 × $4,500 = $90,000
The revenue the station will receive by selling all the advertising slots in advance, R = $90,000
b. The Cost of Overage, = Purchase price - Salvage value
∴ The Cost of Overage, = 4,500 - 2,000 = 2,500
The Cost of Underage, = Selling price - Purchase price
∴ The Cost of Underage, = 8,000 - 4,500 = 3,500
P; 0.03, 0.05, 0.1, 0.15, 0.2, 0.15, 0.1, 0.05, 0.05, 0.05, 0.05, 0.02
With help of MS Excel, we get the following cumulative frequency
F(Q); 0.03, 0.08, 0.18, 0.33, 0.53, 0.68, 0.78, 0.83, 0.88, 0.93, 0.98, 1
The expected monetary value for the decision to sell the 8th slot at the last minute is given by the following relation;
× [1 - F(Q)] - × F(Q)
Where F(Q) = The cumulative frequency for the 8th stock
Therefore, we have;
3,500 × (1 - 0.03) - 2,500 × 0.03 = 3,300
For the 8th slot, the TV station is expected to make an extra $3,300
Using MS Excel, we get;
For the 9th slot, the profit over the purchase price = $3,020
Fot the 10th slot, the profit over the purchase price = $2,420
The profit over the purchase price for the 11th slot = $1,520
The profit over the purchase price for the 12th slot = $320
The profit over the purchase price for the 13th slot = $(-580)
The 14th = $(-1,180)
15th = $(-1,480)
16th = $(-1,780)
17th = $(-2,080)
18th = $(-2,380)
19th = $(-2,500)
Therefore, to maximize profit, only the 8th, 9th, 10th, 11th, and 12th, slots which are 5 slots should be sold late while 20 - 5 = 15 slots should be sold in advance
c. The probability that the total revenue will exceed the identified amount in part 'a' is the cumulative probability at the 12th slot which is given as follows;
P(D ≤ 12) = F(12) = 0.53.