The advertising manager for Roadside Restaurants, Inc., needs to decide whether to spend this month's budget for advertising on print media, television, or a mixture of the two. She estimates that the cost per thousand "hits" (readers or viewers) will vary depending upon the success of the new cable television network she plans to use, as follows:
If she feels that there is a 60 percent chance that the new cable network will be successful, what is her expected cost (per thousand "hits") under certainty?
A) $3.40
B) $4.60
C) $8.00
D) $9.00
E) $10.00

Question

The advertising manager for Roadside Restaurants, Inc., needs to decide whether to spend this month's budget for advertising on print media, television, or a mixture of the two. She estimates that the cost per thousand "hits" (readers or viewers) will vary depending upon the success of the new cable television network she plans to use, as follows: 
If she feels that there is a 60 percent chance that the new cable network will be successful, what is her expected cost (per thousand "hits") under certainty?
A) $3.40 
B) $4.60 
C) $8.00 
D) $9.00 
E) $10.00

Quizplus · Accepted Answer

To find the expected cost per thousand "hits" under certainty, we need to calculate the weighted average of the cost per thousand "hits" based on the probability of the cable network being successful (0.6) and not successful (0.4).

For print media, the cost per thousand "hits" is $3.00.

For television, if the cable network is successful, the cost per thousand "hits" is $3.50. If it's not successful, the cost per thousand "hits" is $6.00.

So, the expected cost per thousand "hits" is:

[(0.6 x $3.50) + (0.4 x $6.00)] = $4.60

Therefore, the best choice is to advertise on television with the new cable network, as it has the lowest expected cost per thousand "hits".

The Advertising Manager for Roadside Restaurants, Inc