Statistics - Calculating probability

Here's the question:
Before negotiating a long-term construction contract, a contractor must carefully estimate the total cost of completing the project. For one particular contract, assume that total cost, X is a normally distributed variable with mean $750,000 & standard deviation $150,000. The revenue (pay) guaranteed to the contractor, R is $925,000.
(a) A contract is profitable if revenue exceeds total cost. What is the probability that this contract is profitable?
(b) What is the probability the project will result in a loss?
(c) Suppose the contractor gets the chance to renegotiate the contract. What value of R (pay) should they ask for (given the distribution of total cost) in order to have a 90% chance of being profitable?

I'm thinking of using z-scores when looking at this but I'm not sure. Any help is appreciated! Thanks.

Torrent -

Yes, you will need to find z-scores. Your goal for items a and b is find a z-score that will solve this equation:

Revenue = Costs

When this is met, the company "breaks even" and neither loses money or gains. So, what standard deviation do we need in order for this equation to work?

Revenue = Costs

$925,000 = $750,000 + 150,000(z) , now solve for z:

z = (925000-750000)/150000 = 1.167, now simply look up this z-value in a Standard Normal table:

(a) P(z > 1.167) = 0.1216

(b) P(z < 1.167) = 1 - 0.1216 = 0.8784

(c) If you want 90% chance of being profitable use the z-value that results in 90%

z-value = 1.282, now plug this back into our formula:

$Revenue = $750,000 + 150,000(z) = $750,000 + 150,000(1.282) = $942,300

Hope that helped

yes - use a z-score approach.
Compute Z=(925,000 - 750,000)/150,000=1.1667
profitable if cost < 925,000; compute Pr(Z < 1.1667) = .88 approx. from normal tables

b) prob of loss =1-.88=.12
c) want (X-750,000)/150,000 = 1.28 (cuts off 90% of the distribution)
X=$942,000