Question

6. Exercise 2.7

The manager of the aerospace division of General Aeronautics has estimated the price it can charge for providing satellite launch services to commercial firms. Her most optimistic estimate (a price not expected to be exceeded more than 10 percent of the time) is $4.0 million. Her most pessimistic estimate (a lower price than this one is not expected more than 10 percent of the time) is $2.0 million. The expected value estimate is $3.0 million. The price distribution is believed to be approximately normal.

The standard deviation of the launch price is_______ million? Choices are: A. 0.88 B. 0.78 C. 0.98

The probability of receiving a price less than $2.4 million is______? Choices are: A. 12% B. 32% C. 22%

Answer 1

Solution:

1. For such normal standard, we know that:

P(x < given X) = P(z = Z) where Z = (X - X(bar))/standard deviation, where x is the price of launch

Then, as we are given that probability of price to be less than $2 is not expected to be more than 10%.

So, P(x < 2) <= 0.1

Using standard normal tables, we can find that 0.1 probability occurs at approximate Z value of -1.28, so

With Z = (X - X(bar))/sd

-1.28 = (2 - 3)/sd

sd = -1/-1.28 = 0.78125 or simply 0.78 approximately

Thus, correct option is (B) 0.78 (please note that similar calculation could have been performed other way around for X = 4: since probability of price exceed 4 is at most 0.1, it means that probability of price not exceeding 4 is at least 1 - 0.1 = 0.9, z value for probability of 0.9 is approximately +1.28, and moving further in similar way as shown).

2. Probability of receiving price less than $2.4m:

P(x < 2.4) = P(z = Z) where

Z = (2.4 - 3)/0.78 = -0.77 (approx)

Using standard normal tables, we know that P(z = -0.77) = 0.2206

So, the required probability is 22% approx, and thus, correct option is (C) 22%.