Question

. In Exercise 16, we saw the following situation: Harada Construction has just bid on a major commercial project. The bid was based on a normal distribution representing Harada’s estimates of possible costs for the project. The distribution is centered on $6.5 million and has a standard deviation of $1.1 million. According to Harada’s estimates:

a. There is a 60% chance that costs will be less than $_______ million.

b. There is a 1% chance that the cost will be greater than $____ million.

c. There is a 45% chance that costs will be between $6.5 million and $______ million. (Make $6.5 million the lower boundary for your interval.)

Answer 1

Let X be the possible costs X follows normal distribution mean μ and standard deviation σ Given H $6.5 million ơ-$1.1 million a) To find X' such that P(X-X)-0.6 Converting to z-scores P(X <X) P(Z <z') 0.6 where (60% chance) 1.1 From standard normal tables or Excel function NORM.S.INV, we find the Z' such that P(Z<) 0.6 z' NORM.S.INV(0.6) z' 0.2533 Thus X-6.5 0.2533 1.1 X' 0.2533 1.16.5 X 6.7786 There is 60% chance that cost will be less than $6.7786 million