1.
a) Assuming that the competitor’s bid x is a random variable that is uniformly distributed between $10,000 and $15,000. than the probability of winning of the bid with $12,000 is equivalent to Competitors bids less then $12,000 P(X<$12,000) = ($12,000­$10,000)/($15,000­$10,000)
=$2,000/$5,000
=0.4 b) P(X<$14,000) = ($14,000­$10,000)/$15,000­$10,000)
=4/5
=0.8 c) $15,000 bids guarantees the property since P(X<=15,000)=1 d) by bidding $15,000 you have a 100% probability of getting the property and are therefore guaranteed $1000 when you sell it for $16,000. 2.
a) z = (x ­ µ) /σ z = (500 ­ 328) /92 = 1.869565 from z­score table
P( x ≤ 500) = 0.969228
P(x > 500) = 1 ­ P( x ≤ 500)
P(x > 500) = 0.0308 b) z = (x ­ µ) /σ z = (250 ­ 328) /92 = ­0.8478261 z = (500 ­ 328) /92 = 1.869565 from z­score table
P( x < 250) = 0.1983 c) z = (300 ­ 328) /92 = ­0.30435
P( x ≤ 300) = 0.380431 z = (400 ­ 328) /92 = 0.782609
P( x ≤ 400) = 0.783072
P( 300 < x < 400 )
= 0.783072 ­ 0.380431
= 0.40 d) 8% = z of 0.84
0.84 = (x – 328)/0.6
77.28 = x – 328

405.28 = x
Therefore the cost is $405.28.

3.
a) dev=0.6
P(X<18)=0.02
Standarize with z=(x­mean)/dev
X=18 ­­> Z=(18­mean)/0.6
P(X<18) = P(Z<(18­mean)/0.6))
We made z=(18­mean)/0.6
We must find the value z that P(Z<z) = 0.02
Table of Normal Standard distribution the value 0.02 we find that z= ­2.05 z=(18­mean)/0.6) ­2.05 = (18­mean)/0.6
18­mean =