a) The probability that the homeowner will get less than $260,000 if he leaves the house on the market for another month is equal to the area under the probability density function (PDF) of the uniform distribution from $245,000$ to $260,000$. Since the distribution is uniform, the PDF is constant over the interval of interest, and its value is $\frac{1}{270000-245000}=\frac{1}{25000}$. Therefore, the probability is:
�
(
selling price
<
260
,
000
)
=
260
,
000
−
245
,
000
25
,
000
=
0.6
P(selling price<260,000)=
25,000
260,000−245,000
=0.6
b) Similarly, the probability that the homeowner will get more than $260,000$ if he leaves the house on the market for another month is equal to the area under the PDF of the uniform distribution from $260,000$ to $270,000$. Therefore, the probability is:
�
(
selling price
>
260
,
000
)
=
270
,
000
−
260
,
000
25
,
000
=
0.4
P(selling price>260,000)=
25,000
270,000−260,000
=0.4
c) The probabilities calculated in parts a) and b) provide a way to assess the risk and potential benefit of leaving the house on the market for another month. If the homeowner is risk-averse and prefers a certain outcome, then he should take the $260,000 offer, since the probability of getting less than $260,000 is higher than the probability of getting more. On the other hand, if the homeowner is willing to take a risk for the potential benefit of a higher selling price, then he should leave the house on the market for another month. Ultimately, the decision will depend on the homeowner's risk preferences and other personal circumstances.