Question

Making high-stakes insurance decisions. The Journal of Economic Psychology (Sept. 2008) published the results of a high-stakes experiment where subjects were asked how much they would pay for insuring a valuable painting. The painting was threatened by fire and theft, hence the need for insurance. To make the risk realistic, the subjects were informed that if it rained on exactly 24 days in July, the painting was considered to be stolen; and, if it rained on exactly 23 days in August, the painting was considered to be destroyed by fire. Although the probability of these two events, "fire" and "theft," was ambiguous for the subjects, the researchers estimated their probabilities of occurrence at .0001. Rain frequencies for the months of July and August were shown to follow a Poisson distribution with a mean of 10 days per month. a. Find the probability that it will rain on exactly 24 days in July. b. Find the probability that it will rain on exactly 23 days in August. c. Are the probabilities, parts a and b good approximations to the probabilities of "fire" and "theft"?

Answer 1

Answer:

a. P(24)=0.00007

b. P(23)=0.00018

c. There is significant difference between the probability of the rainy days and the probabilities of fire and theft.

The probability of theft would be overestimated by 76% and the probability of fire would be subestimated by 27%.

Step-by-step explanation:

The probabilities of two events ("fire"and "theft") are compared to the probabilities of a certain number of days of rain during July.

The probabilities of "fire"and "theft" are around P=0.0001, and we need to calculate if the probability of exactly 23 and exactly 24 days of rain July have approximately the same probability.

Rain frequencies for the months of July and August were shown to follow a Poisson distribution with a mean of 10 days per month.

The parameter then is:

$\lambda=10$

The probability of k days of rain is:

$P(k)=\frac{10^ke^{-10}}{k!}$

For 24 days, the probability is:

$P(24)=\frac{10^{24}e^{-10}}{24!}=\frac{1*10^{24}*4.54*10^{-5}}{6.20*10^{23}} = 0.00007$

The probability of 23 days of rain is 27% less than P=0.0001.

For 23 days of rain, the probability is:

$P(23)=\frac{10^{23}e^{-10}}{23!}=\frac{1*10^{23}*4.54*10^{-5}}{2.59*10^{23}} = 0.00018$

The probability of 23 days of rain is 76% more than P=0.0001.

There is significant difference between the probability of the rainy days and the probabilities of fire and theft.

The probability of theft would be overestimated by 76% and the probability of fire would be subestimated by 27%.