1. A rock concert producer has scheduled an outdoor concert for Saturday, November 6, 1999. If it does not rain, the producer expects to make $20,000 profit from the concert. If it does rain, the producer will be forced to cancel the concert and will lose $12,000 (rock star’s fee, advertising cost, stadium rental, administrative costs, etc.). The producer has learned from the National Weather Service that the probability of rain on November 6 is 0.4.

  1. Find the probability distribution of x=profit of the producer.
  2. Find the producers expected profit from the concert.
  3. For a fee of $1,000 an insurance company has offered to insure the producer against all losses resulting from a rained-out concert. If the producer buys the insurance, what is her expected profit from the concert?
  4. Assuming the forecast is accurate, do you believe the insurance company has charged too much or too little for the policy? Explain and justify your answer.

a.

x

Rain?

p(x)

-12000

Yes

0.4

20000

No

0.6

b. E(profit)=(-12000)0.4+20000(0.6)=7200

 

 

 

 

 

c.

x

Rain?

p(x)

-1000

Yes

0.4

19000

No

0.6

E(profit)=(-1000)0.4+19000(0.6)=11000

d.

y

Rain?

p(y)

-11000

Yes

0.4

1000

No

0.6

E(profit for insurance com.)=

(-11000)0.4+1000(0.6)=-3800 (too little)

2. Pinworm infestation, commonly found in children, can be treated with the drug pyrantel pamoate. According to the Merck Manual, the treatment is effective in 90% of cases. Fifteen children with pinwo rm infestation are given pyrantel pamoate.

  1. What is the probability that less than 12 of them will be cured?
  2. What is the probability that exactly 13 of them will be cured?
  3. (Use Normal approximation to Binomial) What is the probability that more than 350 children out of 400 with pinworm infestation will be cured?

X has a Binomial distribution with n=15 p=0.9

  1. P(X<12)=P(X¾11)=0.056 (from table II)
  2. P(X=13)=P(X¾13)-P(X¾12)=0.451-0.184=0.267
  3. m=400(0.9)=360, s2=400(0.9)(0.1)=36 P(X>350)=P(X„350.5)=P(Z„(350.5-360)/6)=P(Z„-1.58)=0.5+0.4429=0.9429

3. A paper by L. F. Richardson, published in the Journal of the Royal Statistical Society, analyzed the distribution of wars in time. From the data we find that the number of wars that begin during a g iven calendar year has a Poisson distribution with mean 0.7 wars per year.

If a calendar year is selected at random, find the probability that the number of wars that begin during that calendar year will be

  1. zero.
  2. at most 2.

c. Is it likely that the number of wars in a calendar year will exceed two? Explain

  1. P(X=0)=0.497 (from Table III) 0r P(X=0)=e-0.7=0.497
  2. P(X¾2)=0.966
  3. P(X>2)=1-P(X¾2)=1-0.966=0.034, no it is not likely.

4. The length of human pregnancies from conception to birth varies according to a distribution that is normal with mean 266 days and standard deviation 16 days.

  1. What is the probability that a pregnancy will last more than 275 days?
  2. How short are the shortest 2.5% of all pregnancies?
  1. P(X>275)=p(Z>(275-266)/16)=P(Z>0.5625)=0.5-0.2123=0.2877
  2. z=-1.96 x=266-1.96(16)=266-31.36=234.66, less than 234.66 days

5. Some biology students were interested in analyzing the amount of time that bees spend in flower patches gathering nectar (X). They found that the amount of time has an exponential distribution with mean of 230 seconds.

  1. Find the probability that for a randomly selected bee will spend less than 50 seconds in a flower patch gathering nectar.
  2. Suppose that you observe two bees independently. What is the probability that both of them will spend more than 230 seconds in flower patches gathering nectar?
  1. P(X<50)=1-e-50/230=1-e-0.2174=1-0.8046=0.1954
  2. P(X>230)= e2300/230= e-1=0.3679, Probability that both will exceed 230=0.36792=0.13535

6. The U.S. Energy Information Administration collects data on household vehicles and publishes the results in Residential Transportation Energy Consumption Survey, Consumption Patterns of Household Vehicl es. According to this document, the mean monthly fuel expenditure per household vehicle is $58.80. The standard deviation is $30.40.

  1. Describe the sampling distribution of the average fuel expenditure for 100 household, including its mean and standard deviation.
  2. What is the probability that the average fuel expenditure for 100 families will be greater than $60.00?
  3. What is the probability that the average fuel expenditure for 100 families will be between $53 and $65?
  1. By Central Limit Theorem, average fuel expenditure for 100 households will have a Normal distribution with mean $58.80 and standard deviation 30.40/10=3.04.