Question

In western Kansas, the summer density of hailstorms is estimated at about 2.6 storms per 5 square miles. In most cases, a hailstorm damages only a relatively small area in a square mile. A crop insurance company has insured a tract of 8 square miles of Kansas wheat land against hail damage. Let r be a random variable that represents the number of hailstorms this summer in the 8-square-mile tract.
(a) Explain why a Poisson probability distribution is appropriate for r.

Hail storms in western Kansas are a common occurrence. It is reasonable to assume the events are dependent.Hail storms in western Kansas are a rare occurrence. It is reasonable to assume the events are dependent. Hail storms in western Kansas are a common occurrence. It is reasonable to assume the events are independent.Hail storms in western Kansas are a rare occurrence. It is reasonable to assume the events are independent.



What is λ for the 8-square-mile tract of land? Round λ to the nearest tenth so that you can use Table 4 of Appendix II for Poisson probabilities.


(b) If there already have been two hailstorms this summer, what is the probability that there will be a total of four or more hailstorms in this tract of land? Compute P(r≥ 4 | r ≥ 2). (Use 4 decimal places.)


(c) If there already have been three hailstorms this summer, what is the probability that there will be a total of fewer than six hailstorms? Compute P(r < 6 | r ≥ 3). (Use 4 decimal places.)

Answer 1

Answer:

Step-by-step explanation:

mean for 5 square miles is 2.6 storms

mean for 9 square miles is $\frac{2.6}{5}\times 8=4.16=4.2\\\\P(r\geq4|r\geq 2)=\frac{P(r\geq 4nr\geq 2 }{P(r\geq 2)}\\\\=\frac{P(r\geq 4)}{P(r\geq 2)}=0.6557\\\\\\P(r\geq 6|r\geq 3)=\frac{P(r\geq 6nr\geq 3 }{P(r\geq 3)}\\\\=\frac{P(3)+P(4)+P(5)}{1-P(r\geq 2)}\\\\=\frac{0.542905}{0.789762}=0.6874$