Question

In western Kansas, the summer density of hailstorms is estimated at about 1.5 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 9 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 9-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 9-square-mile tract of land? Round λ to the nearest tenth (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

Explanation:

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

(b) The required
`\lambda` = 1.7 * 10/5 = 3.4

(c) P(r ≥ 4 | r ≥ 2)

= P(r ≥ 4)/P(r ≥ 2)

= {1 - P(r < 4)}/{1 - P(r < 2)}

For Poisson distribution, P(X = x)

`= (e^(- \lambda) \lambda ^x)/(x !)`

Thus, P(r < 4) = P(r = 0) + P(r = 1) + P(r = 2) + P(r = 3)

= 0.5584

and P(r < 2) = P(r = 0) + P(r = 1) = 0.1468

Thus, P(r ≥ 4 | r ≥ 2) = (1 - 0.5584)/(1 - 0.1468)

= 0.5177

(d) P(r < 6 | r ≥ 3)

= P(3 ≤ r < 6)/P(r ≥ 3)

P(3 ≤ r < 6) = P(r = 3) + P(r = 4) + P(r = 5) = 0.5308

P(r ≥ 3) = 1 - {P(r = 0) + P(r = 1) + P(r = 2)}

= 0.6603

Thus, P(r < 6 | r ≥ 3) = 0.5308/0.6603 = 0.8309