Suppose that on an average two tornadoes occur in 10 years in a county in Oklaho

Question

Suppose that on an average two tornadoes occur in 10 years in a county in Oklahoma. Further assume that the tornado-generated wind speed can be modeled by a lognormal random variable with a mean of 120 mph and a standard deviation of 12 mph. (a) What is the probability that there will be at least one tornado next year? (b) If a structure in the county is designed for wind speed of 150 mph. what is the probability that the structure will be damaged during such a tornado? (c) What is the probability that the structure will be damaged by tornado next year?

Explanation / Answer

a. Probability that there will be at least one tornado next year

Let A represent the event that a tornado occurs in a year.

We have been given that 2 tornados occur every 10years, thus the probability of having a tornado in a specific year is 2/10 = 0.2

P[having at least one tornado next year] = P[A >= 1]

                                                                                = 1 - P[A < 1]

                                                                              = 1 – P[A=0]

Now note A follows Bernoulli distribution with probability of success 0.2

therefore, P[A=0] = 1 - 0.2 = 0.8

Thus, P[having at least one tornado next year] = 1 - 0.8 = 0.2

b.   Probability that the structure will be damaged.

We have been given that the structure will be damaged if the wind speed is more than 150mph

Let X be the speed of wind observed during a tornado.

X follows Log Normal distribution with mean=120 and std deviation=12 …..given

Note if X follows Log normal distribution then Log(X) follows Normal distribution with the same parameters.

P[structure will be damaged] = P[ X >150 ]

= P[ ln(X) > ln(150) ]    …. Here ln(X) follows Normal distribution

Now we use the central limit theorem,

If Y follows N(mu ,sigma) then (Y-mu)/sigma follows standard normal distribution

Thus P[structure will be damaged]           = P[ ( ln(x) – 120 )/ 12 > (ln(150) -120)/12 ]

= P[ Z > -9.582447 ]

Where , Z = ( ln(x) – 120 )/ 12    and Z follows Standard Normal distribution

P[structure will be damaged]     =             1 – P[ Z < -9.582447 ]

=            1 – 4.7386x10-22

~             1

Thus the probability that the structure will be damaged is almost 1.

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