1. 
(a) f(r) = k * (r-10)*(20-4)  10 < r < 20

For f(r) to be a PDF
Integrate(from 10 to 20) f(r) dr = 1   

Solve for k, gives, k = 3/500
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(b) Median value xm is the value such that the area under the 
PDF until xm is equal to 0.5 (the definition of Median)

Integrate(from 10 to xm)f(r) = 0.5

solve for xm, gives xm = 15
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(c) Mean number of rainstorms in a year for the city is
 = sum(i = 0,1,2,3)x_i * P(x_i)
x_i = 0,1,2,3
P(x_i) = 0.3, 0.4, 0.2, 0.1
Th Mean number of rainstorms = 1.1
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(d) The maximum runoff rate is given as Log Normal with a median of 7cfs
and coefficient of variation Cv = 0.15
Equation 4-22 (Log Normal)
omega = 0.15, theta = ln(7) = 1.946
f(flow) ~ Log Normal (omega = 0.15, theta=1.946)
Flooding occurrs when the flow rate exceeds 8cfs.
Probability of flooding during a storm 
 = 1 - Integrate(0 to 8)f(flow)
1 - qlnorm(8, 1,946, 0.15)  = 0.187
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(e) Rate of typeI tornadoes = 9/18 = 0.5 / year
Rate of typeII tornadoes  = 54/18 = 3/year
N1 = # of type I tornadoes
N2 = # of type II tornadoes
P(two tornadoes in a year) = P(N1=1 and N2=1) + P(N1=2 and N2=0) + P(N1=0 and N2=2)

N1 and N2 are independent so the joint probabilities are equal to their products

P(a typeI tornadoe) = P(N1=1) = exp(-0.5) * (0.5)^1 / 1!
P(two typeI tornadoes) = P(N1=2) = exp(-0.5) * (0.5)^2 / 2!

and so on..
Doing the algebra you get,

P(two tornadoes in a year) = 0.185
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(f)
Probability of ATLEAST 1 tornado of typeI in the next 5 years
= 1- P(NONE of the typeI tornadoes in the 5 years)

= 1 - (1 - P(a typeI tornado in a year))^5

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