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Question 3. Suppose the annual rainfall R in a city is a Gaussian variate with a mean of 50 cm and a coefficient of variation

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Answer #1

3)

a)

According to the question, o= 10

which is the required standard deviation of R.

b)

Now, R~ N(= 50,0% = 10)

So, 30 - 50 P(R<30) = P(Z <0, ) = P(Z < -2) = P(Z > 2) 10

= 1- P(Z < 2) = 1-0.97725 = 0.02275

c)

Now, R~ N(= 50,0% = 10)

So, 60 - 50 P(R>60) = P(Z >> ) = P(Z > 1) = 1 - P(Z <1) 10

=1-0.84134 = 0.15866

d)

Now, R~ N(= 50,0% = 10)

So, P(40<R< 55) = PC 40 - 50 - <Z < 55 - 50 10 -) = P(-1<< < 0.5) ) 10

= P(Z <0.5) - P(Z <-1) = P(Z <0.5) - P(Z > 1)

= P(Z <0.5) - 1 - P(Z <1)] = 0.69146 - [1 -0.84134) = 0.53280

e)

We want to find x0 such that PR > 10) = 0.20

= P(Z > 20 – 50,- 20 - 50 -) = 0.200 - = 0.8416 = 20 = 58.416 100 10

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