Question

1. The annual rainfall is normally distributed with mean = 40.14 inches and standard deviation 8.7 inches.

a) What is the probability this years’ rainfall will exceed 42 inches?

b) What is the probability that the sum of the next two years’ rainfall will exceed 84 inches?

Answer #1

Here we have:

a)μ = 40.14 and σ = 8.7

P(rainfall exceed 42 inches )= P(X>42)

= P(z >
) = P(z > 0.21)
=1-P(z<=.21)=1-0.5832=* 0.4168(*from Z
table)

(b)

For Sum of two years mean and SD (2*40.14 ,
sqrt[2*(8.7)^{2}]) = (80.28 ,12.3)

here μ = 80.28 and σ = 12.3

P(sum of the next 2 years’ rainfall will exceed 84 inches )= P[z
>(84-80.28)/12.3] = P(z > 0.30) =1-P(z<=0.30)=1-
0.6179=**0.3821**

**Note that in both the cases z score is considered to
only 2 decimal place and corresponding probability to four decimal
place.**

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