The annual rainfall (in inches) in a certain region is normally distributed with = 40 and = 4. What is the probability that starting with this year, it will take more than 10 years before a year occurs having a rainfall of more than 50 inches? What assumptions are you making?

0.93970

Step-by-step explanation:

Solution:-

- Denote a random variable "X" The annual rainfall (in inches) in a certain region . The random variable follows a normal distribution with parameters mean ( μ ) and standard deviation ( σ ) as follows:

                          X ~ Norm ( μ , σ^2 )

                          X ~ Norm ( 40 , 4^2 ).

- The probability that it rains more than 50 inches in that certain region is defined by:

                          P ( X > 50 )

- We will standardize our test value and compute the Z-score:

                          P ( Z > ( x - μ )  / σ )

Where, x : The test value

                          P (  Z > ( 50 - 40 )  / 4 )

                          P (  Z > 2.5 )

- Then use the Z-standardize tables for the following probability:

                          P ( Z < 2.5 ) = 0.0062

Therefore,          P ( X > 50 ) = 0.0062

- The probability that it rains in a certain region above 50 inches annually. is defined by:

                           q = 0.0062 ,

- The probability that it rains in a certain region rains below 50 inches annually. is defined by:

                           1 - q = 0.9938

                           n = 10 years   Sample of n years taken

- The random variable "Y" follows binomial distribution for the number of years t it takes to rain over 50 inches.

                          Y ~ Bin ( 0.9938 , 0.0062 )

- The probability that it takes t = 10 years for it to rain:

                         =  10C10* ( 0.9938 )^10 * ( 0.0062 )^0

                         = ( 0.9938 )^10

                         = 0.93970