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