Answer:
0.93970
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