Answer:
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
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![((7x^3)(y^2))^ \frac{2}{7} = \sqrt[7]{((7x^3)(y^2))^2}](https://tex.z-dn.net/?f=%20%28%287x%5E3%29%28y%5E2%29%29%5E%20%5Cfrac%7B2%7D%7B7%7D%20%3D%20%5Csqrt%5B7%5D%7B%28%287x%5E3%29%28y%5E2%29%29%5E2%7D%20)
=![\sqrt[7]{((7x^3)^2(y^2)^2)}](https://tex.z-dn.net/?f=%20%5Csqrt%5B7%5D%7B%28%287x%5E3%29%5E2%28y%5E2%29%5E2%29%7D%20)
=![\sqrt[7]{((7^2x^6)(y^4))}](https://tex.z-dn.net/?f=%20%5Csqrt%5B7%5D%7B%28%287%5E2x%5E6%29%28y%5E4%29%29%7D%20)
=![\sqrt[7]{((49x^6)(y^4))}](https://tex.z-dn.net/?f=%20%5Csqrt%5B7%5D%7B%28%2849x%5E6%29%28y%5E4%29%29%7D%20)
=![\sqrt[7]{49x^6y^4}](https://tex.z-dn.net/?f=%20%5Csqrt%5B7%5D%7B49x%5E6y%5E4%7D%20)
