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26 October, 13:54

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?

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  1. 26 October, 17:22
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    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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