Precipitation Probability - HELP

darthsmozers
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I am cursed with a horrible textbook.
In regards to a normal distribution:

when given a mean and standard deviation, I can compute a probability of precipitation in inches with no problem.

IE: when asked what is the probability that rainfall in a given area will exceed a given # of inches:
I calculate standard score from: (Xi - mean)/standard deviation.
I then use this value and a z score table. I then evaluate the probability value and come up with an answer like "in ___ years out of 100, the annual precipitation in the given region should exceed given amount of inches."


HELP HERE: I am asked "if conditions continue, the region should receive at least ___ inches in 19 out of 20 years". Am i supposed to work the problem backwards?

in short: when only given the # of years out of so many years, and given a mean and standard deviation only, how do i find the amount of precipitation expected?

it sounds like the reverse of the above problem, where you're given the precip. level, the mean, and the SD, and asked to find the # of years out of 100.
 
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How would you solve the problem "the region should receive at least ___ inches in a year"?

Now I think you need either the rain amount or a probability so you can solve for the other.
 
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the question reads: "Considering current conditions, and 30 years of sample observations, the region should receive ___ inches of rainfall in 14 out of 25 years. Given values include mean of 39.95 inches and SD of 7.5 inches." no other data is given.
 
nevermind, got it, thanks anyways.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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