The annual rainfall (in inches) in a certain region is normally distributed with mean 43.2 and variance 20.8. Assume rainfall in different years is independent. Find the probability that out of 15 years, at most 2 have rainfall of more than 50 inches.

92.24% probability that out of 15 years, at most 2 have rainfall of more than 50 inches.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the binomial probability distribution.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation(which is the square root of the variance) , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Binomial probability distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

In this problem, we have that:

Probability that a year has rainfall of more than 50 inches.

pvalue of Z when X = 50. So

has a pvalue of 0.9319

1 - 0.9319 = 0.0681

Find the probability that out of 15 years, at most 2 have rainfall of more than 50 inches.

This is when . So

92.24% probability that out of 15 years, at most 2 have rainfall of more than 50 inches.

+25 = x

step-by-step explanation:

here x is the hypotenuse of a right triangle whose legs are 7 and 24.   by the pythagorean theorem, 7² + 24² must equal x²:

x² = 49 + 576 = 625.   thus, x = +√625 = +25 = x

the answer will be