the average amount of rain per year in Greenville is 49 inches. the standard deviation is 8 inches. find the probability that nest year greeville will receive the following amount of rainfall. assume the variable is normally distributed.

a) At most 55 inches of rain
b) at least 62 inches of rain
c) Between 46 and 55 inches of rain
d) How many inches of rain would you consider to be an extremely wet year?

a) At most 55 inches of rain: 0.8413

b) At least 62 inches of rain: 0.1587
c) Between 46 and 55 inches of rain: 0.6826
d) An extremely wet year would be considered to be any year with more than 65 inches of rain.

To solve these probability problems, we can use the Z-score formula and the standard normal distribution table. The Z-score formula is:

Z = (X - μ) / σ

Where:
X = the given value
μ = the mean (average) rainfall amount
σ = the standard deviation of rainfall amount

Using the given information:
μ = 49 inches
σ = 8 inches

a) To find the probability of at most 55 inches of rain:
We need to find the Z-score for X = 55.

Z = (X - μ) / σ
Z = (55 - 49) / 8
Z = 0.75

Looking up the Z-score of 0.75 in the standard normal distribution table, we find the corresponding probability is 0.7734.

Therefore, the probability of at most 55 inches of rain is 0.7734, or 77.34%.

b) To find the probability of at least 62 inches of rain:
We need to find the Z-score for X = 62.

Z = (X - μ) / σ
Z = (62 - 49) / 8
Z = 1.625

Looking up the Z-score of 1.625 in the standard normal distribution table, we find the corresponding probability is 0.9474.

Since we want the probability of at least 62 inches, we subtract this probability from 1:

1 - 0.9474 = 0.0526

Therefore, the probability of at least 62 inches of rain is 0.0526, or 5.26%.

c) To find the probability of between 46 and 55 inches of rain:
We need to find the Z-scores for X = 46 and X = 55.

Z1 = (X1 - μ) / σ = (46 - 49) / 8 = -0.375
Z2 = (X2 - μ) / σ = (55 - 49) / 8 = 0.75

Using the standard normal distribution table, we find the corresponding probabilities:
P(Z < -0.375) = 0.3531
P(Z < 0.75) = 0.7734

To find the probability between the two values, we subtract the smaller probability from the larger probability:

0.7734 - 0.3531 = 0.4203

Therefore, the probability of between 46 and 55 inches of rain is 0.4203, or 42.03%.

d) To determine what would be considered an extremely wet year, we need to establish a Z-score cutoff. Extreme values are often defined as a Z-score beyond 2 or 3 standard deviations from the mean.

In this case, we'll use a Z-score cutoff of 3.

Z = (X - μ) / σ
3 = (X - 49) / 8

Solving for X:
24 = X - 49
X = 24 + 49
X = 73

Therefore, we would consider a year with more than 73 inches of rain to be an extremely wet year in Greenville.

To answer each part of the question, we will use the standard normal distribution. Since the variable is assumed to be normally distributed, we can convert the rainfall values to z-scores and use the z-table to find the corresponding probabilities.

Step 1: Calculate the z-score for each rainfall value using the formula:
z = (x - mean) / standard deviation

a) To find the probability of at most 55 inches of rain:
First, calculate the z-score:
z = (55 - 49) / 8 = 0.75

Using the z-table or a calculator, find the probability corresponding to a z-score of 0.75. This represents the probability of obtaining a value less than or equal to 55 inches of rain.

b) To find the probability of at least 62 inches of rain:
First, calculate the z-score:
z = (62 - 49) / 8 = 1.625

Using the z-table or a calculator, find the probability corresponding to a z-score of 1.625. This represents the probability of obtaining a value greater than or equal to 62 inches of rain.

c) To find the probability of rainfall between 46 and 55 inches:
First, calculate the z-scores for each end of the range:
z1 = (46 - 49) / 8 = -0.375
z2 = (55 - 49) / 8 = 0.75

Using the z-table or a calculator, find the probability corresponding to a z-score of -0.375 and 0.75. Subtract the probability corresponding to z1 from the probability corresponding to z2 to find the probability of rainfall between 46 and 55 inches.

d) To determine what amount of rainfall would be considered extremely wet, you need to define a threshold. Generally, an extremely wet year can be defined as rainfall exceeding a certain number of standard deviations above the mean.

For example, if we define extremely wet as rainfall exceeding 2 standard deviations above the mean, we can calculate the amount of rainfall by adding 2 standard deviations to the mean:
extremely wet rainfall = mean + (2 * standard deviation)

extremely wet rainfall = 49 + (2 * 8) = 65 inches

So, any year with rainfall exceeding 65 inches would be considered extremely wet based on this definition.

Remember, the exact cutoff for an extremely wet year can vary depending on the chosen threshold.