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The average amount of rain per year in Greenville is 42 inches. The population standard deviation is 6 inches. Find the probability that next year Greenville will receive the following amounts of rainfall. Assume the variable is normally distributed.

a) At most 41 inches of rain

b) At least 49 inches of rain

c) Between 40 and 52 inches of rain

User Pradas
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Final answer:

To find the probability of receiving a certain amount of rainfall, use the z-score formula. For at most 41 inches of rain, the probability is approximately 43.32%. For at least 49 inches of rain, the probability is approximately 87.83%. For rainfall between 40 and 52 inches, the probability is approximately 58.18%.

Step-by-step explanation:

To find the probability of receiving a certain amount of rainfall, we need to convert the rainfall amounts to standard z-scores. The z-score formula is:

z = (x - μ) / σ

where x is the amount of rainfall, μ is the mean, and σ is the standard deviation. Once we have the z-scores, we can use a standard normal distribution table or a calculator to find the corresponding probabilities.

a) Probability of at most 41 inches of rain:

First, we calculate the z-score:

z = (41 - 42) / 6 = -0.1667

Using the standard normal distribution table, we find that the probability of z-score -0.1667 or less is approximately 0.4332. Therefore, the probability of receiving at most 41 inches of rain is 0.4332 or 43.32%.

b) Probability of at least 49 inches of rain:

First, we calculate the z-score:

z = (49 - 42) / 6 = 1.1667

Using the standard normal distribution table, we find that the probability of z-score 1.1667 or more is approximately 0.8783. Therefore, the probability of receiving at least 49 inches of rain is 0.8783 or 87.83%.

c) Probability of rainfall between 40 and 52 inches:

First, we calculate the z-scores:

z1 = (40 - 42) / 6 = -0.3333

z2 = (52 - 42) / 6 = 1.6667

Using the standard normal distribution table, we find that the probability of z-score -0.3333 or less is approximately 0.3707 and the probability of z-score 1.6667 or less is approximately 0.9525. Therefore, the probability of rainfall between 40 and 52 inches is the difference between these two probabilities: 0.9525 - 0.3707 = 0.5818 or 58.18%.

User ZeroSevenTen
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