Final answer:
Using the given mean and standard deviation, we calculate the Z-scores for the respective rainfall amounts and then find the corresponding percentages using the Cumulative Z-Score Table. There is a 0.47% chance of less than 26.44 inches, an 81.59% chance of more than 35.96 inches, and a 76.63% chance of rainfall between 36.24 inches and 51.08 inches annually.
Step-by-step explanation:
To answer the questions about the annual rainfall being normally distributed with a mean of 41 inches and a standard deviation of 5.6 inches, we will first convert the given raw scores of rainfall into Z-scores using the formula:
Z = (X - μ) / σ
where X is the value of interest, μ is the mean, and σ is the standard deviation.
Part a):
For an annual rainfall of less than 26.44 inches:
Z = (26.44 - 41) / 5.6 ≈ -2.60
Looking up the Z-score of -2.60 in the Cumulative Z-Score Table, we find that it corresponds to approximately 0.47%. This means that there is a 0.47% chance of having less than 26.44 inches of rainfall in a year.
Part b):
For an annual rainfall of more than 35.96 inches:
Z = (35.96 - 41) / 5.6 ≈ -0.90
The Z-score table gives the cumulative probability for Z = -0.90 as approximately 18.41%. Since we want the percentage of years with more rainfall, we subtract this from 100%, yielding 81.59%. So, there is an 81.59% chance of having more than 35.96 inches of rainfall in a year.
Part c):
For an annual rainfall of between 36.24 inches and 51.08 inches:
Z for 36.24 inches = (36.24 - 41) / 5.6 ≈ -0.85
Z for 51.08 inches = (51.08 - 41) / 5.6 ≈ 1.80
The cumulative probabilities are approximately 19.78% for Z = -0.85 and 96.41% for Z = 1.80. To find the percentage of years with rainfall in this range, we subtract the two probabilities: 96.41% - 19.78% = 76.63%. There is a 76.63% chance of annual rainfall being between these two values.