Question

the annual rainfall in a certain region is approximately normally distributed with mean 42.4 inches and standard deviation 6.1 inches. round answers to the nearest tenth of a percent.
a) what percentage of years will have an annual rainfall of less than 44 inches? %
b) what percentage of years will have an annual rainfall of more than 39 inches? %
c) what percentage of years will have an annual rainfall of between 38 inches and 43 inches? %
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Explanation:

Step1: Calculate the z - score for part (a)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 42.4\), \(\sigma=6.1\), and \(x = 44\).

$$z=\frac{44 - 42.4}{6.1}=\frac{1.6}{6.1}\approx0.26$$

Using the standard normal table, \(P(Z\lt0.26)\) is the value corresponding to \(z = 0.26\) in the standard - normal distribution table. \(P(Z\lt0.26)=0.6026\) or \(60.3\%\)

Step2: Calculate the z - score for part (b)

For \(x = 39\), \(z=\frac{39 - 42.4}{6.1}=\frac{- 3.4}{6.1}\approx - 0.56\)
\(P(X\gt39)=1 - P(X\leq39)\)
From the standard - normal table, \(P(Z\lt - 0.56)=0.2877\)
\(P(Z\gt - 0.56)=1 - 0.2877 = 0.7123\) or \(71.2\%\)

Step3: Calculate the z - scores for part (c)

For \(x = 38\), \(z_1=\frac{38 - 42.4}{6.1}=\frac{-4.4}{6.1}\approx - 0.72\)
For \(x = 43\), \(z_2=\frac{43 - 42.4}{6.1}=\frac{0.6}{6.1}\approx0.10\)
\(P(38\lt X\lt43)=P(-0.72\lt Z\lt0.10)\)
\(P(Z\lt0.10)=0.5398\), \(P(Z\lt - 0.72)=0.2358\)
\(P(-0.72\lt Z\lt0.10)=0.5398 - 0.2358=0.304\) or \(30.4\%\)

Answer:

a) \(60.3\%\)
b) \(71.2\%\)
c) \(30.4\%\)