A city's annual rainfall totals are normally distributed, and the probability that the city gets more than 43.2 inches of rain in a year is given by $$ P(z \geq 1.5)=0.0668 $$. If the standard deviation of the city's yearly rainfall totals is 1.8 inches, what is the city's mean annual rainfall? 40.5 inches 41.4 inches 45.0 inches 45.9 inches

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Given that the annual rainfall $$X$$ is normally distributed with mean $$\mu$$ and standard deviation $$\sigma = 1.8$$ inches, we have:

$$
z = \frac{X - \mu}{\sigma}
$$

We are told that the probability of getting more than 43.2 inches of rain is:

$$
P(X \geq 43.2) = P\left(z \geq \frac{43.2 - \mu}{1.8}\right) = 0.0668
$$

From the standard normal table, it is given that:

$$
P(z \geq 1.5) = 0.0668
$$

This implies that:

$$
\frac{43.2 - \mu}{1.8} = 1.5
$$

Now, solve for $$\mu$$:

$$
43.2 - \mu = 1.5 \times 1.8
$$

$$
43.2 - \mu = 2.7
$$

$$
\mu = 43.2 - 2.7 = 40.5
$$

Thus, the city's mean annual rainfall is $$ \boxed{40.5\text{ inches}} $$.
The city's mean annual rainfall is 40.5 inches.

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