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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 [tex]$P(z \geq 1.5)=0.0668$[/tex]. If the standard deviation of the city's yearly rainfall totals is 1.8 inches, what is the city's mean annual rainfall?

A. 40.5 inches
B. 41.4 inches
C. 45.0 inches
D. 45.9 inches

Sagot :

To determine the city's mean annual rainfall, we need to understand the relationship between the given variables using the properties of the normal distribution and the z-score formula.

Given:
- The probability that the city gets more than 43.2 inches of rain in a year is [tex]\( P(z \geq 1.5) = 0.0668 \)[/tex].
- The z-score corresponding to this probability is 1.5.
- The standard deviation ([tex]\(\sigma\)[/tex]) of the city's yearly rainfall totals is 1.8 inches.

The z-score formula is:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
where:
- [tex]\( z \)[/tex] is the z-score,
- [tex]\( X \)[/tex] is the given value in the distribution (rainfall amount),
- [tex]\( \mu \)[/tex] is the mean annual rainfall,
- [tex]\( \sigma \)[/tex] is the standard deviation.

We can rearrange this formula to solve for the mean ([tex]\(\mu\)[/tex]):
[tex]\[ \mu = X - z \cdot \sigma \][/tex]

Substituting the given values into the rearranged formula:
- [tex]\( X = 43.2 \)[/tex] inches (the given rainfall amount),
- [tex]\( z = 1.5 \)[/tex] (the given z-score),
- [tex]\( \sigma = 1.8 \)[/tex] inches (the standard deviation),

we get:
[tex]\[ \mu = 43.2 - 1.5 \cdot 1.8 \][/tex]

Calculating inside the parentheses first:
[tex]\[ 1.5 \cdot 1.8 = 2.7 \][/tex]

Subtracting this result from [tex]\( X \)[/tex]:
[tex]\[ \mu = 43.2 - 2.7 = 40.5 \][/tex]

Therefore, the city's mean annual rainfall is 40.5 inches. The correct answer is 40.5 inches.
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