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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.
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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