Question

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 $p(z \geq 1.5) \approx 0.0668$. if the standard deviation of the city’s yearly rainfall totals is 1.8 inches, what is the city’s mean annual rainfall?
\bigcirc 40.5 inches
\bigcirc 41.4 inches
\bigcirc 45.0 inches
\bigcirc 45.9 inches

Explanation:

Step1: Recall z-score formula

The z - score formula is $z=\frac{x - \mu}{\sigma}$, where $z$ is the z - score, $x$ is the value from the dataset, $\mu$ is the mean, and $\sigma$ is the standard deviation.

Step2: Identify known values

We know that $x = 43.2$ inches, $z = 1.5$, and $\sigma=1.8$ inches. We need to solve for $\mu$.

Step3: Rearrange z - score formula for $\mu$

Starting with $z=\frac{x - \mu}{\sigma}$, multiply both sides by $\sigma$: $z\sigma=x - \mu$. Then, rearrange to get $\mu=x - z\sigma$.

Step4: Substitute the known values

Substitute $x = 43.2$, $z = 1.5$, and $\sigma = 1.8$ into the formula: $\mu=43.2-1.5\times1.8$.
First, calculate $1.5\times1.8 = 2.7$. Then, $\mu=43.2 - 2.7=40.5$.

Answer:

40.5 inches