t2.5 the average yearly snowfall in chillyville is approximately normally distributed with a mean of 55…

Explanation:

Step1: Find the z - score

Since the snowfall exceeds 60 inches in 15% of the years, the area to the right of 60 is 0.15. So the area to the left of 60 is $1 - 0.15=0.85$. Looking up in the standard - normal distribution table, the z - score corresponding to an area of 0.85 is approximately $z = 1.04$.

Step2: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 60$ (the value of interest), $\mu = 55$ (the mean), and $\sigma$ is the standard deviation we want to find. Substitute the known values into the formula: $1.04=\frac{60 - 55}{\sigma}$.

Step3: Solve for the standard deviation

Cross - multiply to get $1.04\sigma=60 - 55=5$. Then $\sigma=\frac{5}{1.04}\approx4.81\approx4.83$ inches.

Answer:

A. 4.83 inches