Question

t2.5 the average yearly snowfall in chillyville is approximately normally distributed with a mean of 55 inches. if the snowfall in chillyville exceeds 60 inches in 15% of the years, what is the standard deviation? (a) 4.83 inches (b) 5.18 inches (c) 6.04 inches (d) 8.93 inches (e) the standard deviation cannot be computed from the given information.

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