Question

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? 4.8 inches 5.2 inches 6.0 inches 8.9 inches

Explanation:

Step1: Use the z - score formula

Let \(X\) be the random variable representing the yearly snowfall. We know that \(P(X>60) = 0.15\), so \(P(X\leq60)=1 - 0.15=0.85\).
The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 55\), \(x = 60\), and we need to find \(\sigma\).
From the standard normal distribution table, the \(z\) - value corresponding to a cumulative probability of \(0.85\) is approximately \(z = 1.04\).

Step2: Substitute values into the z - score formula

Substitute \(z = 1.04\), \(x = 60\), and \(\mu = 55\) into \(z=\frac{x-\mu}{\sigma}\).
We get \(1.04=\frac{60 - 55}{\sigma}\).

Step3: Solve for \(\sigma\)

Cross - multiply: \(1.04\sigma=60 - 55\).
Since \(60 - 55 = 5\), then \(\sigma=\frac{5}{1.04}\approx4.8\).

Answer:

4.8 inches