Question

The average yearly snowfall in Chilly Ville is approximately Normally distributed with a mean of 55 inches. If the snowfall in Chilly Ville 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.

Short answer

The correct option is (a) $4.83inches$

Step-by-step solution

Given information

$$\begin{gathered} \mu =55 \\ \sigma =60 \end{gathered}$$

Concept

The formula used:

$z=\frac{x-\mu }{\sigma }$

Calculation

The probability of the snowfall X exceeding $60$ inches is $15\%$

$P(X>60)=15\%=0.15$

So the probability that X does not exceed $60$ is

$$\begin{gathered} P(X\le 60)=1-P(X>60) \\ =1-0.15 \\ =0.85 \end{gathered}$$

The column contains the corresponding $100th$ of the$z-$score $(0.05)$ and the raw contains the proper integer $10th$ of the$z-$score $(0.80),$ therefore the $z-$score is

$Z=1.04$

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ z\sigma =x-\mu \\ \sigma =\frac{x-\mu }{z} \\ \sigma =\frac{60-55}{1.04} \\ =4.8077 \end{gathered}$$

As a result, the standard deviation is $4.8077$ inches, which is quite near to $4.83$ inches.

Hence, the correct option is (a)