Question

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

Answer 1

Answer:

The standard deviation is 4.83 inches.

Explanation:

We are given that the average yearly snowfall in Chillyville is Normally distributed with a mean of 55 inches.  

The snowfall  in Chillyville exceeds 60 inches in 15% of the years, we have to find the standard deviation.

Let X = the average yearly snowfall in Chillyville.

The z-score probability distribution for the normal distribution is given by;

                                 Z = $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = mean amount of rainfall = 55 inches

            $\sigma$ = standard deviation  

Now, it is stated that the snowfall  in Chillyville exceeds 60 inches in 15% of the years, that means;

        P(X > 60 inches) = 0.15

        P( $\frac{X-\mu}{\sigma}$ > $\frac{60-55}{\sigma}$ ) = 0.15

        P(Z > $\frac{60-55}{\sigma}$ ) = 0.15

In the z table, the critical value of z that represents the top 15% of the area is given as  1.0364, that means;

                       $\frac{60-55}{\sigma} = 1.0364$

                       $\sigma} = \frac{5}{1.0364}$ = 4.83 inches

Hence, the standard deviation is 4.83 inches.