Question

the average snow fall in lake Hopatcong is normally distributed. The average snowfall equals 56 inches. the average snow fall in lake Hopatcong exceeds 59 inches in 15% of the year. what is the standard deviation?

Answer 1

Solution: We are given that average snowfall in lake Hopatcong is normally distributed with mean $\mu =56$ inches

The average snowfall in lake Hopatcong exceeds 59 inches in 15% of the year.

Therefore, the z score corresponding to less 59 inches is $z(0.85)=1.0364$

Using the z-score formula, we have:

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

$1.0364=\frac{59-56}{\sigma}$

$\sigma = \frac{59-56}{1.0364}$

$\sigma = \frac{3}{1.0364}$

$\therefore \sigma =2.895$ rounded to 3 decimal places

Hence the standard deviation is 2.895 inches