Question

The average yearly snowfall in chillyville is 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?

Answer 1

The probability of a normally distributed data with mean, μ and standard deviation, σ, exceeds a value, a, is given by

$P(x \ \textgreater \ a) = 1 - P(x \leq a) = 1 - P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)$

Given that the average yearly snowfall in chillyville is normally distributed with a mean of 55 inches. If the snowfall in chillyville exceeds 60 inches in 15% of the years, thus we have:

$1 - P\left(z\ \textless \ \frac{60-55}{\sigma} \right)=0.15 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{5}{\sigma} \right)=1-0.15=0.85 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{5}{\sigma} \right)=P(z\ \textless \ 1.036) \\ \\ \Rightarrow \frac{5}{\sigma}=1.036 \\ \\ \Rightarrow\sigma= \frac{5}{1.036} =4.83$

Therefore, the standard deviation is approximately 5.