The probability of a normally distributed data with mean, μ and standard deviation, σ, exceeds a value, a, is given by
[tex]P(x \ \textgreater \ a) = 1 - P(x \leq a) = 1 - P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)[/tex]
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:
[tex]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[/tex]
Therefore, the standard deviation is approximately 5.
