The 85th percentile is the cutoff time \(t\) such that
\(\mathbb P(X
In other words, the 85th percentile refers to the time needed to belong to the top 15% of the distribution; more generally, the \(n\) percentile is the top \((100-n)\%\) of the distribution.
Anyway, to find this value of \(t\), transform \(X\) to a random variable \(Z\) with the standard normal distribution using
\(Z=\dfrac{X-\mu}\sigma\)
where \(\mu\) is the mean of \(X\) and \(\sigma\) is the standard deviation of \(X\).
\(\mathbb P(X
Here \(t^*\) is used to denote the z-score corresponding to the cutoff time \(t\). Referring to a z-score table, you find that this occurs for \(t^*\approx1.036\). So,
\(\dfrac{t-5.9}{2.1}=t^*\implies t=5.9+2.1t^*\approx8.076\)
