The 15th percentile is the value x such that
P(X ≤ x) = 0.15
Let Z be a random variable following the standard normal distribution with mean 0 and standard deviation 1. Then
X = 57 + 4Z
or
Z = (X - 57)/4
so that
P((X - 57)/4 ≤ (x - 57)/4) = 0.15
P(Z ≤ (x - 57)/4) = 0.15
If F is the inverse CDF for the standard normal distribution, then
(x - 57)/4 = F(0.15) ≈ 0.5596
x - 57 ≈ 2.2385
x ≈ 59.2385
