Find the 36\displaystyle {36} -th percentile for a normal random variable XN(μ=57.3,σ=7.2)\displaystyle {X}\sim{N}{\left(\mu={57.3},\sigma={7.2}\right)}.

The 36\displaystyle {36}-th percentile for X\displaystyle {X} is (Round the answer to 2 decimal places)