The answer for this question is 54
Let Z be a random variable following the standard normal distribution with mean µ = 0 and standard deviation σ = 1.
The 67th percentile of the distribution is the value z that separates the bottom 67% of the distribution from the top 100% - 67% = 33%. In terms of probability, we have
Pr[Z ≤ z] = 0.67
Use the inverse CDF for the normal distribution (or lookup z scores in a table) to find
z = ɸ⁻¹(0.67) ≈ 0.4399
where ɸ(z) is the CDF for the normal distribution.
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Thanks for the points my other answer got deleted lol
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