Kimsellers68id
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Assume the random variable X is normally distributed, with mean = 57 and standard deviation = 4. Find the 15th percentile. ​

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LammettHashid

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

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