Question

Assume that the random variable X is normally distributed with mean y = 60 and standard deviation a = 12. Find the 87th percentile for X

Answer 1

The 87th percentile is the value
`x_(0.87)` such that 87% of the distribution lies below
`X=x_(0.87)`, i.e.

`P(X \le x_(0.87)) = 0.87`

Transform
`X` to the standard normal random variable
`Z`, then solve for
`x_(0.87)` using the inverse CDF of
`Z`.

`P(X\le x_(0.87)) = P\left((X-60)/(12) \le (x_(0.87)-60)/(12)\right) = P\left(Z \le (x_(0.87)-60)/(12)\right) = 0.87`

`\implies (x_(0.87)-60)/(12) = F_Z^(-1)(0.87) \approx 1.12369`

`\implies \boxed{x_(0.87) \approx 73.5167}`