Question

assume that women's heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. find the height of a woman who is at the 76th percentile (this is backwards normal calculation)

Answer 1

65.3658 inches

Explanation:

Let X be the height of a woman randomly choosen. We know tha X have a mean of 63.6 inches and a standard deviation of 2.5 inches. For an x value, the related z-score is given by z = (x-63.6)/2.5. We are looking for a value
`x_(0)` such that
`P(X < x_(0)) = 0.76`, but,
`0.76 = P(X < x_(0)) = P((X-63.6)/2.5 < (x_(0)-63.6)/2.5) = P(Z < (x_(0)-63.6)/2.5)`, i.e.,
`(x_(0)-63.6)/2.5` is the 76th percentile of the standard normal distribution. So,
`(x_(0)-63.6)/2.5 = 0.7063`,
`x_(0) =63.6+(2.5)(0.7063) = 65.3658`. Therefore, the height of a woman who is at the 76th percentile is 65.3658 inches.