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

Answer:

65.3658 inches

Step-by-step 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.