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 [tex]x_{0}[/tex] such that [tex]P(X < x_{0}) = 0.76[/tex], but, [tex]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)[/tex], i.e., [tex](x_{0}-63.6)/2.5[/tex] is the 76th percentile of the standard normal distribution. So, [tex](x_{0}-63.6)/2.5 = 0.7063[/tex], [tex]x_{0} =63.6+(2.5)(0.7063) = 65.3658[/tex]. Therefore, the height of a woman who is at the 76th percentile is 65.3658 inches.
