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)

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 such that , but, , i.e., is the 76th percentile of the standard normal distribution. So, , . Therefore, the height of a woman who is at the 76th percentile is 65.3658 inches.

not sure but it may be b, as i said i'm not sure.