Question 644843
  <pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
Hi,
Assume that the population of heights of female college students is
 approximately normally distributed with mean m of 66.24 inches and standard
 deviation s of 5.07 inches. A random sample of 72 heights is obtained
(A) Find P(x>65.75) = P({{{z > (65.75-66.24)/5.07}}})Find z and then determine corresponding P-value
 (B) Find the mean and standard error of the xbar distribution
 mean = 66.24, SD = {{{5.07/sqrt(72)}}} 
 (C) Find P(<span style="text-decoration: overline">x</span> >65.75)= P({{{t > (65.75 -66.24)/(5.07/sqrt(72))}}})Find t and then determine corresponding P-value 
(D) Why is the formula required to solve (A) different than than (C)?
<span style="text-decoration: overline">x</span> is a sample mean