Question 662745
The diameters of apples in a certain orchard are normally distributed with a mean of 4.77 inches and a standard deviation of 0.43 inches. Show all work. 
(A) What percentage of apples in this orchard is larger than 4.71 inches?
----
z(4.71) = (4.71-4.77)/0.43 = -0.06/0.43 = -0.1395
P(x > 4.71) = P(z > -0.1395) = normalcdf(-0.1395,100) = 0.5555
====================================================================
(B) A random sample of 100 apples is gathered and the mean diameter is calculated. What is the probability that the sample mean is greater than 4.71 inches?
x-bar = 4.71
std for the distribution of sample means when n = 100  = 0.43/sqrt(100) 
-----
z(4.71) = (4.71-4.77)/(0.43/sqrt(100)) = -1.3953
----
P(x-bar > 4.71) = P(z > -1.3953) = normalcdf(-1.3953,100) = 0.9185
===========================
Cheers,
Stan H.
==============