Question 833262: Could anyone help me with this problem?
Suppose that women's heights are normally distributed w/ mean=63.6 inches & standard deviation=2.5 inches. Find the probability that
(a) if 1 woman is randomly selected that her height is less than 60 inches.
(b) if 36 are randomly selected, that they have a mean height less than 60 inches
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Suppose that women's heights are normally distributed w/ mean=63.6 inches & standard deviation=2.5 inches. Find the probability that
(a) if 1 woman is randomly selected that her height is less than 60 inches.
z(60) = (60-63.6)/2.5 = -3.6/2.5 = -1.44
P(x < 60) = P(z < -1.44) = normalcdf(-100,-1.44) = 0.0749
-------------------------------------------------------------
(b) if 36 are randomly selected, that they have a mean height less than 60 inches
z(60) = (60-63.6)/[2.5/sqrt(36) = 6*-1.44 = -8.64
----
P(x-bar < 60) = P(z < -8.64) = normalcdf(-100,-8.64) = 0 to 4 decimals.
---------------
Cheers,
Stan H.
============
|
|
|
