SOLUTION: I have worked this problem and I have it until I get to the last part of how I come up with the answer. z(5.8) = (5.8 -5.55)/0.65 = 0.385 P(x < 5.8) = P(z

SOLUTION: I have worked this problem and I have it until I get to the last part of how I come up with the answer. z(5.8) = (5.8 -5.55)/0.65 = 0.385 P(x < 5.8) = P(z < 0.385) = ? This is

Question 363381: I have worked this problem and I have it until I get to the last part of how I come up with the answer.
z(5.8) = (5.8 -5.55)/0.65 = 0.385
P(x < 5.8) = P(z < 0.385) = ?
This is where I get stuck.

The diameter of apples in a certain orchard are normally distributed with a mean of 5.5 inches and a standard deviation of 0.65 inches. Show all work.
What percentage of the apple in this orchard have diameters less than 5.8 inches?

What percentage of the apples in this orchard are larger than 6.3 inches?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I have worked this problem and I have it until I get to the last part of how I come up with the answer.
z(5.8) = (5.8 -5.55)/0.65 = 0.385
P(x < 5.8) = P(z < 0.385) = 0.6499
====================================================
The diameter of apples in a certain orchard are normally distributed with a mean of 5.5 inches and a standard deviation of 0.65 inches. Show all work.
What percentage of the apple in this orchard have diameters less than 5.8 inches?
----------
z(5.8) = (5.8 -5.55)/0.65 = 0.385
P(x < 5.8) = P(z < 0.385) = 0.6499
------------------
What percentage of the apples in this orchard are larger than 6.3 inches?
z(6.3) = (6.3-5.55)0.65 = 1.1538
P(x> 6.3) = P(z > 1.1538) = 0.1243
======================================
Cheers,
Stan H.
=========
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