A certain type of plywood consists of five layers. The thicknesses of the layers are independent and normally distributed with mean 5 mm and standard deviation 0.2 mm.
a. Find the mean thickness of the plywood.
b. Find the standard deviation of the thickness of the plywood.
c. Find the probability that the plywood is less than 24 mm thick.
Let 
be the thicknesses of the five layers. The thickness of the plywood is 
.
(c) Sis normally distributed with mean 25 and standard deviation 0.4472.
The z-score of 24 is z = (24 ? 25) / 0.4472 = ?2.24. The area to the left of z = ?2.24 is 0.0125.
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