Thicknesses of shims are normally distributed with mean 1.5 mm and standard deviation 0.2 mm. Three shims are stacked, one atop another.
a. Find the probability that the stack is more than 5 mm thick.
b. Find the 80th percentile of the stack thickness.
c. What is the minimum number of shims to be stacked so that the probability that the stack is more than 5 mm thick is at least 0.99?
(a) Let 
be the three thicknesses. Let 
be the thickness of the stack. Then ???? is
normally distributed with mean 
= 3(1.5) = 4.5 and standard deviation 
.
The z-score of 5 is (5 ? 4.5) / 0.34641 = 1.44.
The area to the right of z = 1.44 is 1 ? 0.9251 = 0.0749.
P (S > 5) = 0.0749.
(b) Let 
denote the 80th percentile. The z-score of the 80th percentile is approximately z = 0.84.
Therefore satisfies the equation 
Solving for yields = 4.7910 cm.
(c) Let ???? be the required number. Then 
.
P (S > 5) ? 0.99, so 5 is less than or equal to the 1st percentile of the distribution of S.
The z-score of the 1st percentile is z = ?2.33.
Therefore n satisfies the inequality 
The smallest value of n satisfying this inequality is the solution to the equation
This equation can be rewritten 
Solving for 
with the quadratic formula yields , = 1.9877 so n = 3.95. The smallest integer value of n is therefore n = 4.
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