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?

Step 1 of 3:

We consider the 3 thickness are .

Then the population mean and the standard deviation is 0.2 mm.

Let Y be the thickness of stack.

So Y=.

Our goal is :

a). We need to find the probability that the stack is more than 5 mm thick.

b). We need to find the 80th percentile of the stack thickness.

c). We need to find 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).

Now we have to find the probability that the stack is more than 5 mm thick.

Here .

We know that the sample size n=3.

Then mean .

Therefore mean is 4.5.

Then standard deviation is .

Therefore the standard deviation is 0.3461.

Hence

Then the formula of the z-score is

z=

The z score 5 is

z=

z=

z=.

Therefore z=1.44

Now the probability that the stack is more than 5 mm thick.

P(X>5)=1-P(X<5)

P(X>5)=1-0.9251 (using area under the normal table)

P(X>5)=0.0749.

Therefore the probability that the stack is more than 5 mm thick is 0.0749.

Step 2 of 3:

b).

Now we have find the 80th percentile of the stack thickness.

The z score of the 80th percentile is

z=0.84.

Then the formula of the z score is

z=

0.84=

0.840.3461=- 4.5

0.2907=- 4.5

=0.2907+4.5

=4.79

Therefore the 80th percentile of the stack thickness is 4.79.