Thicknesses of shims are normally distributed with mean

Chapter 4, Problem 15SE

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QUESTION:

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?

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QUESTION:

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?

ANSWER:

Step 1 of 3

We consider the 3 thickness are \(X_{1}, X_{2}, X_{3}\).

Then the population mean \(\mu=1.5 \mathrm{\ mm}\) and the standard deviation is 0.2 mm.

Let Y be the thickness of the stack.

So \(\mathrm{Y}=X_{1}+X_{2}+X_{3}\).

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 \(Y \sim N\left(\mu, \ \sigma^{2}\right)\).

We know that the sample size n=3.

Then mean \(\mu\).

\(\begin{aligned} \mu & =n \mu \\ \mu & =3 \times 1.5 \\ \mu & =4.5 \end{aligned}\)

Therefore mean is 4.5.

The standard deviation is:

\(\begin{array}{l} \sigma=\sqrt{v a r} \\ \sigma=\sqrt{n \sigma^{2}} \\ \sigma=\sigma \times \sqrt{n} \\ \sigma=0.2 \times \sqrt{3} \\ \sigma=0.3461 \end{array}\)

Therefore the standard deviation is 0.3461.

Hence \(Y \sim N(4.5,\ 0.3461)\)

Then the formula of the z-score is

\(z=\frac{(X-\mu)}{\sigma}\)

The z score of 5 is

\(\begin{array}{l} z=\frac{(5-4.5)}{0.3461} \\ z=\frac{(0.5)}{0.3461} \\ z=1.4446  \end{array}\)

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 the 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.

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