The area covered by 1 L of a certain stain is normally

Chapter 4, Problem 18E

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QUESTION:

The area covered by 1 L of a certain stain is normally distributed with mean \(10\ \rm{m}^{2}\) and standard deviation \(0.2\ \rm{m}^{2}\).

a. What is the probability that 1 L of stain will be enough to cover \(10.3\ \rm{m}^{2}\)?

b. What is the probability that 2 L of stain will be enough to cover \(19.9\ \rm{m}^{2}\)?

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QUESTION:

The area covered by 1 L of a certain stain is normally distributed with mean \(10\ \rm{m}^{2}\) and standard deviation \(0.2\ \rm{m}^{2}\).

a. What is the probability that 1 L of stain will be enough to cover \(10.3\ \rm{m}^{2}\)?

b. What is the probability that 2 L of stain will be enough to cover \(19.9\ \rm{m}^{2}\)?

ANSWER:

Step 1 of 2:

(a) In this question, we are asked to find the probability that 1 L of stain will be enough to cover \(10.3 \mathrm{~m}^{2}\) area.

The area covered by 1 L of a certain stain is normally distributed with mean \(10 \mathrm{~m}^{2}\) and standard deviation \(0.2 \mathrm{~m}^{2}\)

 Let X be the area.

We need to find \(P(X<10.3)\).

Here we can write,

\(X \sim N\left(10,0.2^{2}\right)\)

Now we will calculate the Z score because our distribution is approximately normal.

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

The z-score of 10.3 is \(z=\left(\frac{10.3-10}{0.2}\right)=1.5\)

From the z table, the area to the left of 1.5 is 0.9332.

therefore \(P(X<10.3)=0.9332\)

Hence the probability that 1 L of stain will be enough to cover \(10.3 \mathrm{~m}^{2}\) area is 0.9332.

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