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The area covered by 1 L of a certain stain is normally
Chapter 4, Problem 18E(choose chapter or problem)
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 \(19.9\ \rm{m}^{2}\)?
Questions & Answers
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.