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

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

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

Equation transcription:

Text transcription:

m^{2}

Answer:

Step 1 of 2:

(a)

In this question, we are asked to find the probability that 1 of stain will be enough to cover 10.3area.

The area covered by of a certain stain is normally distributed with mean and standard deviation 0.2 .

Let be the area.

We need to find .

Here we can write,

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

The z-score of 10.3 is z =

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

therefore

Hence the probability that 1 of stain will be enough to cover 10.3area is 0.9332.