The fill volume of cans filled by a certain machine is normally distributed with mean 12.05 oz and standard deviation 0.03 oz.

a. What proportion of cans contain less than 12 oz?

b. The process mean can be adjusted through calibration. To what value should the mean be set so that 99% of the cans will contain 12 oz or more?

c. If the process mean remains at 12.05 oz, what must the standard deviation be so that 99% of the cans will contain 12 oz or more?

Answer:

Step 1 of 3:

(a)

In this question, we are asked to find the proportion of cans contain less than 12.

The fill volume of can filled is normally distributed with mean 12.05 and standard deviation .

Let represent the fill volume of can.

We need to find .

Here we can write,

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

The z-score of 12 is z =

From the z table, the area to the left of is 0.0475.

therefore

Hence the proportion of cans contain less than 12is 0.0475.

Step 2 of 3:

(b)

In this question, we are asked to find adjusted mean after calibration so that 99% of the cans will contain 12or more.

We have given .

This is the 1st percentile (1% area of normal curve) of the distribution and corresponding is approximately .

Hence the adjusted mean is 12.07.