A company sells cans of chicken soup that that are advertised to contain 20 oz. However due to variations in the manufacturing process, the volume of soup in the cans is, in fact, Normally distributed with a mean of 20.1 oz and a standard deviation of 0.4 oz.

• What is the probability that a randomly sampled can of soup will contain 19.2 oz or less of soup?
• What is the probability that a randomly sampled can of soup will contain between 19 oz and 21 oz of soup?
• What is the smallest (cut-off) volume of soup that 90% of the cans will contain (or exceed)?

To find the probabilities in the questions, we will use the standard normal distribution with a mean of 20.1 oz and a standard deviation of 0.4 oz.

1. Probability of a randomly sampled can containing 19.2 oz or less of soup:

We need to find the area under the normal distribution curve to the left of 19.2 oz. We can calculate the z-score for 19.2 oz as follows:

z = (19.2 - 20.1) / 0.4 = -2.25

Using a standard normal distribution table or a calculator, we can find the probability corresponding to the z-score of -2.25. The probability is approximately 0.0122 or 1.22%.

Therefore, the probability that a randomly sampled can of soup will contain 19.2 oz or less is approximately 0.0122 or 1.22%.

2. Probability of a randomly sampled can containing between 19 oz and 21 oz of soup:

We need to find the area under the normal distribution curve between 19 oz and 21 oz. We can calculate the z-scores for both values as follows:

z1 = (19 - 20.1) / 0.4 = -2.75
z2 = (21 - 20.1) / 0.4 = 2.25

Using a standard normal distribution table or a calculator, we can find the probabilities corresponding to the z-scores of -2.75 and 2.25. The probability for -2.75 is approximately 0.00339 or 0.339%, and the probability for 2.25 is approximately 0.9878 or 98.78%.

To find the probability between 19 oz and 21 oz, we subtract the probability of 19 oz or less from the probability of 21 oz or less:

0.9878 - 0.00339 ≈ 0.98441 or 98.441%

Therefore, the probability that a randomly sampled can of soup will contain between 19 oz and 21 oz is approximately 0.98441 or 98.441%.

3. Smallest (cut-off) volume of soup that 90% of the cans will contain (or exceed):

We need to find the z-score corresponding to the 90th percentile (or the area to the left of the cut-off volume) of the normal distribution. We can use a standard normal distribution table or a calculator to find the z-score.

The z-score corresponding to the 90th percentile is approximately 1.282.

Now, we can calculate the actual volume of soup using the z-score:

x = z * standard deviation + mean
x = 1.282 * 0.4 + 20.1

x ≈ 20.6

Therefore, the smallest (cut-off) volume of soup that 90% of the cans will contain (or exceed) is approximately 20.6 oz.

To find the probability in each case, we need to use the cumulative distribution function (CDF) of the normal distribution. The CDF calculates the probability that a random variable takes on a value less than or equal to a given number.

In this case, we'll use the mean (μ) of 20.1 oz and the standard deviation (σ) of 0.4 oz to calculate probabilities.

1. To find the probability that a randomly sampled can of soup will contain 19.2 oz or less of soup, we need to calculate the cumulative probability up to 19.2 oz.
The formula for the CDF of a normal distribution is: P(X ≤ x) = Φ((x - μ) / σ)
Plug in the values: P(X ≤ 19.2) = Φ((19.2 - 20.1) / 0.4)
Use a standard normal table or a calculator to find the CDF value corresponding to (19.2 - 20.1) / 0.4. Let's assume it is 0.1587.
Therefore, the probability that a randomly sampled can of soup will contain 19.2 oz or less is approximately 0.1587 or 15.87%.

2. To find the probability that a randomly sampled can of soup will contain between 19 oz and 21 oz of soup, we need to calculate the cumulative probability between 19 oz and 21 oz.
P(19 ≤ X ≤ 21) = P(X ≤ 21) - P(X ≤ 19)
Using the CDF formula and substituting the values:
P(19 ≤ X ≤ 21) = Φ((21 - 20.1) / 0.4) - Φ((19 - 20.1) / 0.4)
Again, use a standard normal table or calculator to find the corresponding CDF values for each part of the equation. Let's assume Φ(0.975) = 0.8413 and Φ(0.025) = 0.1587.
P(19 ≤ X ≤ 21) = 0.8413 - 0.1587 = 0.6826 or approximately 68.26%.

3. To find the smallest (cut-off) volume of soup that 90% of the cans will contain (or exceed), we need to find the z-score corresponding to a cumulative probability of 90% (0.90).
Use the inverse CDF (also known as the Percent-point function) of the normal distribution to find the corresponding z-score. Let's assume the z-score is 1.28.
Rearrange the CDF formula to solve for x: x = (z * σ) + μ
Plug in the values: x = (1.28 * 0.4) + 20.1
x ≈ 20.55
Therefore, the smallest volume of soup that 90% of the cans will contain (or exceed) is approximately 20.55 oz.

Remember that the exact values may vary depending on the standard normal table or calculator used.