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)?

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to the Z scores.

Reverse process with last question, starting with .90.

To find the probabilities and the minimum volume of soup, we can use the information given about the normal distribution of the soup can volumes. Here's how you can calculate each of these values:

1. Probability of 19.2 oz or less of soup:
First, we need to standardize the value of 19.2 oz. To do this, we use the formula:
z = (x - μ) / σ
where z is the standardized score, x is the value of interest, μ is the mean, and σ is the standard deviation.
Plugging in the values, we get:
z = (19.2 - 20.1) / 0.4

Now, we need to find the cumulative probability associated with this z-score. Using a standard normal distribution table or statistical software, we can determine that the cumulative probability is approximately 0.1587. Therefore, the probability that a randomly sampled can of soup will contain 19.2 oz or less is 0.1587 (or 15.87%).

2. Probability of between 19 oz and 21 oz of soup:
Similarly, we need to standardize the values of 19 oz and 21 oz using the same formula.
For 19 oz:
z1 = (19 - 20.1) / 0.4
For 21 oz:
z2 = (21 - 20.1) / 0.4

Again, using the standard normal distribution table or statistical software, we can find the cumulative probabilities associated with these z-scores. The probability associated with z1 is approximately 0.0228, and the probability associated with z2 is approximately 0.9772.
To find the probability between these values, we subtract the smaller cumulative probability from the larger one:
Probability = P(z2) - P(z1) ≈ 0.9772 - 0.0228 ≈ 0.9544 (or 95.44%).

3. Smallest (cut-off) volume of soup that 90% of the cans will contain (or exceed):
We need to find the z-score associated with an upper tail probability of 0.90. This represents the percentage of cans with volumes greater than a certain value.
Using the standard normal distribution table or statistical software, we can find the z-score for this upper tail probability, which is approximately 1.2816.
We can then use the formula to solve for the value of interest:
z3 = (x - μ) / σ
Rearranging the formula gives us:
x = (z3 * σ) + μ
Plugging in the values, we get:
x = (1.2816 * 0.4) + 20.1 ≈ 20.61 oz
Therefore, the smallest (cut-off) volume of soup that 90% of the cans will contain or exceed is approximately 20.61 oz.

Note: In calculations, we have used approximation for convenience. For more accurate results, you should use more precise tables or statistical software.