Final answer:
The probability that a fill volume is less than 12 fluid ounces is approximately 0.000032. The proportion of cans scrapped if they are less than 12.1 or greater than 12.6 ounces can be found by standardizing the values and summing the probabilities. Specifications that include 99% of all cans symmetrically about the mean are approximately (12.14, 12.66) fluid ounces.
Step-by-step explanation:
(a) To find the probability that a fill volume is less than 12 fluid ounces, we need to standardize the value using the given mean and standard deviation. Let X be the fill volume. We can use the z-score formula: z = (X - mean) / standard deviation. Substituting the given values, we have z = (12 - 12.4) / 0.1 = -4. The probability of a fill volume less than 12 fluid ounces can be found using the standard normal table or a calculator, which is approximately 0.000032.
(b) If cans less than 12.1 or greater than 12.6 ounces are scrapped, we need to find the proportions of cans falling outside this range. Let's standardize the values 12.1 and 12.6 using the given mean and standard deviation. For 12.1, z = (12.1 - 12.4) / 0.1 = -3, and for 12.6, z = (12.6 - 12.4) / 0.1 = 2. The proportion of cans scrapped is the sum of the probabilities of fill volumes less than 12.1 or greater than 12.6, which can be found using the standard normal table or a calculator.
(c) To determine specifications that include 99% of all cans symmetrically about the mean, we need to find the z-scores that correspond to the upper and lower tails of 0.005 each. Using the standard normal table or a calculator, we find that the z-score for a tail area of 0.005 is -2.57. Therefore, a specification for 99% of all cans would be (12.4 - 2.57 * 0.1, 12.4 + 2.57 * 0.1), which is approximately (12.14, 12.66) fluid ounces.