Final answer:
a) The proportion of cans containing less than 12oz is 0.04758. b) The mean should be set to 12.07oz for 99% of the cans to contain 12oz or more. c) The standard deviation must be 0.01oz for 99% of the cans to contain 12oz or more.
Explanation:
The normal distribution is a continuous probability distribution that is symmetric about the mean. It is used to represent real-world data sets that are too large to be easily analyzed using other methods. In this question, the fill volume of cans filled by a certain machine is normally distributed with a mean of 12.05oz and a standard deviation of 0.03oz. The proportion of cans containing less than 12oz is calculated using the cumulative probability formula. The cumulative probability is the probability of observing a value less than or equal to a certain point on the normal distribution curve. The formula for calculating the cumulative probability is P(X<=x) = (1/2) * (1 + erf(x-μ/σ√2)). In this formula, μ stands for the mean and σ stands for the standard deviation. Substituting the given values, the cumulative probability is found to be 0.04758. In order to have 99% of the cans contain 12oz or more, the mean should be set to 12.07oz. This is calculated using the inverse cumulative probability formula.
The inverse cumulative probability is the value at which a certain probability is observed on the normal distribution curve. The formula for calculating the inverse cumulative probability is x = μ + (σ√2)*erf^-1(2*P(X<=x)-1). Substituting the given values, the inverse cumulative probability is found to be 12.07oz. If the process mean remains at 12.05oz, the standard deviation must be 0.01oz for 99% of the cans to contain 12oz or more.
This is calculated using the inverse cumulative probability formula. Substituting the given values, the inverse cumulative probability is found to be 0.01oz.