Let the mean amount of soda dispensed be μ. We want to find μ such that P(X > 12.30) = 0.03, where X follows a normal distribution with mean μ and standard deviation 0.12 ounces. This is equivalent to solving for μ with the equation:
P((X - μ) / 0.12 > (12.30 - μ) / 0.12) = 0.03.
To solve for μ, we first need to find the z-score corresponding to a 3% probability in the upper tail of a standard normal distribution. Using a z-table or calculator, we find that the z-score is approximately 1.88. Therefore,
(12.30 - μ) / 0.12 = 1.88
Now, we can solve for μ:
12.30 - μ = 1.88 * 0.12
12.30 - μ = 0.2256
μ = 12.30 - 0.2256 = 12.0744
So, the machine should dispense a mean amount of approximately 12.07 ounces of soda to limit the percentage of cans that need to be cleaned because of spillage to 3%.
The amount of soda a dispensing machine pours into a 12 ounce can of soda follows a normal distribution with a standard deviation of 0.12 ounces. Every can that has more than 12.30 ounces of soda poured into it causes a spill and the can needs to go through a special cleaning process before it can be sold. What is the mean amount of soda the machine should dispense if the company wants to limit the percentage that needs to be cleaned because of spillage to 3%?
1 answer
1 answer