Question

Ketchup A fast-food restaurant has just installed a new automatic ketchup dispenser for use in preparing its burgers. The amount of ketchup dispensed by the machine follows a Normal distribution with mean 1.05 ounces and standard deviation 0.08 ounce.

a) If the restaurant’s goal is to put between 1 and 1.2 ounces of ketchup on each burger, what percent of the time will this happen?

Answer 1

To determine the percentage of time the ketchup dispenser operates within the desired range, convert the desired ketchup quantities to z-scores and find the cumulative probabilities associated with these scores.

Step-by-step explanation:

To calculate the percentage of time the automatic ketchup dispenser will dispense the correct amount of ketchup (between 1 and 1.2 ounces), we can use standard normal distribution principles. Since the amount of ketchup dispensed is normally distributed with a mean (μ) of 1.05 ounces and a standard deviation (σ) of 0.08 ounce, we first need to convert the ketchup amounts into z-scores:

• Z for 1 ounce = (1 - 1.05) / 0.08 = -0.625
• Z for 1.2 ounces = (1.2 - 1.05) / 0.08 = 1.875

Next, we look up these z-scores in a standard normal distribution table or use a calculator capable of providing cumulative distribution function (CDF) values for these z-scores. The percentage of time the machine will dispense between 1 and 1.2 ounces is the area under the normal curve between these two z-scores. Let P1 and P2 be the cumulative probabilities for the z-scores -0.625 and 1.875, respectively.

The final answer is found by calculating P2 - P1 to get the area between the z-scores. This will give us the percentage of time the dispenser applies the correct amount of ketchup within the desired range.