Question

A botling plant fills 12-ounce cans of soda by an automated filing process that can be adjusted to any mean fll volume and that wil fir cans acoondng to a normal distebtion. However, nok at cans will contain the same volume due to variation in the filing process. Historical records show that regardless of what the mean is sot at, the stancard devation in fil will be o. 02s ounse. Ocerations managers at the plant know that it they put too much soda in a can, the company loses money. If loo litle is put in the can, customers are short changed, and the 5 twe Department of Wighes and Measures may fine the company, Complete parts a and b below. a. Suppose the industry standards for fill volume call for each 12-ounce can to contain between 11.98 and 12.02 ounces. Aswming that the manyper sets the mean 2x at 12 ounces. what is the probability that a can will contain a volume of product that falls in the desired range? The probability is 0.5762. (Round to four decimal places as needed.) b. Assume that the manager is focused on an upcoming audit by the Department of Weights and Measures she knows the process is to sect one can at rardom and hat f 2 conlake ies tran 11,97 ounces, the company will be reprimanded and potentaly fined. Assuming that the maragor wants at most a 5 N chance of this happening, at what ierel theud she set the mein ta min? Comenent on the ramifications of this step, assuming that the company fils tens of thousands of cans each week Set the mean fill ievel at ounces. (Round to two decimal places as needed.)

Answer 1

a. The probability that a can falls within the desired range is 0.5762.

b. To ensure less than a 5% chance of being reprimanded, the manager should set the mean fill level at 12.03 ounces. However, this increases the risk of overfilling and potential financial loss for the company.

Step-by-step explanation:

In order to find the probability that a can will contain a volume of product that falls in the desired range, we need to calculate the z-scores for the lower and upper limits of the range and then find the probability using the standard normal distribution table.

a. The lower limit z-score can be calculated as (11.98 - 12) / 0.02 = -1. The upper limit z-score can be calculated as (12.02 - 12) / 0.02 = 1.

The probability of the volume falling within the desired range is the area between these two z-scores, which can be found in the standard normal distribution table as 0.5762.

b. To find the interval at which the manager should set the mean to ensure less than a 5% chance of being reprimanded, we need to find the z-score that corresponds to a 5% right-tail probability.

This z-score can be found in the standard normal distribution table as 1.645. We can then find the corresponding volume by rearranging the z-score formula:

z= (x - mean) / 0.02

= 1.645.

Solving for x, we get

x = (1.645 * 0.02) + mean

x= 12.0329.

Setting the mean fill level at 12.03 ounces would minimize the chance of being reprimanded by the Department of Weights and Measures. However, it is important to consider the ramifications of this step. By setting the mean fill level higher, there is a higher chance of cans containing more soda than the desired range, leading to potential loss of money for the company.