Question

A Coca-Cola bottling plant's product line includes 12-ounce cans of Coke products. The

cans are filled by an automated filling process that can be adjusted to any mean filll volume and that

will ll cans according to a normal distribution. However, not all cans will contain the same volume

due to variation in the filling process. Historical records show that regardless of what the mean is set

at, the standard deviation in ll will be 0.035 ounce. Operations managers at the plant know that if

they put too much Coke in a can, the company loses money. If too little is put in the can, customers

are short changed, and the State Department of Weights and Measures may fine the company.

(a) Suppose the industry standards for fill volume call for each 12-ounce can to contain between

11.98 and 12.02 ounces. Assuming that the manager sets the mean fill at 12 ounces, what is the

probability that a can will contain a volume of Coke product that falls in the desired range?

(b) Assume that the manager is focused on an upcoming audit by the Department of Weights and

Measures. She knows the process is to select one Coke can at random and that if it contains less

than 11.97 ounces, the company will be reprimanded and potentially fined. Assuming that the

manager wants at most a 5% chance of this happening, at what level should she set the mean fill

level?

Answer 1

Given: Std.deviation, $\sigma$ =0.035 ounce

(a)

Mean, $\mu$ =12 ounces

X1 =11.98 ounces

Z1 =(X1 - $\mu$​​​​​​)/$\sigma$ =(11.98 - 12)/0.035 = -0.5714

X2 =12.02 ounces

Z2 =(X2 - $\mu$​​​​​​)/$\sigma$ =(12.02 - 12)/0.035 =0.5714

The probability that a can will contain a volume of Coke product that falls in the desired range =P(11.08 $\leq$ X $\leq$ 12.02) =P(-0.5714 $\leq$ Z $\leq$ 0.5714) =0.4323

(b)

It is required that P(X < 11.97) $\leq$ 0.05

$\implies$ P[Z < (11.97 - $\mu$​​​​​​)/0.035)] $\leq$ 0.05

We know that, P(Z < - 1.645) =0.05

So, (11.97 - $\mu$​​​​​​)/0.035 = - 1.645 $\implies$$\mu$ =11.97 + 1.645(0.035) =12.027575

So, the required mean fill level must be 12.027575 ounces or more. So, $\mu$$\geq$ 12.027575 ounces.

Therefore, she should set the mean fill level at 12.027575 ounces or more. (For example, she can set at 12.03 ounces).

[If $\mu$ increases from 12.027575, then the probability, P decreases from 0.05 (which is desirable as at most 5% chance is required, that is, 5% or less).

If $\mu$ decreases from 12.027575, then the probability, P increases from 0.05 (which is not desirable as we need at most 0.05 probability, that is 0.05 or less but not more than 0.05).

Hence, $\mu$ should be 12.027575 or more ounces].