A brewery has a beer dispensing machine that dispenses beer into the company's 12 ounce bottles. The distribution for the amount of beer dispensed by the machine follows a normal distribution with a standard deviation of 0.13 ounce. The company can control the mean amount of beer dispensed by the machine. What value of the mean should the company use if it wants to guarantee that 98.5% of the bottles contain at least 12 ounces (the amount on the label)?

To determine the value of the mean that the company should use to guarantee that 98.5% of the bottles contain at least 12 ounces of beer, we need to use the standard deviation and the concept of the normal distribution.

The normal distribution is a bell-shaped curve that describes the probability distribution of a continuous random variable. In this case, we are interested in the amount of beer dispensed by the machine, which follows a normal distribution with a standard deviation of 0.13 ounce.

We want to find the mean value such that 98.5% of the bottles contain at least 12 ounces. In other words, we are looking for the cutoff point where only 1.5% of the bottles will have less than 12 ounces of beer.

To find this cutoff point, we need to determine the Z-score associated with the left tail of the normal distribution that corresponds to the 1.5% probability. The Z-score represents the number of standard deviations a particular value is from the mean.

Using a Z-table or a statistical software, we can find that the Z-score corresponding to 1.5% probability is approximately -2.17 (this corresponds to the area to the left of the Z-score).

Once we have the Z-score, we can use the formula for a Z-score:

Z = (x - μ) / σ

Where Z is the Z-score, x is the value we are interested in (12 ounces in this case), μ is the mean, and σ is the standard deviation.

Rearranging the formula, we get:

μ = x - (Z * σ)

Plugging in the values, we can calculate:

μ = 12 - (2.17 * 0.13)

Calculating this, we find:

μ ≈ 11.7169

Therefore, the company should use a mean value of approximately 11.7169 ounces to guarantee that 98.5% of the bottles contain at least 12 ounces of beer.

To find the value of the mean, we need to determine the corresponding z-score for a given percentile. In this case, we want to find the z-score that corresponds to the 98.5th percentile.

Step 1: Determine the standard deviation (σ) and the desired percentile (P).

σ = 0.13 ounces
P = 98.5%

Step 2: Find the z-score corresponding to the desired percentile using a standard normal distribution table or a calculator.

The z-score corresponding to the 98.5th percentile is approximately 2.17.

Step 3: Use the formula for z-score to find the mean (μ).

z = (x - μ) / σ

Solving for μ:

2.17 = (12 - μ) / 0.13

Multiply both sides by 0.13:

0.13 * 2.17 = 12 - μ

0.281 = 12 - μ

Subtract 12 from both sides:

-11.719 = -μ

Multiply by -1 to isolate μ:

11.719 = μ

The company should use a mean value of 11.719 ounces if it wants to guarantee that 98.5% of the bottles contain at least 12 ounces.

The value of the mean should be 12, of course. Is that really the question you want to ask?