Problem 76 - Applied Statistics and Probability for Engineers

Question

The fill volume of an automated filling machine used for filling cans of carbonated beverage is normally distributed with a mean of 12.4 fluid ounces and a standard deviation of 0.1 fluid ounce. (a) What is the probability that a fill volume is less than 12 fluid ounces? (b) If all cans less than 12.1 or more than 12.6 ounces are scrapped, what proportion of cans is scrapped? (c) Determine specifications that are symmetric about the mean that include \(99 \%\) of all cans.

Step-by-Step Solution

Answer

(a) Probability of less than 12 ounces is approximately 0. (b) Proportion of scrapped cans is 2.41%. (c) Specifications for 99% of cans are [12.1424, 12.6576].

Step 1: Understanding the Problem Statement

We're given a normal distribution with a mean (\( \mu \)) of 12.4 fluid ounces and a standard deviation (\( \sigma \)) of 0.1 fluid ounces. We must find probabilities related to this distribution.

Step 1: Find Probability Less Than 12 Ounces

To find the probability \( P(X < 12) \), we first calculate the z-score. The formula for the z-score is \( z = \frac{{X - \mu}}{\sigma} \). Here, \( X = 12 \), \( \mu = 12.4 \), and \( \sigma = 0.1 \). So, \( z = \frac{12 - 12.4}{0.1} = -4 \). Using the standard normal distribution table, \( P(Z < -4) \) is practically 0 because -4 is extremely low.

Step 2: Calculate Proportion of Scrapped Cans

Cans are scrapped if their fill is less than 12.1 ounces or more than 12.6 ounces. Calculate the z-scores for 12.1 and 12.6. For 12.1: \( z = \frac{12.1 - 12.4}{0.1} = -3 \). From the table, \( P(Z < -3) \approx 0.00135 \).For 12.6: \( z = \frac{12.6 - 12.4}{0.1} = 2 \). From the table, \( P(Z < 2) \approx 0.97725 \). The probability for this part \( P(X > 12.6) = 1 - P(Z < 2) = 1 - 0.97725 = 0.02275 \).The total scrapped proportion is \( 0.00135 + 0.02275 = 0.0241 \).

Step 3: Develop 99% Symmetric Specification

Find the z-scores for the middle 99% of the distribution. From the z-table, the z-value that captures 99% in the middle is around \( z = \pm 2.576 \).Calculate the interval around the mean:\[X = \mu \pm z \cdot \sigma = 12.4 \pm 2.576 \cdot 0.1\]So the specification range is \( [12.1424, 12.6576] \).

Key Concepts

Understanding Z-score Calculations

Before diving into solving questions based on normal distribution, understanding the concept of a z-score is crucial. The z-score measures how many standard deviations a data point is from the mean. It gives us an idea of how unusual or typical a value is compared to the average.

To calculate a z-score, you can use the formula:

• The formula is: \( z = \frac{X - \mu}{\sigma} \)

Where:

• \( X \) is the value you are examining,
• \( \mu \) is the mean,
• \( \sigma \) is the standard deviation.

For instance, if we want to know how far 12 fluid ounces is from the mean of 12.4 fluid ounces, with a standard deviation of 0.1, we compute:

• \( z = \frac{12 - 12.4}{0.1} = -4 \)

This tells us that 12 fluid ounces is 4 standard deviations below the mean. Such a score indicates that 12 ounces is exceptionally lower than the average value.

Probability Calculation in Normal Distribution

Once you have the z-score, determining probability involves using a standard normal distribution table (or Z-table). This table tells us the probability that a standard normal random variable will be less than a particular value.

For example, to find the probability that a can holds less than 12 fluid ounces, we calculate the z-score first, as illustrated with \( z = -4 \). Then, we consult the Z-table to find \( P(Z < -4) \), which is so low it's practically 0. This means it's extremely unlikely to pour less than 12 ounces.

When dealing with specification ranges, such as when cans are scrapped if outside 12.1 to 12.6 ounces, calculate each limit's z-scores:

• For 12.1 ounces: \( z = -3 \). \( P(Z < -3) \approx 0.00135 \)
• For 12.6 ounces: \( z = 2 \). \( P(Z < 2) \approx 0.97725 \)

The proportion of scrapped cans considers probabilities outside this range. Adding these, the total is 0.0241, or 2.41% of the cans, showing they are rare but existent.

Setting Specification Limits for Quality Control

In manufacturing or quality assurance, setting specification limits centered around the mean is vital to ensuring most products meet standards. For a normal distribution like ours, it involves establishing a range where a specific percentage of values lie.

For instance, a common industry practice is to capture 99% of all possible outcomes. Using the Z-table, a z-score of about \( \pm 2.576 \) corresponds to the middle 99% of values around the mean. Therefore, we calculate the limits as follows:

• Since \( \mu = 12.4 \) and \( \sigma = 0.1 \), then
  \[ X = 12.4 \pm 2.576 \times 0.1 \]

This results in a specification range of approximately [12.1424, 12.6576].

This setup highlights the balance between variance and control, as it ensures products fall within a tight range, minimizing defects and maintaining product quality. By employing these calculated limits, you're ensuring reliability and consistency in production output.