Problem 70 - Applied Statistics and Probability for Engineers

Question

An automated filling machine fills soft-drink cans, and the standard deviation is 0.5 fluid ounce. Assume that the fill volumes of the cans are independent, normal random variables. (a) What is the standard deviation of the average fill volume of 100 cans? (b) If the mean fill volume is 12.1 oz, what is the probability that the average fill volume of the 100 cans is less than 12 oz? (c) What should the mean fill volume equal so that the probability that the average of 100 cans is less than 12 oz is \(0.005 ?\) (d) If the mean fill volume is 12.1 oz, what should the standard deviation of fill volume equal so that the probability that the average of 100 cans is less than 12 oz is \(0.005 ?\)

Step-by-Step Solution

Answer

(a) 0.05 oz. (b) 0.0228. (c) 12.12875 oz. (d) 0.38834 oz.

Step 1: Understanding the Problem

We are dealing with a normal distribution problem involving filling machine and cans, given the standard deviation and average fill volume. We will need to compute different statistical measures like the standard deviation of the average of samples, and use properties of normal distribution to find probabilities.

Step 1: Standard Deviation of Average Fill Volume

The standard deviation of the average fill volume for a sample size of 100 cans can be calculated using the formula for the standard deviation of the sample mean: \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the standard deviation of the original distribution and \( n \) is the sample size. Here, \( \sigma = 0.5 \) fluid ounce and \( n = 100 \).\[ \sigma_{\bar{x}} = \frac{0.5}{\sqrt{100}} = \frac{0.5}{10} = 0.05 \] The standard deviation of the average fill volume of 100 cans is 0.05 fluid ounce.

Step 2: Probability of Average Fill Volume Less Than 12 oz

To find this probability, firstly transform the problem into a standard normal distribution (Z-score). The Z-score is calculated as \( Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} \), where \( \bar{x} = 12 \), \( \mu = 12.1 \) oz, and \( \sigma_{\bar{x}} = 0.05 \).\[ Z = \frac{12 - 12.1}{0.05} = \frac{-0.1}{0.05} = -2 \] Now, look up this Z-score in the standard normal distribution table or use a calculator to find the corresponding probability. \( P(Z < -2) \approx 0.0228 \). The probability that the average fill volume of 100 cans is less than 12 oz is approximately 0.0228.

Step 3: Determine Mean Fill Volume for Probability 0.005

We need to adjust the mean so that the probability of the average being less than 12 oz is 0.005. Using the Z-score tables, we find that \( P(Z < -2.575) \approx 0.005 \). Substitute \( Z = -2.575 \) back into the Z-score formula to find \( \mu \): \( -2.575 = \frac{12 - \mu}{0.05} \). Solve for \( \mu \): \[ \mu = 12 + 2.575 \times 0.05 = 12.12875 \] The mean fill volume should be approximately 12.12875 oz.

Step 4: Determine Standard Deviation for Probability 0.005

Given \( \mu = 12.1 \) oz, we need to determine what \( \sigma \) should be for \( P(\bar{x} < 12) = 0.005 \). Using \( Z = -2.575 \) again, set up the equation: \( -2.575 = \frac{12 - 12.1}{\sigma_{\bar{x}}} \), and solve for \( \sigma_{\bar{x}} \):\[ \sigma_{\bar{x}} = \frac{-0.1}{-2.575} \approx 0.038834 \] Now find the new \( \sigma \) using \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{100}} \). Thus, \( \sigma \approx 0.038834 \times 10 = 0.38834 \). The standard deviation should be approximately 0.38834 fluid ounces.

Key Concepts

Standard Deviation

The concept of standard deviation is central to understanding how data varies. It's a measure of the spread or dispersion in a set of data points. When we're dealing with the normal distribution, the standard deviation tells us how much the values deviate from the mean. For example, in the context of a soft drink filling machine, a standard deviation of 0.5 fluid ounces means that the fill amounts for individual cans typically vary by that amount from the average fill amount.

• The standard deviation is symbolized by \( \sigma \).
• It helps in assessing the consistency and reliability of the filling process.

To calculate the standard deviation of the average fill volume when considering a sample (like 100 cans), we use the formula for the standard deviation of the sample mean: \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \), where \( n \) is the number of samples. This formula shows us that as the sample size increases, the standard deviation of the sample mean decreases, indicating the averages become more reliable.

Sample Mean

The sample mean is the average of a set of measurements from a sample rather than the entire population. Calculating the sample mean provides an estimate of the population mean, giving us a measure of the central tendency specifically for the sampled data.

• In the example, the average fill volume is given as 12.1 oz, which is the sample mean.
• It's critical in comparing individual sample measurements against overall trends.

When evaluating fill volumes, the sample mean helps predict the typical contents of any can from the machine. If many samples (like 100 cans) are considered, the mean of these samples will generally approximate the population mean, especially in normally distributed data. This forms the foundation for many probabilistic calculations involving large data sets.

Probability Calculation

Probability calculation teaches us to quantify the likelihood of an event happening. With normally distributed data, this involves determining where a specific value lies within the distribution curve.

• For instance, calculating the probability that the average fill volume of 100 cans is less than 12 oz requires converting the sample mean into a Z-score.
• Z-score tells us how many standard deviations away a value is from the mean.

Using this, we can consult standard normal distribution tables or computational tools to find probabilities. For this exercise, the calculated Z-score was -2, indicating that the average fill volume of 12 oz is two standard deviations below the mean of 12.1 oz. By checking a Z-table, we found the probability of this occurrence to be approximately 0.0228, demonstrating the scenarios in which the automated system might not perform as expected.

Z-score Computation

A Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is measured in terms of standard deviations from the mean.

• To compute a Z-score, use: \( Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} \).
• It converts a normal distribution to a standard normal distribution with a mean of 0 and a standard deviation of 1.

In this exercise, for a sample mean of 12 oz, the Z-score calculation allowed us to understand how far below the given mean of 12.1 oz this particular average was. Since the Z-score was -2, it indicated that the sample mean was unlikely under standard conditions. This numeric value helps engineers adjust parameters to achieve targeted probabilities, such as setting the mean fill volume accurately to decrease the excess rejection rate or unnecessary waste.