Problem 17 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

QUALITY CONTROL A toy manufacturer makes hollow rubber balls. The thickness of the outer shell of such a ball is normally distributed with mean \(0.03\) millimeter and standard deviation \(0.0015\) millimeter. What is the probability that the outer shell of a randomly selected ball will be less than \(0.025\) millimeter thick?

Step-by-Step Solution

Answer

The probability is approximately 0.0004.

Step 1 - Understand the Problem

We need to find the probability that the thickness of the outer shell of a randomly selected ball will be less than 0.025 millimeters. The thickness is normally distributed with mean \(\mu = 0.03\) millimeters and standard deviation \(\sigma = 0.0015\) millimeters.

Step 2 - Set Up the Z-Score Formula

The Z-score formula is used to standardize a value in a normal distribution. The formula is: \[Z = \frac{X - \mu}{\sigma}\] where \(X\) is the value we are standardizing, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 3 - Calculate the Z-Score

Plug in the values into the Z-score formula: \[Z = \frac{0.025 - 0.03}{0.0015} = \frac{-0.005}{0.0015} = -3.33\]

Step 4 - Find the Probability from Z-Score Table

Look up the Z-score of -3.33 in a standard normal distribution table or use a calculator that provides cumulative probabilities for normal distributions. The corresponding probability value for Z = -3.33 is approximately 0.0004.

Key Concepts

Z-score calculation

To understand quality control in the context of normal distribution, it's essential to get familiar with Z-score calculation. The Z-score helps us determine how many standard deviations a data point is from the mean.

The formula for calculating the Z-score is: \(Z = \frac{X - \mu}{\sigma}\) where:

• X is the data point (in this case, the shell thickness of 0.025 mm)
• \(\mu\) is the mean (0.03 mm)
• \(\sigma\) is the standard deviation (0.0015 mm)

In our example, substituting the values in gives: \(Z = \frac{0.025 - 0.03}{0.0015} = \frac{-0.005}{0.0015} = -3.33\). This result tells us that the thickness of 0.025 mm is 3.33 standard deviations below the mean.

Understanding Z-scores is essential because they allow us to use standard normal distribution tables to find probabilities.

Probability theory

Probability theory provides the frameworks to understand and evaluate events' likelihood. When a data set follows a normal distribution, it means we can use Z-scores and related tables to find probabilities associated with specific ranges of data.

For instance, in our quality control scenario with the toy manufacturer, we calculated a Z-score of -3.33 for a ball thickness of 0.025 mm. With this Z-score, we can determine the probability of this event using a standard normal distribution table.

By looking up -3.33 in the Z-table, we find a cumulative probability of approximately 0.0004. This small probability (0.04%) indicates it is quite unlikely to find a ball with a shell thickness less than 0.025 mm.

This insight allows manufacturers to assess quality control measures effectively and ensures products stay within specified tolerances.

Standard deviation

Standard deviation is a measure of the dispersion or spread of a set of values in a data set. It describes how much individual data points differ from the mean. The formula is: \[\sigma = \sqrt{\frac{\sum (X_i - \mu)^2}{N}}\] where:

• X_i is each data point
• \(\mu\) is the mean
• N is the number of data points

In the context of quality control for the toy manufacturer, a standard deviation of 0.0015 mm indicates minor variability in the shell thickness of hollow rubber balls. This small standard deviation shows the precision in production, providing confidence that most balls produced will have a thickness close to the mean of 0.03 mm.

Accurate standard deviation helps in setting control limits and ensures the manufacturing process adheres to our quality standards.