Problem 212 - Applied Statistics and Probability for Engineers

Question

The diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.002 inch. (a) Suppose that the specifications require the dot diameter to be between 0.0014 and 0.0026 inch. If the probability that a dot meets specifications is to be \(0.9973,\) what standard deviation is needed? (b) Assume that the standard deviation of the size of a dot is 0.0004 inch. If the probability that a dot meets specifications is to be \(0.9973,\) what specifications are needed? Assume that the specifications are to be chosen symmetrically around the mean of 0.002 .

Step-by-Step Solution

Answer

(a) \( \sigma = 0.0002 \). (b) Specifications: \([0.0008, 0.0032]\).

Step 1: Understanding the Problem

We're given a normally distributed random variable for dot diameter with mean \( \mu = 0.002 \) inch. We need to determine the standard deviation \( \sigma \) for a probability of \( 0.9973 \) between the diameters of \( 0.0014 \) and \( 0.0026 \) inches (part a). For part b, we have \( \sigma = 0.0004 \) inch and need to find the range of the specifications that result in the same probability \( 0.9973 \).

Step 2: Understanding the Normal Distribution and Z-scores

The probability of \(0.9973\) corresponds to being within \(3\) standard deviations (approximately) of the mean in a normal distribution because \(0.9973\) probability equates to a \(z\) value of about \( \pm 3 \).

Step 3: Solving Part (a)

We use the equation for \(z\): \( z = \frac{x - \mu}{\sigma} \). For \( z = 3 \), and diameters \(0.0014\) (lower) and \(0.0026\) (upper), you have \(\frac{0.0026 - 0.002}{\sigma} = 3\). Solving \( \sigma = \frac{0.0026 - 0.002}{3} = 0.0002 \).

Step 4: Solving Part (b)

With \( \sigma = 0.0004 \) inch and a probability of \(0.9973\), the specifications are \(3\sigma\) from the mean in both directions. So calculate \( \mu \pm 3\sigma \), where \(\mu = 0.002 \). The limits are \(0.002 \pm 3(0.0004) \), giving \([0.0008, 0.0032]\).

Step 5: Conclusion: Writing the Final Answer

For part (a), the required standard deviation is \(0.0002\). For part (b), the specification limits should be \([0.0008, 0.0032]\).

Key Concepts

Standard Deviation

The standard deviation is a measure of how spread out the numbers in a data set are. It tells us how much the individual measurements vary from the mean, or average, value. In our printer dot diameter problem, we're told the mean diameter is 0.002 inches. But not every dot will be exactly that size. The standard deviation helps quantify how much the sizes can differ from this mean.

In statistics, a smaller standard deviation means the values are closely clustered around the mean. A larger one means there's more variation. For part (a) of our problem, knowing the probability that a dot's size will be in the range from 0.0014 to 0.0026 inches as 0.9973, we calculated that the standard deviation needed is 0.0002 inches. This small value indicates that most dots will have sizes quite close to the mean, with little variation.

Understanding standard deviation is key to working with normally distributed data, where data points are symmetrically arranged around the mean. This is often visualized with the famous bell curve.

Z-scores

Z-scores are a crucial part of working with normal distributions. A z-score measures how many standard deviations a data point is from the mean. This score allows us to standardize different data sets, making them comparable. The z-score is calculated with the formula:

• \( z = \frac{x - \mu}{\sigma} \)

where \( x \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

In our exercise, the z-score helps us understand the position of the dot diameters within the normal distribution. The probability of 0.9973 corresponds to a z-score of approximately \( \pm 3 \). This means that most of the diameter measurements fall within 3 standard deviations from the mean. This is why for the specified probability, the range of standard deviations is of interest in defining whether a dot meets the specifications.

Probability

Probability is a measure of how likely an event is to occur. It can range from 0 to 1, where 0 means the event cannot happen and 1 means it is certain to happen. In a normal distribution, probabilities are used to calculate the chance of a data point falling within a certain range.

In this exercise, the specified probability of 0.9973 is very high, meaning we're interested in almost all the data points in the distribution. This probability translates to the range covering three standard deviations away from the mean on either side. Having such a high probability ensures that specifications encompass nearly all dimensions of the printer dots, preventing defects. In practical terms, invoking a high probability helps to minimize errors within the printing process.

Specifications

Specifications in the context of quality control often refer to the acceptable limits for a product feature. For our printer problem, the specifications are the acceptable range of diameters for the printer's dots. They ensure that the product meets a certain standard for quality.

For part (b) of the exercise, with a given standard deviation of 0.0004 inches, the goal was to determine the specification limits that guarantee a high probability (0.9973) that any chosen dot will be within acceptable size. Using the concept of z-scores, we find that these limits are symmetrically placed around the mean diameter of 0.002 inches, as \( \mu \pm 3\sigma \). Thus, the result is a specification range between 0.0008 and 0.0032 inches, assuring that 99.73% of dots meet the quality requirement.

Understanding specifications in normal distributions helps designers and engineers ensure that their processes produce products within threshold limits, maintaining consistency and reducing the likelihood of defects.