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Chapter 2: Problem 75

One of the assumptions underlying the theory of control charting (see Chapter 16) is that successive plotted points are independent of one another. Each plotted point can signal either that a manufacturing process is operating correctly or that there is some sort of malfunction. Even when a process is running correctly, there is a small probability that a particular point will signal a problem with the process. Suppose that this probability is \(.05\). What is the probability that at least one of 10 successive points indicates a problem when in fact the process is operating correctly? Answer this question for 25 successive points.

Short Answer

Expert verified
The probability of at least one problematic point in 10 points is approximately 0.40126, and in 25 points, it is approximately 0.72224.

Step by step solution

01

Understanding the Scenario

We have a sequence of plotted points, each with a probability of 0.05 of incorrectly signaling a problem when the process operates correctly. We need to calculate the probability that at least one of these points signals a problem over a series of 10 and 25 points.
02

Finding Complementary Probability

To find the probability that at least one indicator shows a problem, it is easiest to first calculate the probability that no point indicates a problem. If each point's error probability is 0.05, the probability of a correct signal for each point is 0.95.
03

Probability of No Problem in 10 Points

The probability that none of the 10 points signals a problem is the product of the probabilities of each behaving correctly: \[P(\text{no problem in 10 points}) = 0.95^{10}\]
04

Calculate Complement for 10 Points

Using the complement rule, the probability that at least one point indicates a problem (when in fact none should) in 10 points is: \[ P(\text{at least one problem in 10 points}) = 1 - 0.95^{10} \]
05

Probability of No Problem in 25 Points

Similarly, the probability that none of the 25 points signals a problem is: \[P(\text{no problem in 25 points}) = 0.95^{25}\]
06

Calculate Complement for 25 Points

Using the complement rule, the probability that at least one point indicates a problem (when in fact none should) in 25 points is:\[ P(\text{at least one problem in 25 points}) = 1 - 0.95^{25} \]
07

Compute Final Probabilities

Calculate the above expressions for numerical results. For 10 points, compute:\[ P(\text{at least one problem in 10 points}) = 1 - 0.95^{10} = 1 - 0.59874 \approx 0.40126 \]For 25 points, compute:\[ P(\text{at least one problem in 25 points}) = 1 - 0.95^{25} = 1 - 0.27776 \approx 0.72224 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Theory
Probability theory forms the foundation for understanding how likely certain events are to occur. In our case, we are exploring the probability that a process signals a problem under specified conditions. This probability is a fraction where the numerator represents the number of favorable outcomes, and the denominator represents all possible outcomes.
  • If we say an event has a probability of 0.05, it means that out of 100 trials, on average, we expect the event to happen 5 times.
  • This concept guides us in predicting the likelihood of different outcomes based on historical data or assumptions.
Probability theory is crucial because it helps us with decision-making processes in uncertain situations. To solve problems like identifying process faults, we rely heavily on understanding how likely these events are based on the probabilities given.
Independent Events
Independent events in probability are those events whose occurrence does not affect each other. When we talk about two events being independent, it means the outcome of one does not influence the outcome of the other.
  • In our context, each plotted point on the control chart operates independently of the others, allowing us to analyze them individually.
  • To say events A and B are independent mathematically, the formula is: \[ P(A \text{ and } B) = P(A) \times P(B) \]
Applying this to our exercise, the independence of each point allows us to multiply their probabilities to determine the likelihood of several points acting in tandem (e.g., none indicating a problem). This is crucial for calculating the probability of at least one point showing an incorrect signal.
Statistical Quality Control
Statistical quality control (SQC) uses statistical methods to monitor and control a process. Its aim is to ensure that the process operates at its best performance, producing items with minimal defects.
  • A control chart is a tool within SQC that plots data over time to identify any signs of volatility or deviation from normal performance.
  • Each point on the chart offers data about the process at a particular time, letting us visually assess process performance and take corrective actions when needed.
The assumption is that if the process is statistically controlled, any point falling outside of the prescribed limits likely indicates a malfunction or variance from expected operations. By applying probability concepts, we can assess the risk of false alarms and maintain effective monitoring.
Complementary Probability
Complementary probability involves calculating the likelihood of an event not happening. For example, if there's a 5% chance for an event to occur, there's a 95% chance it won't.
  • We find the probability of at least one event happening by first calculating the complement: the probability of none occurring.
  • The formula is: \[ P(\text{at least one event}) = 1 - P(\text{none occur}) \]
This technique reduces complexity. We often find calculating zero occurrences is simpler, especially when dealing with multiple trials. Effectively, complementary probability allows us to manage and interpret events that involve understanding risk and occurrences over repeated trials.

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Most popular questions from this chapter

Each of a sample of four home mortgages is classified as fixed rate \((F)\) or variable rate \((V)\). a. What are the 16 outcomes in \(s\) ? b. Which outcomes are in the event that exactly three of the selected mortgages are fixed rate? c. Which outcomes are in the event that all four mortgages are of the same type? d. Which outcomes are in the event that at most one of the four is a variable- rate mortgage? e. What is the union of the events in parts (c) and (d), and what is the intersection of these two events? f. What are the union and intersection of the two events in parts (b) and (c)?

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