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During a manufacturing process 15 units are randomly selected each day from the production line to check the percent defective. From historical information it is known that the probability of a defective unit is \(0.05 .\) Any time that two or more defectives are found in the sample of \(15,\) the process is stopped. This procedure is used to provide a signal in case the probability of a defective has increased. (a) What is the probability that on any given day the production process will be stopped? (Assume \(5 \%\) defective.) (b) Suppose that the probability of a defective has increased to \(0.07 .\) What is the probability that on any given day the production process will not be stopped?

Short Answer

Expert verified
The probability that the manufacturing process will be stopped on a given day given 5% defect rate is calculated in Step 2, while the probability that the manufacturing process won't be stopped given 7% defect rate is calculated in Step 3.

Step by step solution

01

Identifying parameters and probability calculation

Identify the binomial distribution parameters: the number of experiments n=15, and the probability of success p1=0.05 for part (a) and p2=0.07 for part (b). The probability that a process will be stopped on any given day, is the probability that there are two or more defective units. Using the binomial pmf, we can express it as the sum of the probabilities of finding exactly 2, 3, ..., or 15 defectives. However, it’s faster to calculate the complementary event (finding 0 or 1 defective units), and subtract the result from 1. Let's denote r as the number of 'successes' (defective units). (a)The probability that 2 or more units are defective: \(P(r\geq2) = 1 - (P(r=0) + P(r=1)). \)
02

Probability calculation for part (a)

Use the probability mass function of the binomial distribution to calculate the probabilities. Recall the PMF of a binomial distribution is: \(P(r=k)= C(n,k) * (p^k) * (1-p)^(n-k)\). Plug in known values to the formula for k=0 and k=1 cases: \(P(r=0)= C(15,0)* 0.05^0 * (1-0.05)^{15-0} \) And, \(P(r=1)= C(15,1)* 0.05^1 * (1-0.05)^{15-1}\). Then use these to get the final result.
03

Probability calculation for part (b)

Here, we have to find the probability that the manufacturing process is not stopped, or equivalently, the probability that we have zero or exactly one defective unit when the probability of defect has increased to 0.07. The calculation will be direct in this case, unlike in part (a). With similar calculation as in Step 2: \(P(r=0)\) and \(P(r=1)\) when \(p=0.07\). Add up these two probabilities to get the answer.

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Key Concepts

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

Probability Mass Function (PMF)
When we talk about probability, we often want to know how likely a certain outcome is, especially when dealing with random processes. This is where the probability mass function (PMF) comes into play. The PMF is a function that gives the probability that a discrete random variable is exactly equal to some value. For instance, if you toss a fair coin, the probability of getting a 'heads' is 0.5, and the probability of getting 'tails' is also 0.5.

In the context of binomial distribution, the PMF gets a bit more intricate. The binomial distribution models the number of 'successes' in a fixed number of independent experiments, with each experiment having two possible outcomes - typically termed as 'success' or 'failure'. The PMF for a binomial distribution is expressed as:
\[P(r=k)= {n \choose k} p^k (1-p)^{n-k}\]
Here, \(P(r=k)\) is the probability of getting exactly \(k\) successes in \(n\) trials, \(p\) is the probability of success on a single trial, and \(n \choose k\) represents the binomial coefficient, which calculates the number of ways to choose \(k\) successes from \(n\) trials. Let's not forget that the \(1-p\) is the probability of failure, which when raised to the power of \((n-k)\) represents the probability of the remaining trials resulting in failure.
Complementary Events in Probability
Understanding complementary events in probability is incredibly handy when solving problems where calculating the direct probability is complex or time-consuming. A complementary event is basically the opposite of the event of interest. For any event \(A\), its complement \(A^c\) includes all outcomes that are not in \(A\). The probability of an event and its complement always add up to 1, since between them, they cover all possible outcomes.

For instance, in a quality control scenario, if we're interested in the event that a defective item is found \(A\), the complement \(A^c\) would be the event that no defective items are found. Mathematically, we can write:
\[P(A^c) = 1 - P(A)\]
This is immensely helpful when the probability of \(A\) is arduous to calculate. Instead, we find \(P(A^c)\) and then easily derive \(P(A)\). In the binomial distribution exercise, the use of complementary events simplifies the calculation by avoiding the need to compute multiple probabilities for each possible outcome of two or more defectives.
Binomial Parameter Calculation
In a binomial distribution, parameter calculation is critical to determine the likelihood of different outcomes. Two parameters define this distribution: \(n\), the number of trials or experiments, and \(p\), the probability of success on any given trial. The exercise posed illustrates this concept by taking a fixed number of trials (15 units checked per day) and known probabilities of finding a defective unit (0.05 and later 0.07).

The parameter \(n\) is usually determined by the design of the experiment or the process in question, often representing a fixed quantity that is controlled by the experimenter or the nature of the process. The parameter \(p\), on the other hand, is a measure of how likely the 'success' (in this case, a defective unit) is to occur. Comprehending how to adjust and use these parameters is vital in calculating probabilities in diverse scenarios, including quality control, as they dictate the behavior of the binomial distribution.
Quality Control Statistics
Statistics are essential in quality control processes, providing a way to understand and manage variability in manufacturing. In the exercise, we're dealing with a quality control technique utilized to monitor the production line and identify when the defects exceed a certain threshold. This is an application of statistical process control, which employs statistical methods to monitor and control a process.

By selecting a random sample and calculating the probability of defectives, companies can decide whether the process is deviating from performance standards. The binomial distribution is particularly useful in quality control because it can model the number of defectives in a sample. If the calculated probability of finding an unacceptable number of defectives exceeds a pre-set threshold, this signals a potential increase in the defect rate, prompting further investigation or process adjustments.

Using binomial distribution probability, companies can accurately determine their process capability and maintain control over production quality, ensuring that the final products meet the required specifications and customer expectations. It's a fundamental tool for maintaining high standards in manufacturing and avoiding the costs associated with poor quality.

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