Problem 82 - An Introduction to Mathematical Statistics and Its Applications

Question

When a machine is improperly adjusted, it has probability \(0.15\) of producing a defective item. Each day, the machine is run until three defective items are produced. When this occurs, it is stopped and checked for adjustment. What is the probability that an improperly adjusted machine will produce five or more items before being stopped? What is the average number of items an improperly adjusted machine will produce before being stopped?

Step-by-Step Solution

Answer

There is a \(P(X \leq 2)\) probability that the machine will produce five or more items before being stopped, and on average, it will produce \(\frac{1}{0.15}\) items before being stopped.

Step 1: Define the parameters

Let's denote \( p \) as the probability of producing a defective item and \( q \) as the probability of producing a non-defective item. Thus, we have \( p = 0.15 \) and \( q = 1 - p = 0.85 \). The number \( k \) of items produced until the machine is stopped is a random variable with a geometric distribution.

Step 2: Compute the probability of producing five or more items

The probability of producing five or more items before being stopped is the same as the probability of producing only 0, 1 or 2 defective items during the first 4 trials. We can compute this probability using the binomial distribution formula \( P(X = k) = C(n, k) \cdot p^k \cdot q^{n-k} \), where \( C(n,k) \) is the number of combinations of \( n \) items taken \( k \) at a time. The cumulative probability is then \( P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) \).

Step 3: Calculate the average number of items produced

For a geometric distribution, the expected value or average is given by \(\frac{1}{p} \), where \( p \) is the probability of 'success' (producing a defective item in this case). Thus, on average, the number of items an improperly adjusted machine will produce before being stopped is \(\frac{1}{p}\).

Key Concepts

Geometric Distribution

The geometric distribution is a useful concept when dealing with scenarios where you are interested in the number of trials until a success occurs. Here, a 'success' refers to getting a defective item from the machine. The geometric distribution focuses on the probability of achieving a success on a specific trial.

In the given problem, the machine's probability of producing a defective item is denoted as \( p = 0.15 \). To find the probability of needing five or more items to hit three defective ones, we consider scenarios needing less than three defective items in fewer trials. This ties the geometric distribution to the task, as it will stop following the third defective item.

• Success: Producing a defective item (probability \( p \))
• Failure: Producing a non-defective item (probability \( q = 1 - p \))

So, in the geometric distribution, the key feature is that each trial is independent, and the probability \( p \) remains consistent for each trial.

Binomial Distribution

The binomial distribution helps calculate the probability of obtaining a fixed number of successes in a fixed number of independent and identical trials. In the exercise, the trials are the first four machine operations producing items, while a success is a defective item.

The setup uses the binomial distribution to calculate the probability of having fewer than three defective items in the first four operations. The binomial formula is:\[P(X = k) = C(n, k) \cdot p^k \cdot q^{n-k}\]where:

• \( n \): number of trials (first four items)
• \( k \): number of successes (defective items within these trials)
• \( p \): probability of success (defective item, \( 0.15 \))
• \( q \): probability of failure (\( 1 - p \))

By calculating \( P(X \leq 2) \), we determine the likelihood the machine runs for at least five items before needing a stop for adjustment.

Expected Value

Expected value is an essential concept in probability theory, allowing us to calculate the long-term average of random variables. It provides a sense of the "center" of the probability distribution.

For a geometric distribution, the expected value is straightforward. It indicates the average number of trials until the first success. The formula for the expected value \( E(X) \) of a geometric distribution when success probability is \( p \) is given by:\[E(X) = \frac{1}{p}\]In the context of the problem, this translates to the average number of items produced by an improperly adjusted machine before it stops. Given \( p = 0.15 \), the expected number of items is \( \frac{1}{0.15} \), approximately 6.67 items. This means, on average, an improperly adjusted machine will produce around 6 to 7 items before producing three defective ones and being stopped.