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An electronic scale in an automated filling operation stops the manufacturing line after three underweight packages are detected. Suppose that the probability of an underweight package is 0.001 and each fill is independent. (a) What is the mean number of fills before the line is stopped? (b) What is the standard deviation of the number of fills before the line is stopped?

Short Answer

Expert verified
(a) 3000 fills. (b) Standard deviation is approximately 1731.0.

Step by step solution

01

Define the Problem

We are dealing with a scenario where an operation stops after three underweight packages are detected, given that each fill is independent and the probability of an underweight package is 0.001.
02

Identify the Distribution

The scenario can be modeled using the negative binomial distribution because we want the number of trials before getting the third success (where 'success' corresponds to an underweight package) with probability of success on each trial being 0.001.
03

Calculate the Mean

The mean of a negative binomial distribution, where we want to count the number of trials before achieving the r-th success, is given by \( \mu = \frac{r}{p} \). Here, \( r = 3 \) and \( p = 0.001 \), so the mean is \( \frac{3}{0.001} = 3000 \).
04

Calculate the Variance

The variance of a negative binomial distribution is given by \( \sigma^2 = \frac{r(1-p)}{p^2} \). Substituting \( r = 3 \) and \( p = 0.001 \), we get \( \sigma^2 = \frac{3(1-0.001)}{(0.001)^2} = 2997000 \).
05

Calculate the Standard Deviation

The standard deviation is the square root of the variance. Thus, \( \sigma = \sqrt{2997000} \approx 1731.0 \).

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Key Concepts

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

Mean of a Distribution
The mean of a distribution helps us understand the central tendency or the average of a set of outcomes in random trials. For the negative binomial distribution, used when counting trials until a certain number of 'successes', the mean is calculated using the formula:
  • \( \mu = \frac{r}{p} \)
Here, \( r \) is the number of successes desired, and \( p \) is the probability of success in a single trial. In our case, we want to find the average number of fills before the line stops, triggered by three underweight packages (successes). With \( r = 3 \) and \( p = 0.001 \), the mean becomes:
  • \( \mu = \frac{3}{0.001} = 3000 \)
This tells us on average it takes 3000 fills before the line is stopped due to three underweight packages.
Variance Calculation
Variance measures how much the results deviate from the mean, offering insights into the distribution's spread. In a negative binomial distribution, variance is given by:
  • \( \sigma^2 = \frac{r(1-p)}{p^2} \)
For our exercise, substituting \( r = 3 \) and \( p = 0.001 \), we find:
  • \( \sigma^2 = \frac{3(1-0.001)}{(0.001)^2} = 2997000 \)
This considerable variance indicates that while the mean is 3000 fills, the actual number of fills can deviate significantly.
Standard Deviation
Standard deviation is a crucial statistical measure that provides a clear idea of the spread or dispersion of a distribution. It is the square root of the variance. For our situation:
  • \( \sigma = \sqrt{2997000} \approx 1731.0 \)
This value suggests that the number of fills before stopping may vary by about 1731 fills from the mean (3000). The substantial standard deviation implies a wide variability, meaning that actual results can substantially stray from the average.
Probability Theory
Probability theory forms the foundation of statistical analysis and allows us to model real-world situations mathematically. When dealing with independent events, like our scenario of package filling, each event's probability remains constant across trials.
  • The given probability of a package being underweight is 0.001.
Using probability theory, we choose the negative binomial distribution for modeling because we're counting how many trials are needed to reach a fixed number of unsuccessful events (underweight packages in this case). The application of probability theory in this context ensures that our results are based on rigorous mathematical principles, providing reliable insights into the operational challenge of maintaining package weight consistency.

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