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A machine fills containers with a particular product. The standard deviation of filling weights is known from past data to be .6 ounce. If only \(2 \%\) of the containers hold less than 18 ounces, what is the mean filling weight for the machine? That is, what must \(\mu\) equal? Assume the filling weights have a normal distribution.

Short Answer

Expert verified
The mean filling weight is 19.23 ounces.

Step by step solution

01

Understanding the Problem

We are given that the standard deviation \((\sigma)\) of the filling weights is 0.6 ounces, and we are looking for the mean filling weight \((\mu)\) such that only 2% of the containers hold less than 18 ounces. The filling weights follow a normal distribution.
02

Identifying the Z-value

Since 2% of the containers hold less than 18 ounces, we need to find the Z-score that corresponds to this cumulative probability in a standard normal distribution. From the Z-table, the Z-score for the 2% or 0.02 cumulative probability is approximately -2.05.
03

Using the Z-score Formula

The Z-score is calculated using the formula \[Z = \frac{{X - \mu}}{{\sigma}}\]where \(X\) is 18 ounces, \(\mu\) is the mean we are trying to find, and \(\sigma = 0.6\) ounces. We know \(Z = -2.05\).
04

Solving for the Mean \(\mu\)

Plug the values into the Z-score formula and solve for \(\mu\):\[-2.05 = \frac{{18 - \mu}}{{0.6}}\]Multiply both sides by 0.6 to isolate the equation:\[-2.05 \times 0.6 = 18 - \mu\]This simplifies to:\[-1.23 = 18 - \mu\]Add \(\mu\) to both sides and then add 1.23 to both sides to get:\[\mu = 18 + 1.23 = 19.23\]
05

Conclusion

The mean filling weight \((\mu)\) for the machine must be 19.23 ounces for only 2% of the containers to hold less than 18 ounces.

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Key Concepts

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

Standard Deviation
In the world of statistics, the standard deviation is a crucial measure that helps us understand how much variation exists in a set of data.
It tells us how much the individual data points differ from the mean of the dataset. When the standard deviation is small, this indicates that the data points are closely packed around the mean. Conversely, a large standard deviation shows that the data points are spread out over a wide range of values.
In practical terms, for our problem, each time the machine fills a container, the weight might vary slightly. The standard deviation of 0.6 ounces shows how much these weights typically vary from the mean weight. Understanding standard deviation in a normal distribution allows us to predict how likely it is for a certain weight to occur.
  • Helpful for quality control processes.
  • Essential for determining process consistency.
Knowing the standard deviation enables us to assess the reliability and precision of the machine's filling process.
Z-Score
The Z-score is a key concept in statistics that quantifies the number of standard deviations a data point is from the mean. This standard score is pivotal when comparing data point positions relative to the rest of the data.
In a standard normal distribution, the mean is 0, and the standard deviation is 1. Using Z-scores helps simplify comparisons across different datasets or distributions.
  • Z-scores below 0 indicate values less than the mean.
  • Z-scores above 0 indicate values greater than the mean.
In our exercise, we know that only 2% of the containers hold less than 18 ounces.
From the standard normal distribution table, this corresponds to a Z-score of approximately -2.05, meaning 18 ounces is 2.05 standard deviations below the mean. Calculation using the Z-score helps in translating this percentage into meaningful weight information.
Mean Calculation
Calculating the mean, denoted by the Greek letter \(\mu\), is the process of finding the average of a dataset. In our context, the mean represents the average filling weight that the machine should aim for.
Finding the mean is pivotal to ensuring only 2% of containers hold less than 18 ounces, ensuring that the machine operates within the desired range.To calculate the mean using a Z-score, we rearrange the Z-score formula: \(Z = \frac{(X - \mu)}{\sigma}\).
In our exercise, translating this formula to solve for \(\mu\) comes from substituting the given values:
  • \(X = 18 \text{ ounces}\),
  • \(\sigma = 0.6 \text{ ounces}\),
  • \(Z = -2.05\)
By isolating \(\mu\) and solving, we derive the mean filling weight. This process ensures the machine is calibrated correctly to avoid underfilling.
Cumulative Probability
Cumulative probability is a tool in statistics that calculates the probability that a random variable will obtain a value less than or equal to a specified value. It's particularly useful in contexts where we need to know the likelihood of a process outcome occurring below a certain threshold.
In our problem, we deal with a cumulative probability of 2%, or 0.02, which represents the proportion of containers holding less than 18 ounces.
By referring to a standard normal distribution table, we find corresponding Z-scores for such probabilities. Understanding cumulative probability helps in setting standards and maintaining quality.
  • It aids in decision making for production processes.
  • Useful for setting acceptable limits in quality control.
Thus, probabilities like these ensure the machine consistently fills containers correctly, limiting any operational errors.

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