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Chapter 5: Problem 74

Business: coffee production. Suppose the amount of coffee beans loaded into a vacuum-packed bag has a mean weight of \(\mu\) ounces, which can be adjusted on the filling machine. Also, the amount dispensed is normally distributed with \(\sigma=0.2 \mathrm{oz}\). What should \(\mu\) be set at to ensure that only 1 bag in 50 will have less than 16 oz?

Short Answer

Expert verified
Set \(\mu\) to 16.41 oz to ensure only 1 in 50 bags weighs less than 16 oz.

Step by step solution

01

Understand the problem

We need to find the mean weight \(\mu\) such that only 1 out of 50 bags will have a weight less than 16 oz. The weight distribution is normal with a standard deviation \(\sigma = 0.2\) oz.
02

Identify the statistical requirement

A probability of 1 out of 50 corresponds to 2% or 0.02 of the distribution. In the standard normal distribution, this probability is represented by a specific z-score, which can typically be found using a z-table or statistical calculator.
03

Find the Z-score for 2%

The z-score corresponding to the bottom 2% of a standard normal distribution is approximately \(-2.05\). This means that 2% of the data lies to the left of this z-score.
04

Use the Z-score formula

The formula to convert from a z-score to a raw score is: \[ X = \mu + Z \cdot \sigma \] where \(X\) is the raw score, \(Z\) is the z-score, and \(\sigma\) is the standard deviation. Here, \(X = 16\) and \(Z = -2.05\).
05

Solve for \(\mu\)

Plug the values into the formula: \[ 16 = \mu - 2.05 \times 0.2 \] Simplifying gives: \[ 16 = \mu - 0.41 \] Thus, \[ \mu = 16.41 \]
06

Conclusion

To ensure that only 1 out of 50 bags weighs less than 16 oz, set \(\mu\) (the average fill weight) to 16.41 oz.

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Key Concepts

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

Standard Deviation
In the realm of statistics, the standard deviation is a crucial measure that shows how spread out numbers are from the mean. It provides an insight into the amount of variability or dispersion in a set of data.
  • If the data points are close to the mean, then the standard deviation will be low.
  • If many data points are far from the mean, it indicates a higher standard deviation.
This metric is pivotal in understanding normal distribution, a bell-shaped distribution where data tends to cluster around the mean or average. For example, in our coffee production scenario, we know the amount dispensed is normally distributed with a standard deviation, \( \sigma = 0.2 \) ounces. This indicates the typical deviation of a bag's weight from the mean we set. A low standard deviation, such as 0.2 oz, implies that most of the coffee weights will be close to this mean, ensuring consistency in bag filling.
Z-score
The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values, measured in terms of standard deviations. If a Z-score is 0, it indicates that the data point's score is identical to the mean score. Z-scores may be positive or negative, revealing the number of standard deviations a data point is from the mean.
  • A positive Z-score indicates the data point is above the mean.
  • A negative Z-score indicates it is below the mean.
For instance, when ensuring only 1 out of 50 coffee bags has less than 16 oz, we need to identify the Z-score that corresponds to the bottom 2% of the distribution. From tables or a statistical calculator, this Z-score is found to be -2.05. This value helps us adjust our mean to make sure only the desired proportion of bags is below the specified weight.
Probability
Probability quantifies the likelihood of an event occurring, measured on a scale from 0 to 1. In our coffee example, we are concerned with the probability of bags weighing less than 16 oz. We want this to occur in just 2% of cases, meaning there's a 0.02 probability of selecting a bag that is less than 16 oz.
  • A probability of 0 means an event will not happen.
  • A probability of 1 means an event will surely happen.
Knowing the exact probability allows us to tie it with a Z-score in a normal distribution鈥攚hich further helps us adjust the mean. This way, we maintain a balance, ensuring precision in the production process so that only a stipulated number of bags are underweight.
Statistical Calculator
A statistical calculator is a vital tool for dealing with complex statistical calculations, such as finding probabilities and Z-scores for normal distributions. It speeds up data analysis, especially when dealing with large data sets, by providing accurate results quickly.
  • They can convert raw data into Z-scores for easier interpretation.
  • They can be used to directly find the probability values for specific Z-scores, skipping extensive manual calculations.
For our example, a statistical calculator helps determine the Z-score that corresponds to the 2% probability we need. In large-scale operations like coffee production, the use of a statistical calculator streamlines the process of setting appropriate mean values quickly and efficiently, ensuring the desired quality and standards are met consistently.

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Most popular questions from this chapter

The validity of the WeberFechner Law has been the subject of great debate among psychologists. An alternative model, $$ \frac{d R}{d S}=k \cdot \frac{R}{S} $$ where \(k\) is a positive constant, has been proposed. Find the general solution of this equation. (This model has also been referred to as the Power Law of Stimulus-Response.)

How would you explain the concepts of present value and accumulated present value to a friend who has not studied this chapter?

(a) write a differential equation that models the situation, and (b) find the general solution. If an initial condition is given, find the particular solution. Recall that when \(y\) is directly proportional to \(x,\) we have \(y=k x\), and when \(y\) is inversely proportional to \(x,\) we have \(y=k / x,\) where \(k\) is the constant of proportionality. In these exercises, let \(k=1\). The rate of change of \(y\) with respect to \(x\) is directly proportional to the quotient of \(x\) divided by \(y\), and \(y=2\) when \(x=-1\).

For each probability density function, over the given interval, find \(\mathrm{E}(x), \mathrm{E}\left(x^{2}\right),\) the mean, the variance, and the standard deviation. $$ f(x)=\frac{1}{\ln 4} \cdot \frac{1}{x}, \quad[0.8,3.2] $$

Approximate the integral $$\int_{-\infty}^{\infty} e^{-x^{2}} d x$$
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