Chapter 4: Problem 47
There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters that are normally distributed with mean \(3 \mathrm{~cm}\) and standard deviation \(.1 \mathrm{~cm}\). The second machine produces corks with diameters that have a normal distribution with mean \(3.04 \mathrm{~cm}\) and standard deviation \(.02 \mathrm{~cm}\). Acceptable corks have diameters between \(2.9\) and \(3.1 \mathrm{~cm}\). Which machine is more likely to produce an acceptable cork?
Short Answer
Step by step solution
Understand the Problem
Calculate Z-scores for Machine 1
Calculate Probability for Machine 1
Calculate Z-scores for Machine 2
Calculate Probability for Machine 2
Compare Probabilities
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Z-scores
\[Z = \frac{X - \mu}{\sigma}\]
Where:
- \(X\) is the value being measured
- \(\mu\) is the mean of the distribution
- \(\sigma\) is the standard deviation
Probability
In our cork cutting case, once Z-scores are calculated for the boundaries, the probability for the range of acceptable corks can be found. Machine 1, with its boundaries of Z-scores between -1 and 1, shows a probability of 68.27%, indicating how many corks fall within this acceptable range. Machine 2, with scores from -7 to 3, suggests a much higher likelihood at up to 99.87% probability for corks fitting the required size.
Standard Deviation
Mathematically, it is defined as:
\[\sigma = \sqrt{\frac{1}{N} \sum (X_i - \mu)^2}\]
Where:
- \(N\) is the number of data points
- \(X_i\) is each individual data point
- \(\mu\) is the mean
Cumulative Probability
In terms of normal distribution, cumulative probability is read from Z-tables and represents the fraction of observations expected to fall below a given Z-score.
In the example: Machine 1’s cumulative probability from Z-scores -1 to 1 leads to a 68.27% chance of corks being the right size. For Machine 2, as the Z-score from -7 to the desired range is beyond standard Z-tables, we consider the cumulative probability up to a Z-score of 3, making it more likely within the acceptable cork diameter range.
Statistical Comparison
In the cork example, both machines' products were compared by evaluating probabilities derived from their Z-scores. By comparing these statistical indicators:
- Machine 1 produces acceptable corks with a 68.27% probability.
- Machine 2 shows a higher probability, at approximately 99.87% when compared over a significant Z-score range.