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Assume that cans of Coke are filled so that the actual amounts are normally distributed with a mean of \(12.00 \mathrm{oz}\) and a standard deviation of \(0.11\) oz. a. Find the probability that a single can of Coke has at least \(12.19 \mathrm{oz}\). b. The 36 cans of Coke in Data Set 26 "Cola Weights and Volumes" in Appendix \(\mathrm{B}\) have a mean of \(12.19\) oz. Find the probability that 36 random cans of Coke have a mean of at least \(12.19 \mathrm{oz}\) c. Given the result from part (b), is it reasonable to believe that the cans are actually filled with a mean equal to \(12.00\) oz? If the mean is not equal to \(12.00 \mathrm{oz}\), are consumers being cheated?

Short Answer

Expert verified
a) 0.0418 b) 鈮 0 c) The mean is likely higher than 12.00 oz; consumers are not being cheated.

Step by step solution

01

Calculating the Z-Score for Part (a)

Use the Z-score formula to determine the probability. The formula is: \[ Z = \frac{x - \mu}{\sigma} \]where \( x = 12.19 \) oz, \( \mu = 12.00 \) oz, and \( \sigma = 0.11 \) oz.Substitute the values into the formula:\[ Z = \frac{12.19 - 12.00}{0.11} = \frac{0.19}{0.11} \approx 1.73 \]
02

Finding the Probability for Part (a)

Using the Z-score obtained (1.73), look up the corresponding probability in the Z-table for a standard normal distribution. The Z-table value for 1.73 is approximately 0.9582. This represents the probability of a can containing less than 12.19 oz.The probability of a can containing at least 12.19 oz is:\[ P(X \geq 12.19) = 1 - 0.9582 = 0.0418 \]
03

Calculating the Z-Score for Part (b)

For 36 cans, the standard deviation of the mean is calculated by \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \), where \( n = 36 \). So, \( \sigma_{\bar{x}} = \frac{0.11}{\sqrt{36}} = \frac{0.11}{6} = 0.0183 \).Now apply the Z-score formula for the sample mean:\[ Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{12.19 - 12.00}{0.0183} = \frac{0.19}{0.0183} \approx 10.38 \]
04

Finding the Probability for Part (b)

Using the Z-score (10.38), look up the corresponding probability in the Z-table. Since the value is exceedingly high, the probability is very close to 1. For practical purposes, we can consider it as 1. This signifies that \[ P(\bar{X} \geq 12.19) \approx 0 \]
05

Analyzing the Result for Part (c)

The result from part (b) shows that the probability of obtaining a sample mean of 12.19 oz or more is extremely low if the true population mean is 12.00 oz. Thus, it is highly unlikely that the actual mean is 12.00 oz. Consequently, it suggests that the mean fill of the cans might be different from 12.00 oz (likely higher), and consumers are not being cheated since they are getting more than 12.00 oz on average.

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Key Concepts

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

normal distribution
The normal distribution is a continuous probability distribution that is symmetrical about the mean. This means the data near the mean are more frequent in occurrence than data far from the mean. Many real-world datasets form a bell-shaped curve when plotted. This curve represents the normal distribution.
Key characteristics of a normal distribution include:
  • The mean, median, and mode of the data are all equal.
  • It is perfectly symmetric around the mean.
  • It follows the empirical rule: about 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations.
Understanding the normal distribution is crucial for solving many statistical problems, including the one in our exercise regarding the filling of Coke cans.
Z-score calculation
A Z-score tells us how far a data point is from the mean in terms of standard deviations. The formula for calculating the Z-score is:
\[ Z = \frac{x - \mu}{\sigma} \]
Where:
  • \( x \) is the value of the data point
  • \( \mu \) is the mean of the dataset
  • \( \sigma \) is the standard deviation of the dataset
The Z-score shows how many standard deviations a particular value lies above or below the mean. For instance, in our exercise, we calculated the Z-score to determine the probability of a Coke can containing at least 12.19 oz of soda.
By converting raw data points to Z-scores, we can use the standard normal distribution table (Z-table) to find probabilities associated with those scores.
sample mean
The sample mean is the average of a set of observations. It provides an estimate of the population mean. The formula to calculate the sample mean \( \bar{x} \) is:
\[ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \]
Where:
  • \( \sum \) signifies the sum of all observations
  • \( x_i \) represents each individual observation
  • \( n \) is the number of observations in the sample
In our exercise with 36 cans of Coke, the sample mean is used to assess the probability of getting a sample mean of at least 12.19 oz. Ensuring the sample size is sufficiently large is essential for the reliability of the sample mean as an estimator.
standard deviation
Standard deviation measures the dispersion or spread of a dataset relative to its mean. It quantifies how much the values in the dataset differ from the mean. The formula for standard deviation \( \sigma \) is:
\[ \sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \mu)^2}{n}} \]
Where:
  • \( x_i \) is each individual observation
  • \( \mu \) is the mean of the dataset
  • \( n \) is the number of observations
A smaller standard deviation indicates that data points are close to the mean, whereas a larger standard deviation indicates that data points are spread out over a large range of values.
In our exercise, measuring the standard deviation of the can volumes helps us assess the variability of the fill amount and calculate probabilities based on the normal distribution.
probability analysis
Probability analysis involves calculating the likelihood of various outcomes. Using the normal distribution and Z-scores, we can determine the probability of a single event occurring within a certain range.
In our example, we calculated the probability of a single Coke can containing at least 12.19 oz of soda and found it using the Z-table. For 36 cans, we used the sample mean and adjusted the standard deviation to find the probability of having a sample mean of at least 12.19 oz.
This type of analysis allows us to consider whether it is reasonable to believe the claimed population mean. As demonstrated, the result provided insights into whether the average fill amount was as stated or different, which can be vital for quality control and understanding consumer expectations.

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Most popular questions from this chapter

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