91Ó°ÊÓ

Express all z scores with two decimal places. Data Set 31 "Garbage Weight" in Appendix B lists weights (lb) of plastic discarded by households. The highest weight is \(5.28 \mathrm{lb}\), the mean of all of the weights is \(\bar{x}=\) \(1.911 \mathrm{lb}\), and the standard deviation of the weights is \(s=1.065 \mathrm{lb}\). a. What is the difference between the weight of \(5.28 \mathrm{lb}\) and the mean of the weights? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the weight of \(5.28 \mathrm{lb}\) to a \(z\) score. d. If we consider weights that convert to \(z\) scores between \(-2\) and 2 to be neither significantly low nor significantly high, is the weight of \(5.28 \mathrm{lb}\) significant?

Step by step solution

01

- Calculate the Difference

To find the difference between the weight of 5.28 lb and the mean weight, subtract the mean from the given weight: Difference = Weight - Mean Difference = 5.28 - 1.911 = 3.369

02

- Compute the Number of Standard Deviations

Divide the difference found in Step 1 by the standard deviation to determine how many standard deviations the difference represents: Number of Standard Deviations = Difference / Standard Deviation = 3.369 / 1.065 = 3.16

03

- Convert to a Z Score

The z score is defined as the number of standard deviations a data point is from the mean. Use the formula: z = (Weight - Mean) / Standard Deviation z = (5.28 - 1.911) / 1.065 z = 3.16

04

- Assess Significance

To determine if the weight of 5.28 lb is significant, check if the z score falls within the range of -2 to 2. Since the z score of 3.16 is outside this range, the weight of 5.28 lb is considered significant.

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Key Concepts

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

standard deviation

The concept of standard deviation is crucial in statistics. Standard deviation measures the spread or dispersion of a set of data points. It's calculated using the formula: \[ s = \sqrt{ \frac{ \sum (x - \bar{x})^2 }{ n - 1 } } \] \Here, \(s\) stands for the standard deviation, \(x\) represents each data point, \(\bar{x}\) is the mean of those data points, and \(n\) is the number of data points. A smaller standard deviation indicates that data points are close to the mean. Conversely, a larger standard deviation shows more spread out data points. In our problem, we used a standard deviation (\(s\)) of 1.065 lb to understand the dispersion of garbage weights around the mean weight of 1.911 lb.

mean weight calculation

The mean is another fundamental concept in statistics. It represents the average value of a dataset and is calculated by summing all data points and then dividing by the number of points. The formula for the mean is: \[ \bar{x} = \frac{ \sum x }{ n } \] \Where \(\sum x\) is the sum of all data points and \(n\) is the total number of points. In the exercise, the mean weight (\bar{x}) of plastic discarded by households was calculated as 1.911 lb. The mean helps us to understand what a 'typical' value in the dataset looks like. When comparing other data points, like the 5.28 lb weight, we can see how far they deviate from this average.

significance in statistics

Significance in statistics determines whether an observed effect or result is meaningful or due to random chance. A common way to assess significance is by using z scores. A z score tells us how many standard deviations a data point is from the mean. If we define a range for typical values (usually between -2 and 2 standard deviations), any value falling outside this range is considered significant. In our exercise, the weight of 5.28 lb converts to a z score of 3.16, which is outside the typical range. Thus, it is considered a significant value.

garbage weight analysis

Analyzing garbage weight can provide insights into consumption patterns and waste management efficiencies. For the garbage weight data, we consider each household's discarded plastic weight. By calculating the mean and standard deviation, we can identify outliers and understand overall trends. In the exercise, we applied these calculations to determine how the highest weight of 5.28 lb compares to others. By converting this to a z score, an index showing how extreme a data point is, we found it to be significant as the z score was 3.16, indicating it's much higher than the average.