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The data listed here are the weights (in pounds) of 27 packages of ground beef in a supermarket meat display: $$ \begin{array}{rrrrrrr} 1.08 & .99 & .97 & 1.18 & 1.41 & 1.28 & .83 \\ 1.06 & 1.14 & 1.38 & .75 & .96 & 1.08 & .87 \\ .89 & .89 & .96 & 1.12 & 1.12 & .93 & 1.24 \\ .89 & .98 & 1.14 & .92 & 1.18 & 1.17 & \end{array} $$ a. Construct a stem and leaf plot or a relative frequency histogram to display the distribution of weights. Is the distribution relatively moundshaped? b. Find the mean and standard deviation of the data set. c. Find the percentage of measurements in the intervals \(\bar{x} \pm s, \bar{x} \pm 2 s,\) and \(\bar{x} \pm 3 s\) d. How do the percentages obtained in part compare with those given by the Empirical Rule? Explain. e. How many of the packages weigh exactly 1 pound? Can you think of any explanation for this?

Step by step solution

01

a. Constructing a Stem and Leaf Plot and Determining Mound-shaped Distribution

A stem and leaf plot combines categorical and numerical values to create a graphical representation of the data. Organize the data by tens (0.0 - 0.9, 1.0 - 1.9) and list them vertically: $$ \begin{array}{c|l} 0.0-0.9 & 397967899879\\ 1.0-1.9 & 0814112861472378184 \end{array} $$ The distribution appears slightly mound-shaped, with a lower frequency on the ends and the highest frequency in the middle.

02

b. Finding Mean and Standard Deviation

To find the mean of the data set, we can use the formula: \(\bar{x} = \frac{\sum_{i=1}^n x_i}{n}\). Calculate the sum of all the weights and divide by the total number of packages, 27, to get the mean: $$ \bar{x} = \frac{1.08 + 0.99 + 0.97+\cdots +1.17}{27} = \frac{30.16}{27} \approx 1.12 $$ To find the standard deviation, we can use the formula: \(s = \sqrt{\frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n-1}}\). Find the difference between each weight and the mean, square each difference, and then sum those squared differences. Divide by \(n-1\) and take the square root of the result: $$ s = \sqrt{\frac{2(0.08)^2 + (0.01)^2 + \cdots + (0.05)^2}{26}} \approx \sqrt{0.353} \approx 0.19 $$ So, the mean weight of a package is approximately 1.12 pounds, and the standard deviation is approximately 0.19 pounds.

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c. Finding the Percentage of Measurements within Intervals Around the Mean

To find the percentage of measurements within the given intervals, count the number of weights within each range, divide by the total number of packages (27), and multiply by 100. We need to use the mean and standard deviation found in step b: - \(\bar{x} \pm s\): (0.93, 1.31) - There are 18 packages within this range. - \(\bar{x} \pm 2s\): (0.74, 1.50) - There are 25 packages within this range. - \(\bar{x} \pm 3s\): (0.55, 1.69) - There are 27 packages within this range. Calculate the percentages for each range: 1. \(\frac{18}{27} \times 100 \approx 66.7\%\) 2. \(\frac{25}{27} \times 100 \approx 92.6\%\) 3. \(\frac{27}{27} \times 100 = 100\%\)

04

d. Comparing Percentages with the Empirical Rule

The Empirical Rule states that, for a mound-shaped (normal) distribution, approximately: - 68% of the measurements are in the interval \(\bar{x} \pm s\) - 95% of the measurements are in the interval \(\bar{x} \pm 2s\) - 99.7% of the measurements are in the interval \(\bar{x} \pm 3s\) Comparing our percentages found in step c with the Empirical Rule: - 66.7% vs. 68% for \(\bar{x} \pm s\) - 92.6% vs. 95% for \(\bar{x} \pm 2s\) - 100% vs. 99.7% for \(\bar{x} \pm 3s\) The percentages are close to those given by the Empirical Rule, but not exact. This may be due to slight deviations from mound-shape in our distribution.

05

e. Number of Packages Weighing Exactly 1 Pound and Possible Explanation

To find the packages weighing exactly 1 pound, look at the stem and leaf plot. There are two packages with a weight of 1.00 pounds. A possible explanation for why there are not more packages weighing exactly 1 pound could be due to variations in the ground beef consistency or the weighing and packaging process used, causing a natural variation in weights.

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Key Concepts

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

Mean and Standard Deviation

The mean is a measure of central location, commonly referred to as the average. It provides us with a single value that represents the entire dataset. To find the mean, we sum all the values and divide by the number of values. In our beef package weight example, the mean weight is approximately 1.12 pounds. This tells us that, on average, each package of ground beef weighs about 1.12 pounds.

The standard deviation, on the other hand, gives us an idea of how spread out the weights are around this mean. A larger standard deviation indicates that the weights of the packages vary widely. In contrast, a smaller standard deviation means they are closely packed around the mean. For the ground beef, the standard deviation is approximately 0.19 pounds, indicating that while most packages are close to the average, there is some variation.

• Formula for Mean: \[ \bar{x} = \frac{\sum_{i=1}^n x_i}{n} \]
• Formula for Standard Deviation: \[ s = \sqrt{\frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n-1}} \]
• A smaller \(s\) value suggests data points are closer to the mean.

Stem and Leaf Plot

A stem and leaf plot is a helpful way to visualize data distribution while preserving the original data values. This plot helps you to see the shape of your data and to identify any outliers or clusters relatively easily. Each number in the dataset is split into a stem (the initial digits) and a leaf (the final digits). For our weights example, numbers like 0.99, 1.08, and 1.41 are separated into stems of 0.9 and 1.0, while the leaves are the remaining digits.

The stem and leaf plot of the weights allows us to quickly assess whether the distribution of weights is mound-shaped. In our case, the plot shows a higher frequency of weights around the center, with fewer weights at the extremes, indicating a slightly mound-shaped distribution.

• Stems often represent the integer part of a value, while leaves show the fractional part.
• This method combines characteristics of bar graphs and histograms, providing a clear view of data frequency.
• Helps identify median values and outliers easily.

Empirical Rule

The Empirical Rule is a statistical guideline for normal distributions, describing the data's tendency to drift around the mean. It states that for a mound-shaped distribution, approximately
- 68% of data lies within one standard deviation (\(\bar{x} \pm s\)) of the mean,
- 95% within two standard deviations (\(\bar{x} \pm 2s\)),
- a remarkable 99.7% within three standard deviations (\(\bar{x} \pm 3s\)).

For the package weights, we found that roughly 66.7% of the data fell within one standard deviation, about 92.6% within two, and all data points were within three standard deviations. These results are close to the expectations from the Empirical Rule, but they don't match perfectly due to slight deviations from a perfect normal distribution.

• Use in Practice: Provides a quick way to understand data distribution behavior.
• Helps in predicting the likelihood of future data points falling within a certain range.
• Useful for identifying and anticipating outliers or anomalies.