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The following data represent the weights (in grams) of a simple random sample of \(50 \mathrm{M} \& \mathrm{M}\) plain candies. $$ \begin{array}{lllllll} \hline 0.87 & 0.88 & 0.82 & 0.90 & 0.90 & 0.84 & 0.84 \\ \hline 0.91 & 0.94 & 0.86 & 0.86 & 0.86 & 0.88 & 0.87 \\ \hline 0.89 & 0.91 & 0.86 & 0.87 & 0.93 & 0.88 & \\ \hline 0.83 & 0.95 & 0.87 & 0.93 & 0.91 & 0.85 & \\ \hline 0.91 & 0.91 & 0.86 & 0.89 & 0.87 & 0.84 & \\ \hline 0.88 & 0.88 & 0.89 & 0.79 & 0.82 & 0.83 & \\ \hline 0.90 & 0.88 & 0.84 & 0.93 & 0.81 & 0.90 & \\ \hline 0.88 & 0.92 & 0.85 & 0.84 & 0.84 & 0.86 & \\ \hline \end{array} $$ Determine the shape of the distribution of weights of M\&Ms by drawing a frequency histogram. Find the mean and median. Which measure of central tendency better describes the weight of a plain M\&M?

Step by step solution

01

Organize the Data

Sort the given data in ascending order to make it easier to analyze: 0.79, 0.81, 0.82, 0.82, 0.83, 0.83, 0.84, 0.84, 0.84, 0.84, 0.84, 0.85, 0.85, 0.86, 0.86, 0.86, 0.86, 0.86, 0.86, 0.87, 0.87, 0.87, 0.87, 0.87, 0.88, 0.88, 0.88, 0.88, 0.88, 0.88, 0.88, 0.89, 0.89, 0.90, 0.90, 0.90, 0.90, 0.91, 0.91, 0.91, 0.91, 0.91, 0.92, 0.93, 0.93, 0.93, 0.94, 0.95

02

Create a Frequency Distribution Table

Form intervals (bins) and count the frequency of data points within each interval. Intervals could be [0.79-0.81], [0.82-0.84], [0.85-0.87], [0.88-0.90], and [0.91-0.95].

03

Draw the Frequency Histogram

On the x-axis, mark the bin intervals and on the y-axis, mark the frequency. Draw bars for each interval representing the frequency of data points.

04

Calculate the Mean

The mean is calculated by summing all the data points and dividing by the number of points. \[ \text{Mean} = \frac{0.87 + 0.88 + 0.82 + \text{...} + 0.84 + 0.86}{50} \] Perform the addition and division.

05

Calculate the Median

The median is the middle value of the sorted data. For 50 data points, the median is the average of the 25th and 26th values in the sorted list. Find these values and calculate the average.

06

Compare Mean and Median

Discuss whether the mean or the median better describes the central tendency. This usually depends on whether the data is symmetric or skewed.

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Key Concepts

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

frequency histogram

To comprehend the distribution of the weights of M&Ms, we first create a frequency histogram. A histogram gives a pictorial representation of data distribution by illustrating the number of data points within specified ranges of values (called bins).

First, we sort the data in ascending order. Then, we form appropriate bin intervals, such as [0.79-0.81], [0.82-0.84], [0.85-0.87], [0.88-0.90], and [0.91-0.95]. Each interval represents a range of weights.

• Next, we tally how many data points fall within each bin interval.
• On graph paper or using a graphing tool, we mark these intervals on the x-axis and the frequency (number of counts) on the y-axis.
• We then draw bars for each interval, with the height of the bar corresponding to the frequency of data points within that bin.

The resulting histogram offers an immediate visual insight into the data distribution, showing how weights are spread across the bins.

mean calculation

The mean, or average, provides a central location for your data. It is calculated by summing all the data points and then dividing by the number of points. For our 50 M&Ms, the mean \(\text{Mean} \approx \frac{\sum \text{weights}}{50}\)

We sum all the weights: 0.87 + 0.88 + 0.82...+0.84+0.86 = 43.92.

Next, we divide by the number of data points: \(\text{Mean} \approx \frac{43.92}{50} = 0.8784\)

Thus, the mean weight of the M&Ms is approximately 0.8784 grams.

median calculation

The median is another measure of central tendency and is the middle value in a sorted dataset. For an even number of data points, the median is the average of the two central numbers.

With 50 M&Ms, the median is the average of the 25th and 26th values in the sorted list.

From our sorted list, the 25th and 26th values are both 0.88. Hence, \(\text{Median} = \frac{0.88 + 0.88}{2} = 0.88\)

This median of 0.88 grams gives us another perspective on the central tendency of the data.

data distribution analysis

Analyzing the distribution helps us understand the dataset more thoroughly. We look at the histogram, mean, and median to gain insights.

The histogram might show a symmetric or skewed distribution.

• If the histogram is symmetric, the mean and median will be close to each other, suggesting a normal distribution.
• If it’s skewed left or right, one measure (either mean or median) might be more accurate in representing the central tendency.
• In our dataset, the mean is 0.8784 and the median is 0.88 showing that the distribution is quite symmetric.

This analysis indicates that both measures of central tendency describe the M&Ms weights well, but typically, the median is more robust to outliers and skewed data distributions.