Problem 8 - Algebra 2 and Trigonometry

Question

In \(7-9,\) find the mean, median, range, and interquartile range for each set of data to the nearest tenth. $$ \begin{array}{|c|c|}\hline x_{i} & {f_{i}} \\ \hline 10 & {2} \\ {9} & {4} \\\ {8} & {6} \\ {7} & {9} \\ {6} & {3} \\ {5} & {3} \\ {4} & {2} \\ {3} & {0} \\ {2} & {1} \\ \hline\end{array} $$

Step-by-Step Solution

Answer

Mean: 7.0, Median: 7.5, Range: 8, IQR: 3.

Step 1: Calculate the Mean

To find the mean, sum the product of each value and its frequency, then divide by the total frequency. First, calculate each product: \(10 \times 2 = 20\), \(9 \times 4 = 36\), \(8 \times 6 = 48\), \(7 \times 9 = 63\), \(6 \times 3 = 18\), \(5 \times 3 = 15\), \(4 \times 2 = 8\), \(2 \times 1 = 2\). Next, find the total sum: \(20 + 36 + 48 + 63 + 18 + 15 + 8 + 2 = 210\). The total frequency is \(2 + 4 + 6 + 9 + 3 + 3 + 2 + 1 = 30\). Thus, the mean is \( \frac{210}{30} = 7.0\).

Step 2: Arrange Data for Median

List all values considering their frequencies in ascending order: \(4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10\).

Step 3: Calculate the Median

With 30 data points, the median is the average of the 15th and 16th values in the list. The 15th and 16th terms are both \(7\) and \(8\), thus the median is \( \frac{7 + 8}{2} = 7.5\).

Step 4: Determine the Range

The range is the difference between the highest and lowest values. Here, that's \(10 - 2 = 8\).

Step 5: Find the Interquartile Range

To find the interquartile range (IQR), first calculate the 1st quartile (Q1) and the 3rd quartile (Q3). For Q1 (the 7.5th position), between the 7th and 8th values is \(6\). Q3 (the 22.5th position), between the 22nd and 23rd value is \(9\). The IQR is \(Q3 - Q1 = 9 - 6 = 3\).

Key Concepts

Understanding the Mean

In statistical terms, the mean, often referred to as the average, represents a central value of a data set. To find the mean, we follow a straightforward process:

• Multiply each data value by its frequency (how often it appears).
• Sum up all these products to get a total.
• Finally, divide the total sum by the sum of frequencies.

This gives us the mean. For example, with the data provided, these steps calculated a mean of 7.0. The mean provides a quick snapshot into the overall behavior of the data, acting like a single value summary that balances the data on both sides.

Exploring the Median

The median is another measure of central tendency, representing the middle value of a dataset. Here's how to find it:

• Firstly, list the data values in ascending order.
• Identify the middle point of the dataset, which is straightforward if there's an odd number of values.
• If there is an even number of values, the median is the average of the two middle numbers.

In this example, with 30 values, the median is calculated as the average of the 15th and 16th values, giving a median of 7.5. This measure helps in understanding the dataset by showing where half the data lies below and half above this point.

The Range Explained

The range is one of the simplest measures of spread in statistics, showing how spread out the values in a dataset are.

• To calculate the range, subtract the smallest value in the dataset from the largest value.
• This gives you a sense of the extent of variability within the dataset.

For the data presented, the highest value is 10 and the lowest is 2, hence, the range is 10 - 2 = 8. This tells us that the values in the dataset span a distance of 8 units. While it is useful for quick insights, remember the range can be sensitive to outliers, or unusually low or high values.

Unpacking the Interquartile Range

The interquartile range (IQR) offers a broader picture of data spread by focusing on the middle portion of a data set, which is more stable since it's less affected by outliers. To calculate the IQR:

• First, determine the lower quartile (Q1), which is the value at the 25th percentile.
• Then, determine the upper quartile (Q3), or the value at the 75th percentile.
• The IQR is then calculated as the difference between Q3 and Q1.

For the given exercise, Q1 was found to be 6 and Q3 was 9, resulting in an IQR of 3. The IQR is especially helpful for understanding the variation in the central portion of a dataset, thus, providing a clear picture of the data without distortion by extremes.