Problem 15 - Algebra 2 and Trigonometry

Question

The 14 students on the track team recorded the following number of seconds as their best time for the 100 -yard dash: $$ \begin{array}{lllllll}{13.5} & {13.7} & {13.1} & {13.0} & {13.3} & {13.2} & {13.0} \\ {12.8} & {13.4} & {13.3} & {13.1} & {12.7} & {13.2} & {13.5}\end{array} $$ Find the range and the interquartile range.

Step-by-Step Solution

Answer

Range: 1.0, Interquartile Range: 0.4.

Step 1: Arrange the Data

First, arrange the times in ascending order: \(12.7, 12.8, 13.0, 13.0, 13.1, 13.1, 13.2, 13.2, 13.3, 13.3, 13.4, 13.5, 13.5, 13.7\).

Step 2: Find the Range

The range is the difference between the maximum and minimum values in the data set. Here, the maximum value is 13.7 and the minimum is 12.7. Thus, the range is \(13.7 - 12.7 = 1.0\).

Step 3: Determine the Median

To find the interquartile range, we first need the median. With 14 values, the median is the average of the 7th and 8th values (13.2 and 13.2). Therefore, the median is \( \frac{13.2+13.2}{2} = 13.2\).

Step 4: Calculate Q1

Q1 (the first quartile) is the median of the first half of the data (first 7 values: 12.7, 12.8, 13.0, 13.0, 13.1, 13.1, 13.2). The median of these values (4th position) is 13.0.

Step 5: Calculate Q3

Q3 (the third quartile) is the median of the second half of the data (last 7 values: 13.2, 13.3, 13.3, 13.4, 13.5, 13.5, 13.7). The median of these values (11th position) is 13.4.

Step 6: Find the Interquartile Range (IQR)

The interquartile range is the difference between Q3 and Q1. Here, the IQR is \(13.4 - 13.0 = 0.4\).

Key Concepts

Range

The range is the simplest measure of dispersion in a data set. It tells us how far apart the smallest and largest numbers are from each other. To calculate the range, you simply subtract the smallest value from the largest value in the set.

In the track team's data, the fastest time is 12.7 seconds and the slowest is 13.7 seconds. Hence, the range is:

• Maximum time (slowest): 13.7 seconds
• Minimum time (fastest): 12.7 seconds
• Range = 13.7 - 12.7 = 1.0 second

This tells us that within the recorded times, the difference between the fastest and slowest run is just 1 second.

Interquartile Range

The interquartile range (IQR) provides insight into the spread of the middle half of the data set. It eliminates outliers by focusing on the central portion, providing a clearer picture of variability among typical values.

To obtain the IQR, we subtract the first quartile (Q1) from the third quartile (Q3). This range is:

• Q3 represents the 75th percentile and is 13.4 seconds.
• Q1 represents the 25th percentile and is 13.0 seconds.
• IQR = Q3 - Q1 = 13.4 - 13.0 = 0.4 seconds

This means that the middle 50% of the team's dash times are spread over a range of 0.4 seconds.

Median

The median is the value that separates a data set into two equal halves. It is considered a more robust measure of central tendency compared to the average, especially in skewed distributions. For our data set of 14 times, the median is the average of the 7th and 8th values when arranged in order.

Here, both the 7th and 8th values are 13.2 seconds. Thus, the median is:

• Median = (13.2 + 13.2) / 2 = 13.2 seconds

This tells us that half of the students ran faster than 13.2 seconds while the other half ran slower.

Quartiles

Quartiles divide your data set into four equal parts. They help to understand the data's distribution and variability. The first quartile (Q1) is the median of the first half, and the third quartile (Q3) is the median of the second half.

For the track team, we determine the following:

• Q1: Median of first half (12.7, 12.8, 13.0, 13.0, 13.1, 13.1, 13.2) is 13.0 seconds.
• Q3: Median of second half (13.2, 13.3, 13.3, 13.4, 13.5, 13.5, 13.7) is 13.4 seconds.

Thus, the quartiles tell us that the middle 50% of data lies between 13.0 and 13.4 seconds, providing a more detailed view of the distribution compared to only looking at the range.