Question

for exercises 5 and 6, use the dot plot to answer the questions. the dot plot shows the number of days each cast member needs to memorize the lines for the school play. 5. what is the median number of days it takes the cast members to memorize their lines? what is the mode? which measure do you think better represents the center? why?

Explanation:

Step1: Count the number of data points

First, we count the number of dots (data points) in the dot plot. Let's list out the number of dots for each day:

• Day 2: 1 dot
• Day 3: 1 dot
• Day 4: 1 dot
• Day 5: 1 dot
• Day 6: 1 dot
• Day 7: 2 dots
• Day 8: 1 dot
• Day 9: 2 dots
• Day 11: 1 dot
• Day 13: 3 dots
• Day 14: 1 dot

Now, sum them up: \(1 + 1+1 + 1+1 + 2+1 + 2+1 + 3+1= 15\) data points.

Step2: Find the median

The median is the middle value when the data is ordered. For \(n = 15\) (odd number of data points), the median is the \(\frac{n + 1}{2}=\frac{15+ 1}{2}=8\)-th value when ordered from least to greatest.

Let's order the data by counting the cumulative number of dots:

• Days 2 - 6: Each has 1 dot, so cumulative up to day 6: \(1\times5 = 5\) dots.
• Day 7: 2 dots, cumulative: \(5 + 2=7\) dots.
• Day 8: 1 dot, cumulative: \(7+ 1 = 8\) dots. So the 8 - th value is at day 8. Thus, the median is 8.

Step3: Find the mode

The mode is the value that appears most frequently. Looking at the counts:

• Day 13 has 3 dots, which is more than any other day (others have at most 2 dots). So the mode is 13.

Step4: Determine the better measure of center

The median is often a better measure of the center when there are outliers or the data is skewed. In this dot plot, the mode (13) is higher than most of the data points (since there are only 3 dots at 13 and 1 at 14, while most are between 2 - 9 and 11), so the data is skewed right. The median is not affected by extreme values (outliers) as much as the mode (or mean, if we calculated it) and represents the middle of the data better.

Answer:

Median: 8; Mode: 13; The median better represents the center because the data is skewed right (with a higher - frequency value at 13 which is an outlier - like value relative to the main cluster of data), and the median is less affected by extreme values (or skewness) than the mode.