Let's start by computing the mean and the median for the given set of data:

Data:

82, 93, 26, 56, 75, 73, 80, 61, 79, 90, 94, 93, 100, 71, 100, 60

Step 1: Mean

The mean (or average) is the sum of all the numbers divided by the total number of values.

Formula for the mean:

$\text{Mean} = \frac{\sum \text{All scores}}{\text{Number of scores}}$

Let's calculate the sum of the data: $82 + 93 + 26 + 56 + 75 + 73 + 80 + 61 + 79 + 90 + 94 + 93 + 100 + 71 + 100 + 60 = 1023$

Now, count the number of data points: $\text{Number of scores} = 16$

Now, calculate the mean: $\text{Mean} = \frac{1023}{16} = 63.9375$

So, the mean is approximately 63.94.

Step 2: Median

The median is the middle number in a data set when the numbers are arranged in ascending order. If there is an even number of data points, the median is the average of the two middle values.

Step 2.1: Arrange the data in ascending order:

26, 56, 60, 61, 71, 73, 75, 79, 80, 82, 90, 93, 93, 94, 100, 100

Step 2.2: Find the middle values

Since we have 16 data points (an even number), the median will be the average of the 8th and 9th values in the ordered list.

The 8th and 9th values are:

• 8th value: 79
• 9th value: 80

Now, calculate the median: $\text{Median} = \frac{79 + 80}{2} = \frac{159}{2} = 79.5$

So, the median is 79.5.

Step 3: Why are the mean and median different?

The mean and median are different because the data set is skewed by the extreme values at the higher end (the two 100s). These high values push the mean higher than the median. In contrast, the median is less affected by extreme values and more reflective of the central position of the data.

Summary:

• Mean: 63.94
• Median: 79.5

Would you like any further details or have any questions?

Here are some related questions to explore:

1. What is the effect of outliers on the mean and median?
2. How would the mean and median change if the highest score (100) were removed?
3. What is the mode of this data set?
4. How do you calculate the range of a data set?
5. How can the standard deviation help in understanding data dispersion?

Tip: Always consider both the mean and median when analyzing data, especially if there are outliers, as they can tell you different stories about the distribution.