Write the mean as $\dfrac{1 \cdot a + 2 \cdot 2a + 3 \cdot 3a + 4 \cdot 2a + 5 \cdot a}{a + 2a + 3a + 2a + a}$. What happens to the $a$?

Instead of listing every number, count how many values are less than 3, equal to 3, and greater than 3. Which value ends up in the middle when the data are ordered?

Once you know the mean and the median as specific numbers, subtract the median from the mean to answer "How much greater is the mean than the median?"

In Desmos, define a positive integer value for $a$ (for example, type a = 1). Then enter the expression (1*a + 2*2a + 3*3a + 4*2a + 5*a) / (a + 2a + 3a + 2a + a) to see the numerical value of the mean.

For the same value of $a$, list out the data. For example, if a = 1, the data set is [1, 2, 2, 3, 3, 3, 4, 4, 5]. Type median([1,2,2,3,3,3,4,4,5]) in Desmos to see the median for that case. Try another value (like a = 2) by expanding the list to match the frequencies and check that the median stays the same.

Once you have the mean and median from Desmos, type an expression like mean_value - median_value (replacing with the actual numbers or earlier expressions you found) to see how much greater the mean is than the median.

The table tells you how many times each value appears.

• The value 1 appears $a$ times.
• The value 2 appears $2a$ times.
• The value 3 appears $3a$ times.
• The value 4 appears $2a$ times.
• The value 5 appears $a$ times.

So the total number of data points is

The mean is

Compute the sum of all values using the frequencies:

• From the 1s: $1 \cdot a = a$
• From the 2s: $2 \cdot 2a = 4a$
• From the 3s: $3 \cdot 3a = 9a$
• From the 4s: $4 \cdot 2a = 8a$
• From the 5s: $5 \cdot a = 5a$

Total sum:

So the mean is

The $a$ cancels, so the mean is 3 for any positive integer $a$.

To find the median, think about how many values are below, equal to, and above 3.

• Below 3: values 1 and 2

  • 1 appears $a$ times
  • 2 appears $2a$ times
  • Total below 3: $a + 2a = 3a$

• Equal to 3:

  • 3 appears $3a$ times

• Above 3: values 4 and 5

  • 4 appears $2a$ times
  • 5 appears $a$ times
  • Total above 3: $2a + a = 3a$

So there are $3a$ values below 3, $3a$ values equal to 3, and $3a$ values above 3, for a total of $9a$ values.

Half of the data is $\dfrac{9a}{2} = 4.5a$. The number of values at or below 3 is $3a + 3a = 6a$, which is more than half of the data; the number of values at or above 3 is also $6a$, which is more than half.

That means 3 is the middle value of the ordered list (the median), because at least half of the data are $\le 3$ and at least half are $\ge 3$. So the median is 3.

From the earlier steps:

• Mean $= 3$
• Median $= 3$

The question asks, "How much greater is the mean of the set of data than the median?"

Compute the difference:

So the mean is greater than the median by 0.