When you compute the mean, both the numerator (sum) and the denominator (number of values) will include $a$. Think about what happens to $a$ when you simplify the fraction.

There are $10a$ data points. For an even number of points, which two positions do you average to get the median? Use cumulative frequencies (adding up counts as you go down the table) to see which value those positions land on.

Once you know the mean and the median, subtract the median from the mean and match that difference to the closest answer choice.

In Desmos, type the expression for the mean:

Desmos will simplify this expression to a constant decimal; that decimal is the mean of the data set. Note its value.

Set $a=1$ in Desmos (for example, type a = 1). Then define a list of data reflecting the table for $a=1$, such as:

L = {1, 2, 2, 3, 3, 3, 4, 4, 5, 10}

Now type median(L); the value Desmos returns is the median for this distribution (and it will be the same for any positive integer $a$). Note this value.

In Desmos, type a final expression that subtracts the median from the mean, using the values you observed (for example, mean_value - median(L) if you stored the mean). The resulting decimal is how much greater the mean is than the median; compare this number with the answer choices to select the correct one.

Add the frequencies to confirm how many data points are in the set:

So there are $10a$ data values in the set (as stated in the problem).

Multiply each value by its frequency and add:

• Value $1$ appears $a$ times: contributes $1 \cdot a = a$.
• Value $2$ appears $2a$ times: contributes $2 \cdot 2a = 4a$.
• Value $3$ appears $3a$ times: contributes $3 \cdot 3a = 9a$.
• Value $4$ appears $2a$ times: contributes $4 \cdot 2a = 8a$.
• Value $5$ appears $a$ times: contributes $5 \cdot a = 5a$.
• Value $10$ appears $a$ times: contributes $10 \cdot a = 10a$.

Now add these contributions:

So the sum of all data values is $37a$.

The mean is

The $a$ in the numerator and denominator cancels:

So the mean of the data set is $3.7$.

There are $10a$ data values, which is an even number. For $10a$ values, the median is the average of the $5a$-th and $(5a+1)$-st values in order.

List the cumulative counts as you go up through the values:

• All the 1s occupy positions $1$ through $a$.
• All the 2s then occupy positions $a+1$ through $3a$ (because $a + 2a = 3a$).
• All the 3s then occupy positions $3a+1$ through $6a$ (because $3a + 3a = 6a$).

The $5a$-th and $(5a+1)$-st positions are between $3a+1$ and $6a$. Since this whole block is made of the value $3$, both the $5a$-th and $(5a+1)$-st values are $3$.

From the previous step, both middle positions are $3$, so the median is

The mean is $3.7$, so the difference "mean minus median" is

Therefore, the mean of the data set is $0.7$ greater than the median, which corresponds to choice C.