In a mathematics class, half of the students scored 40 on an achievement test. With the exception of a few students who scored 80, the remaining students scored 58. Which statement about the distribution is true: the mean equals the median, the mean is less than the mode, the mean is less than the median, or the mean is greater than the median?

Step-by-step solution

1. 1

   Set up notation instead of reasoning by feel. Let $N$ be the class size (even), so $\tfrac{N}{2}$ students scored $40$, $k$ students scored $80$ with $k$ small and $k \ge 1$, and the remaining $\tfrac{N}{2} - k$ scored $58$. Everything below follows from these counts alone, so the conclusion cannot depend on how "a few" is interpreted.

2. 2

   Locate the median exactly. Sorted ascending, positions $1$ through $\tfrac{N}{2}$ all hold the value $40$; positions $\tfrac{N}{2}+1$ onward hold $58$ then $80$. For even $N$ the median averages positions $\tfrac{N}{2}$ and $\tfrac{N}{2}+1$:

   $\text{median} = \frac{40 + 58}{2} = 49$

   This is independent of $k$ — a key point, since it means only the mean can move.

3. 3

   Compute the mean in closed form.

   $\text{mean} = \frac{40\cdot\tfrac{N}{2} + 58\left(\tfrac{N}{2} - k\right) + 80k}{N} = \frac{49N + 22k}{N} = 49 + \frac{22k}{N}$

4. 4

   Compare. Since $k \ge 1$, the term $\tfrac{22k}{N}$ is strictly positive, so

   $\text{mean} = 49 + \frac{22k}{N} > 49 = \text{median}$

   The distribution is skewed right: a small group far above the bulk drags the mean past the median while leaving the median fixed. Option D is correct.

5. 5

   Rule out the other choices. A is false because the mean strictly exceeds $49$. C reverses the inequality. B fails because the mode is $40$ (half the class) while the mean is above $49$, so the mean is greater than the mode, not less.

6. 6

   Check with concrete numbers. With $N = 20$ and $k = 3$: ten scores of $40$, three of $80$, seven of $58$. Mean $= \tfrac{400 + 240 + 406}{20} = 52.3$, median $= 49$, mode $= 40$. With $N = 40$, $k = 5$: mean $= 51.75$, median $= 49$. Both match the formula $49 + \tfrac{22k}{N}$.