In a mathematics class, half of the students scored 78 on an achievement test. With the exception of a few students who scored 55, the remaining students scored 74. Which statement about the distribution is true?

Step-by-step solution

1. 1

   Write the counts symbolically. For an even class size $N$: $\tfrac{N}{2}$ students scored $78$, $k \ge 1$ scored $55$, and $\tfrac{N}{2} - k$ scored $74$. The answer must hold for every admissible $k$, so keeping it as a symbol is the safe route.

2. 2

   Determine the median. Sorted low to high, the $55$s come first ($k$ of them), then the $74$s ($\tfrac{N}{2} - k$ of them), then the $78$s ($\tfrac{N}{2}$ of them). The two central positions $\tfrac{N}{2}$ and $\tfrac{N}{2}+1$ hold $74$ and $78$:

   $\text{median} = \frac{74 + 78}{2} = 76$

3. 3

   Compute the mean.

   $\text{mean} = \frac{78\cdot\tfrac{N}{2} + 55k + 74\left(\tfrac{N}{2}-k\right)}{N} = \frac{76N - 19k}{N} = 76 - \frac{19k}{N}$

   Each student who scored $55$ instead of $74$ removes $19$ points from the total, which is exactly the $-19k$ in the numerator.

4. 4

   Compare. Because $k \ge 1$,

   $\text{mean} = 76 - \frac{19k}{N} < 76 = \text{median}$

   so option D is correct: the low outliers move the mean but not the median.

5. 5

   Eliminate the alternatives. B and C contradict the strict inequality just proved. A is false because the mode is $78$ (the most frequent score, held by half the class) and the mean is below $76 < 78$.

6. 6

   Numerical check. $N = 20$, $k = 3$: ten $78$s, three $55$s, seven $74$s. Mean $= \tfrac{780 + 165 + 518}{20} = 73.15$, median $= 76$, mode $= 78$. The formula predicts $76 - \tfrac{57}{20} = 73.15$ — an exact match.