2005 AMC 10B Problem 19

The remaining share is $100 - (10 + 25 + 20 + 15) = 30\%$, who scored $95$.

Mean. Weight each score by its percent:
$$0.10(70) + 0.25(80) + 0.20(85) + 0.15(90) + 0.30(95) = 7 + 20 + 17 + 13.5 + 28.5 = 86.$$

Median. Order the scores from low to high and accumulate the percents: $70$ covers the bottom $10\%$, $80$ brings it to $35\%$, and $85$ brings it to $55\%$. The $50\%$ mark falls inside the $85$ group, so the median is $85$. (With an even class size the two middle students both score $85$.)

The difference is $86 - 85 = 1$.

The answer is $\boxed{\textbf{(B)}\ 1}$.

Percentages are weights: the mean is the dot product of scores with weights, and the median is found by walking up the cumulative distribution until half the class is covered. No actual class size is needed, but choosing one (say $20$ students) is a fine way to make the counting concrete.

Pretend there are $20$ students: $2, 5, 4, 3, 6$ of them with the five scores. The 10th and 11th scores are both $85$, and the total $1720$ divided by $20$ is $86$.

• Taking the median to be the middle listed score (85 happens to be it) without checking cumulative percents, or averaging the five scores unweighted.
• Forgetting that the "rest" is $30\%$, not $20\%$, which shifts the mean.