Question

Difficulty: Very hardMeasures of Central Tendency (Mean, Median, Mode)

A dataset SS consists of 99 distinct integers. The median of SS is 2020, and the mean of SS is 2424. The mean of the 44 smallest integers in SS is 1010. If the largest integer in SS is removed to form a new dataset RR consisting of 88 integers, what is the maximum possible median of dataset RR?

  1. A
    18.518.5
  2. B
    19.019.0
  3. 19.519.5Answer
  4. D
    20.020.0
  5. E
    20.520.5

Answer

The maximum possible median of dataset RR is 19.519.5.
In the sorted set of 99 distinct integers, the median is the 55 th element, which is 2020. Removing the largest value leaves 88 integers, whose median is the average of the 44 th and 55 th elements. Because all integers are distinct, the 44 th element can be at most 1919. The maximum median is therefore 19+202=19.5\frac{19 + 20}{2} = 19.5.

Step-by-Step Solution

1
Identify the position of the median in the original dataset SS.
In a dataset of 99 sorted distinct integers x1<x2<x3<x4<x5<x6<x7<x8<x9x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7 < x_8 < x_9, the median is the 55 th element, so x5=20x_5 = 20.
For an odd number of ordered elements n=9n=9, the median is at index 9+12=5\frac{9+1}{2} = 5.
2
Determine the formula for the median of the modified dataset RR.
Removing the largest element x9x_9 leaves 88 ordered elements x1,x2,x3,x4,x5,x6,x7,x8x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8. The median of RR is x4+x52=x4+202\frac{x_4 + x_5}{2} = \frac{x_4 + 20}{2}.
For an even number of ordered elements n=8n=8, the median is the average of the two middle terms at indices 44 and 55.
3
Maximize the value of x4x_4.
Since all integers are distinct and x5=20x_5 = 20, x4x_4 must be an integer strictly less than 2020. Thus, the maximum integer value x4x_4 can take is 1919.
Maximizing x4x_4 maximizes the average x4+202\frac{x_4 + 20}{2}.
4
Verify that x4=19x_4 = 19 is achievable under the given mean constraint.
The sum of the smallest 44 integers is 4×10=404 \times 10 = 40. With x4=19x_4 = 19, we require x1+x2+x3=21x_1 + x_2 + x_3 = 21. Choosing distinct integers such as x1=1x_1 = 1, x2=2x_2 = 2, and x3=18x_3 = 18 satisfies 1+2+18=211 + 2 + 18 = 21 and x1<x2<x3<x4x_1 < x_2 < x_3 < x_4.
Confirming feasibility ensures the upper bound 1919 is valid.
5
Calculate the maximum median of dataset RR.
19+202=19.5\frac{19 + 20}{2} = 19.5.
Substitute the maximum value of x4=19x_4 = 19 into the median expression.

Key Concept

Properties of median for odd vs. even datasets and optimization under distinctness constraints
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