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Difficulty: Very hardDescriptive Statistics and Data Representations

A set of 1010 distinct positive integers has a median of 2525 and a range of 3030. What is the greatest possible value of the mean of these 1010 integers?

  1. A
    25
  2. B
    29
  3. C
    31
  4. 33Answer
  5. E
    37

Answer

The greatest possible value of the mean of the 10 integers is 33.
The correct answer is 33. To find the greatest possible mean of the 10 distinct positive integers, we must maximize their sum. Let the sorted integers be x1<x2<x3<x4<x5<x6<x7<x8<x9<x10x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7 < x_8 < x_9 < x_{10}. The median is the average of the 5th and 6th terms, so x5+x62=25\frac{x_5 + x_6}{2} = 25, or x5+x6=50x_5 + x_6 = 50. Since the integers are distinct, the maximum value for x5x_5 is 24, which forces x6=26x_6 = 26. To maximize the sum of the first five terms, they should be consecutive integers ending at 24: x1=20,x2=21,x3=22,x4=23,x5=24x_1 = 20, x_2 = 21, x_3 = 22, x_4 = 23, x_5 = 24. Since the range is 30, the maximum value is x10=x1+30=20+30=50x_{10} = x_1 + 30 = 20 + 30 = 50. To maximize the remaining terms in the upper half, we choose the largest possible distinct integers less than 50: x7=47,x8=48,x9=49x_7 = 47, x_8 = 48, x_9 = 49. The maximum sum is 20+21+22+23+24+26+47+48+49+50=33020 + 21 + 22 + 23 + 24 + 26 + 47 + 48 + 49 + 50 = 330, yielding a maximum mean of 330/10=33330 / 10 = 33.

Step-by-Step Solution

1
Define the variables and apply the median constraint.
Let the 10 sorted distinct positive integers be x1<x2<x3<x4<x5<x6<x7<x8<x9<x10x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7 < x_8 < x_9 < x_{10}. The median is the average of the 5th and 6th terms: x5+x62=25\frac{x_5 + x_6}{2} = 25, which means x5+x6=50x_5 + x_6 = 50.
Since the number of terms is even, the median is the average of the two middle terms.
2
Maximize the first five integers to find the maximum value of the first term.
Since the integers are distinct, we must have x5<x6x_5 < x_6. With x5+x6=50x_5 + x_6 = 50, the maximum possible integer value for x5x_5 is 24 (which makes x6=26x_6 = 26). To maximize the sum, we make the preceding terms as large as possible: x4=23x_4 = 23, x3=22x_3 = 22, x2=21x_2 = 21, and x1=20x_1 = 20.
To maximize the mean, we must maximize the sum of all terms, which requires making each term as large as possible within the distinct integer constraints.
3
Apply the range constraint to find the maximum value of the last term.
The range is 30, so x10−x1=30x_{10} - x_1 = 30. Since the maximum value of x1x_1 is 20, the maximum possible value for x10x_{10} is 20+30=5020 + 30 = 50.
The range of a dataset is the difference between the maximum and minimum values.
4
Maximize the remaining terms in the upper half of the dataset.
We have x6=26x_6 = 26. The remaining terms must satisfy 26<x7<x8<x9<x10=5026 < x_7 < x_8 < x_9 < x_{10} = 50. To maximize the sum, we choose the largest possible distinct integers for these slots: x9=49x_9 = 49, x8=48x_8 = 48, and x7=47x_7 = 47.
This maximizes the sum of the upper half of the dataset under the constraint that the maximum value is 50.
5
Calculate the maximum sum and the resulting maximum mean.
Sum = 20+21+22+23+24+26+47+48+49+50=33020 + 21 + 22 + 23 + 24 + 26 + 47 + 48 + 49 + 50 = 330. Mean = 33010=33\frac{330}{10} = 33.
The mean is calculated by dividing the sum of the elements by the number of elements.

Key Concept

Maximizing the mean of a bounded dataset using median, range, and distinctness constraints.
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Topics

Descriptive Statistics and Data Representations

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