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Difficulty: Very hardMean, Median, and Mode

An analytics team tracked the monthly recurring revenue (in thousands of dollars) for 10 distinct client accounts of a SaaS company. The arithmetic mean of the monthly revenues across these 10 accounts was $45\$45 thousand, and the median monthly revenue was $38\$38 thousand. If the revenue for each account was a distinct positive integer (in thousands of dollars), what is the maximum possible monthly revenue, in thousands of dollars, generated by the single highest-performing account?

Answer: 241 thousand dollars

Answer

The maximum possible monthly revenue generated by the single highest-performing account is 241 thousand dollars.
The total sum of the 10 distinct positive integer revenues is 10×45=45010 \times 45 = 450 thousand dollars. To maximize the largest term (x10x_{10}), the sum of the remaining 9 terms (x1x_1 through x9x_9) must be minimized. The smallest 4 positive integers are 1,2,3,41, 2, 3, 4 (sum = 1010). Given a median of 38, the two central terms must satisfy x5+x6=76x_5 + x_6 = 76. Because all values are distinct integers and x5<x6x_5 < x_6, the smallest possible integer value for x6x_6 is 39 (making x5=37x_5 = 37). To minimize x7,x8,x9x_7, x_8, x_9, they should be the smallest distinct integers greater than 39, which are 40,41,4240, 41, 42. Summing these minimal values gives 10+76+40+41+42=20910 + 76 + 40 + 41 + 42 = 209. Subtracting this sum from the total sum yields 450−209=241450 - 209 = 241 thousand dollars.

Step-by-Step Solution

1
Calculate the total revenue across all 10 client accounts.
Total sum = 10×45=45010 \times 45 = 450 thousand dollars.
The sum of a data set is equal to the number of elements multiplied by the arithmetic mean.
2
Use the median to determine the sum of the 5th and 6th terms.
Since n=10n = 10, the median is x5+x62=38\frac{x_5 + x_6}{2} = 38, so x5+x6=76x_5 + x_6 = 76.
In an ordered set of an even number of elements, the median is the average of the two central terms.
3
Formulate the minimization strategy to maximize the largest term (x10x_{10}).
To maximize x10=450−(x1+x2+⋯+x9)x_{10} = 450 - (x_1 + x_2 + \dots + x_9), minimize the sum of the first 9 terms.
For a fixed sum, maximizing one element requires minimizing all other elements in the set.
4
Minimize the first four terms (x1,x2,x3,x4x_1, x_2, x_3, x_4).
Assign x1=1,x2=2,x3=3,x4=4x_1 = 1, x_2 = 2, x_3 = 3, x_4 = 4, which sum to 1010.
The revenues are distinct positive integers, so the smallest possible values are the first 4 positive integers.
5
Minimize x6x_6 to minimize the subsequent terms (x7,x8,x9x_7, x_8, x_9).
Since x5<x6x_5 < x_6 and x5+x6=76x_5 + x_6 = 76, the smallest integer value for x6x_6 is 3939 (which sets x5=37x_5 = 37). Then set x7=40,x8=41,x9=42x_7 = 40, x_8 = 41, x_9 = 42.
Minimizing x6x_6 allows x7,x8,x9x_7, x_8, x_9 to take the smallest possible distinct integer values greater than x6x_6.
6
Calculate the maximum value for x10x_{10}.
x10=450−(10+76+40+41+42)=450−209=241x_{10} = 450 - (10 + 76 + 40 + 41 + 42) = 450 - 209 = 241.
Subtracting the minimal sum of the first 9 elements from the total sum gives the theoretical maximum for the 10th term.

Key Concept

Extremal optimization of distinct statistics data sets given mean and median constraints
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Topics

Mean, Median, and Mode

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