A dataset consists of 15 positive integers. The median of the dataset is 24, and the mean … — SAT

A dataset consists of $15$ positive integers. The median of the dataset is $24$ , and the mean is $20$ . The dataset has a unique mode of $32$ , which occurs exactly $5$ times. What is the maximum possible range of the dataset?

Answer: 81

Answer

The maximum possible range of the dataset is 81.

The total sum of the 15 elements is 300. With the median (8th term) equal to 24, we minimize the first 7 terms to 10 by using four 1s and three 2s (since no element other than 32 can repeat 5 or more times). Setting the 9th term to 24 and the 10th through 14th terms to the mode of 32 minimizes the sum of the first 14 terms to 218. This leaves a maximum possible value of 82 for the 15th term. The range is the difference between the maximum and minimum values: 82 - 1 = 81.

Step-by-Step Solution

Calculate the total sum of the dataset.

The sum of all 15 elements is

300

Since the mean of 15 elements is 20, the sum must be

15 \times 20 = 300

Identify the median and set up the sorted terms.

The 8th term

x_8 = 24

For 15 sorted elements, the median is the 8th term.

Minimize the sum of the first 7 terms.

The minimum sum of

x_1

through

x_7

10

, with

x_1 = 1

To minimize the sum, we use the smallest positive integers. However, no value other than 32 can appear 5 or more times. Thus, we can have at most four 1s and three 2s:

4(1) + 3(2) = 10

Minimize the 9th term.

x_9 = 24

Since the dataset is sorted,

x_9 \geq x_8 = 24

. To minimize the sum of the other terms and maximize

x_{15}

, we set

x_9 = 24

Account for the mode of 32.

x_{10} = x_{11} = x_{12} = x_{13} = x_{14} = 32

The mode 32 appears exactly 5 times. Since it is greater than the median 24, these 5 occurrences must be in the upper half of the sorted list.

Calculate the maximum possible value of the 15th term and the range.

x_{15} = 82

, and the range is

81

Subtracting the minimum sum of the first 14 terms from the total sum:

300 - (10 + 24 + 24 + 160) = 82

. The range is

x_{15} - x_1 = 82 - 1 = 81

Key Concept

Calculating measures of center (mean, median, mode) and variability (range) under constrained datasets.

Estimated Time:3m 0s

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