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01 Jun 2026, 11:26
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Maximum value of D: 6
Minimum value of d: 1
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40%
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Let x and y be elements of the set {5, 12, 19, 33, 61, 82}.Consider the set S = {31, 14, 64, 22, 43, 66} with mean mₛ, the set T = {x, 31, 14, 64, 22, 43, 66} with mean mₜ, and the set U = {y, 31, 14, 64, 22, 43, 66} with mean mu . The value of x maximizes D = |mₛ - mₜ| and the value of y minimizes d = |mₛ - mu| . Select the maximum value of D and select the minimum value of d. Make two selections, one in each column.
| Maximum value of D | Minimum value of d | |
| 1 | ||
| 2 | ||
| 3 | ||
| 4 | ||
| 5 | ||
| 6 |
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Official Answer
Maximum value of D: 6
Minimum value of d: 1
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01 Jun 2026, 11:34
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Official answer :
The mean mₛ is (31 + 14 + 64 + 22 + 43 + 66)/6 = 240/6 = 40.
Notice that the set T is the set S with one member x added, and the set U is the set S with one member y added. Thus, T and U each have 7 members, including all 6 members of S.
So mₜ = (240 + x)/7, and mu = (240 + y)/7.
The value of x that maximizes D = |mₛ - mₜ| is the element of {5, 12, 19, 33, 61, 82} that maximizes |40 - mₜ|, or in other words the element whose value is farthest from 40.
That element is 82, which gives us mₜ = (240 + 82)/7 = 46, so that the maximum value of D is |40 - 46| = 6. Likewise, the value of y that minimizes d = |mₛ - mu| is the element of {5, 12, 19, 33, 61, 82} that minimizes |40 - mu|, or in other words the element whose value is nearest 40.
That element is 33, which gives us mu = (240 + 33)/7 = 39, so that the minimum value of d = |40 - 39| = 1.
First column: The correct answer is 6.
Second column: The correct answer is 1.
The mean mₛ is (31 + 14 + 64 + 22 + 43 + 66)/6 = 240/6 = 40.
Notice that the set T is the set S with one member x added, and the set U is the set S with one member y added. Thus, T and U each have 7 members, including all 6 members of S.
So mₜ = (240 + x)/7, and mu = (240 + y)/7.
The value of x that maximizes D = |mₛ - mₜ| is the element of {5, 12, 19, 33, 61, 82} that maximizes |40 - mₜ|, or in other words the element whose value is farthest from 40.
That element is 82, which gives us mₜ = (240 + 82)/7 = 46, so that the maximum value of D is |40 - 46| = 6. Likewise, the value of y that minimizes d = |mₛ - mu| is the element of {5, 12, 19, 33, 61, 82} that minimizes |40 - mu|, or in other words the element whose value is nearest 40.
That element is 33, which gives us mu = (240 + 33)/7 = 39, so that the minimum value of d = |40 - 39| = 1.
First column: The correct answer is 6.
Second column: The correct answer is 1.
02 Jun 2026, 03:00
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A shortcut once you see that 82 and 33 are the numbers that are farthest from and closest to 40 is to divide the difference between each and 40 by the number of elements in the set, 7, to get the differences between the averages.
(82 - 40)/7 = 6
(40 - 33)/7 = 1
(82 - 40)/7 = 6
(40 - 33)/7 = 1
