Definition D1
The data arranged in order of size are written x₍₁₎ ≤ ⋯ ≤ x₍ₙ₎, and when n = 2m + 1 the value x₍ₘ₊₁₎ is called the median, written Me.
Proof.
(1)Σaᵢ denotes a₁ + ⋯ + aₙ, and u ≤ v denotes u < v or u = v.convention
(2)By trichotomy and transitivity such an arrangement exists, and the value in the j-th position is uniquely determined.A7, A8
(3)Sorting merely rearranges the data, so Σx₍ⱼ₎ = Σxᵢ.(2), A1, A2
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Definition D2
The number of data points equal to a value v is called the frequency of v, written f(v). A value v where f(v) is largest is called a mode.
Proof.
This definition does not say the mode is unique. If several values attain the largest f, there are several modes.convention
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Definition D3
The data are called symmetric about c when for some c, x₍ⱼ₎ + x₍ₙ₊₁₋ⱼ₎ = 2c holds for every j.
Proof.
This means that when the sorted data are paired from both ends, the sum of each pair is always the same.D1, convention
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Theorem T1
If n = 2m + 1 and the data are symmetric about c, then x̄ = c and Me = c.
Proof.
(1)By hypothesis, for every j, x₍ⱼ₎ + x₍ₙ₊₁₋ⱼ₎ = 2cA10, D3
(2)Summing these equations for j = 1 to n, the left side is Σx₍ⱼ₎ + Σx₍ₙ₊₁₋ⱼ₎.(1), A3
(3)As j runs from 1 to n, n + 1 − j also runs through 1 to n exactly once, so Σx₍ₙ₊₁₋ⱼ₎ = Σx₍ⱼ₎A1, A2
(4)Sorting merely rearranges, so Σx₍ⱼ₎ = ΣxᵢD1
(5)Summing the right sides gives Σ2c = 2ncA4
(6)Therefore 2Σxᵢ = 2nc(2), (3), (4), (5), E
(7)Cancelling 2 gives Σxᵢ = nc, and by the definition of the mean x̄ = c(6), A5, A6
(8)For j = m + 1 we have n + 1 − j = m + 1, so (1) becomes 2x₍ₘ₊₁₎ = 2c.(1), N
(9)Cancelling 2 gives x₍ₘ₊₁₎ = c, and by the definition of the median Me = c(8), A5, D1
(10)Therefore x̄ = Me = c(7), (9)
∎
Theorem T2
On non-symmetric data the three can diverge, and the mean lies toward the isolated value.
Proof.
(1)Consider the data 1, 2, 2, 3, 92. The sum is 100 and n = 5, so x̄ = 20A6, N
(2)The data are already sorted, so Me = 2, and f(2) = 2 is the largest frequency, so the mode is also 2.(1), D1, D2, N
(3)x₍₁₎ + x₍₅₎ = 93 and x₍₂₎ + x₍₄₎ = 5, so the condition that the pair sums agree fails. These data are symmetric about no c.(1), D3, N
(4)The mean 20 exceeds the median 2, and lies toward the isolated large value 92.(1), (2), N
(5)Consider the data 8, 97, 98, 98, 99. The sum is 400, so x̄ = 80, Me = 98, and the mode is also 98.A6, D1, D2, N
(6)Here the mean 80 is below the median 98, and lies toward the isolated small value 8.(5), N
(7)The mean uses the sizes of the values through Σxᵢ, while the median uses only position and the mode only frequency.A6, D1, D2
(8)Therefore a value far off to one side moves only the mean toward it, and the direction of the split is set by which way the data are skewed.(4), (6), (7)
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Theorem T3
There exist non-symmetric data whose mean and median are equal.
Proof.
(1)Consider the data 0, 2, 4, 5, 9. The sum is 20 and n = 5, so x̄ = 4A6, N
(2)The data are already sorted, so the third value is the median and Me = 4. That is, x̄ = Me(1), D1, N
(3)But x₍₁₎ + x₍₅₎ = 9 and x₍₂₎ + x₍₄₎ = 7, so these data are not symmetric.(1), D3, N
(4)Therefore the statement 'if the mean and median are equal, the data are symmetric' is false.(2), (3), A9
(5)Symmetry cannot be read back from the equality of mean and median.(4)
∎
Theorem T4
The mode may fail to be a single number.
Proof.
(1)In the data 1, 1, 2, 2 we have f(1) = 2, f(2) = 2, and no larger frequency.D2, N
(2)Therefore the modes are both 1 and 2.(1), D2
(3)In the data 0, 2, 4, 5, 9 every value has frequency 1, so all five values are modes.D2, N
(4)The mean is determined as a single number by n x̄ = Σxᵢ, and when n is odd the median is also determined by a single position.A6, D1
(5)But the mode is not a value: it is the set of values with the largest frequency, and that set may have several elements.(2), (3), D2
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Corollary C
The condition for the three to diverge is not the presence or absence of symmetry, but the fact that they read different things from the data.
Proof.
(1)If the data are symmetric about c, the mean and median coincide at the single value c.T1
(2)When symmetry breaks they can diverge, and the mean shifts toward the skew.T2
(3)But the mean and median can be equal without symmetry, so the split cannot be diagnosed by symmetry alone.T3
(4)The mode may fail to be a single number in the first place, so the three cannot be compared on equal footing.T4
(5)The mean reads the sizes of the values, the median reads position, and the mode reads frequency. Since they read different things, diverging is the norm and coinciding is the special event.(2), (3), (4), A6, D1, D2
(6)Symmetry is merely one condition that gathers the three together; it is not the cause of the split.(1), (5)
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