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08 Apr 2020, 18:46
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
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Question Stats:
56% (02:18) correct
44%
(02:16)
wrong
based on 234
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\(s\) is the standard deviation of \(3\) real numbers \(x, y\), and \(z\), where \(a = x - y, b = y – z\) and \(c = z - x\). What is the expression of \(s^2\) in terms of \(a, b,\) and \(c\)?
A. \(a^3 + b^3 + c^3\)
B.\((abc)^2\)
C. \(0\)
D. \(a^2 + b^2 + c^2\)
E. \(\frac{1}{9}(a^2 + b^2 + c^2)\)
A. \(a^3 + b^3 + c^3\)
B.\((abc)^2\)
C. \(0\)
D. \(a^2 + b^2 + c^2\)
E. \(\frac{1}{9}(a^2 + b^2 + c^2)\)
09 Apr 2020, 01:55
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\(Variance = SD^2 = \frac{Σ(x_i-x_j)^2}{N^2}\)
\(s^2 = \frac{(x-y)^2+(y-z)^2 + (z-x)^2}{3^2}\)
\(s^2 = \frac{a^2+b^2+c^2}{9}\)
OR
\(Mean = \frac{x+y+z}{3}\)
variance= s^2= \(([\frac{x+y+z}{3} -x]^2 + [\frac{x+y+z}{3} -y]^2 + [\frac{x+y+z}{3} -z]^2)/3\)
\(s^2 = \frac{6(x^2+y^2+z^2 -xy-yz-zx)}{ 3^3}\)
\(s^2 = \frac{(2x^2+2y^2+2z^2 -2xy-2yz-2zx)}{ 3^2}\)
\(s^2 = \frac{(x^2+y^2-2xy +y^2+z^2-2yz+z^2+x^2-2zx)}{ 3^2}\)
\(s^2 = \frac{(x-y)^2 + (y-z)^2 +(z-x)^2}{ 3^2}\)
\(s^2 = \frac{a^2 + b^2 +c^2}{ 3^2}\)
\(s^2 = \frac{(x-y)^2+(y-z)^2 + (z-x)^2}{3^2}\)
\(s^2 = \frac{a^2+b^2+c^2}{9}\)
OR
\(Mean = \frac{x+y+z}{3}\)
variance= s^2= \(([\frac{x+y+z}{3} -x]^2 + [\frac{x+y+z}{3} -y]^2 + [\frac{x+y+z}{3} -z]^2)/3\)
\(s^2 = \frac{6(x^2+y^2+z^2 -xy-yz-zx)}{ 3^3}\)
\(s^2 = \frac{(2x^2+2y^2+2z^2 -2xy-2yz-2zx)}{ 3^2}\)
\(s^2 = \frac{(x^2+y^2-2xy +y^2+z^2-2yz+z^2+x^2-2zx)}{ 3^2}\)
\(s^2 = \frac{(x-y)^2 + (y-z)^2 +(z-x)^2}{ 3^2}\)
\(s^2 = \frac{a^2 + b^2 +c^2}{ 3^2}\)
MathRevolution
10 Apr 2020, 17:15
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=>
Assume \(m = (\frac{1}{3})(x + y + z)\), which is the average of \(x, y,\) and \(z.\)
\(x – m = x – (\frac{1}{3})(x + y + z) = (\frac{1}{3})(3x – x – y - z) = (\frac{1}{3})(2x – y - z) = (\frac{1}{3})(x – y - (z - x)) = (\frac{1}{3})(a - c).\)
We have \(y – m = (\frac{1}{3})(b - a)\) and \(z – m = (\frac{1}{3})(c - b)\) similarly.
Then, we have a + b + c = (x - y) + (y - z) + (z - x) = 0.
\(s^2 = (\frac{1}{3})[(x - m)^2 + (y - m)^2 + (z - m)^2]\)
\(= (\frac{1}{3})(\frac{1}{3})[(a - c)^2 + (b - a)^2 + (c - b)^2]\)
\(= (\frac{1}{27})[(a - c)^2 + (b - a)^2 + (c - b)^2]\)
\(= (\frac{1}{27})[(a – c)(a – c) + (b – a)(b -a ) + (c – b)(c – b)]\)
\(= (\frac{1}{27})(a^2 – 2ac + c^2 + b^2 – 2ab + a^2 + c^2 – 2bc + b^2)\)
\(= (\frac{1}{27})[2(a^2 + b^2 + c^2) - 2(ab + bc + ca)]\)
\(= (\frac{1}{27})[3(a^2 + b^2 + c^2) - (a + b + c)^2]\)
\(= (\frac{1}{27})[3(a^2 + b^2 + c^2)],\) since \(a + b + c = 0\)
\(= (\frac{1}{9})(a^2 + b^2 + c^2)\)
Therefore, E is the answer.
Answer: E
Assume \(m = (\frac{1}{3})(x + y + z)\), which is the average of \(x, y,\) and \(z.\)
\(x – m = x – (\frac{1}{3})(x + y + z) = (\frac{1}{3})(3x – x – y - z) = (\frac{1}{3})(2x – y - z) = (\frac{1}{3})(x – y - (z - x)) = (\frac{1}{3})(a - c).\)
We have \(y – m = (\frac{1}{3})(b - a)\) and \(z – m = (\frac{1}{3})(c - b)\) similarly.
Then, we have a + b + c = (x - y) + (y - z) + (z - x) = 0.
\(s^2 = (\frac{1}{3})[(x - m)^2 + (y - m)^2 + (z - m)^2]\)
\(= (\frac{1}{3})(\frac{1}{3})[(a - c)^2 + (b - a)^2 + (c - b)^2]\)
\(= (\frac{1}{27})[(a - c)^2 + (b - a)^2 + (c - b)^2]\)
\(= (\frac{1}{27})[(a – c)(a – c) + (b – a)(b -a ) + (c – b)(c – b)]\)
\(= (\frac{1}{27})(a^2 – 2ac + c^2 + b^2 – 2ab + a^2 + c^2 – 2bc + b^2)\)
\(= (\frac{1}{27})[2(a^2 + b^2 + c^2) - 2(ab + bc + ca)]\)
\(= (\frac{1}{27})[3(a^2 + b^2 + c^2) - (a + b + c)^2]\)
\(= (\frac{1}{27})[3(a^2 + b^2 + c^2)],\) since \(a + b + c = 0\)
\(= (\frac{1}{9})(a^2 + b^2 + c^2)\)
Therefore, E is the answer.
Answer: E
