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--- lehrkraefte:blc:miniaufgaben:kw08-2020 [2020/02/16 19:43] – created Ivo Blöchliger
+++ lehrkraefte:blc:miniaufgaben:kw08-2020 [2020/08/09 13:40] (current) – external edit 127.0.0.1
@@ Line 1: / Line 1: @@
+<PRELOAD>
+miniaufgabe.js
+</PRELOAD>
+==== 17. Februar 2020 bis 21. Februar 2020 ====
+=== Montag 17. Februar 2020 ===
+Berechnen Sie den Mittelwert $\mu$ und den Median $\tilde x$ für die folgenden zwei Wertereihen.
+<JS>miniAufgabe("#exomeanmed","#solmeanmed",
+[["a) 12, 18, 11, 11, 8 &nbsp; b) 5, 8, 4, 10, 8, 31 &nbsp; ", "a) $\\mu = 12$, $\\tilde x = 11$ &nbsp; b) $\\mu = 11$, $\\tilde x = 8$ &nbsp; "], ["a) 24, 15, 22, 26, -2 &nbsp; b) 10, 8, 15, 6, 12, 27 &nbsp; ", "a) $\\mu = 17$, $\\tilde x = 22$ &nbsp; b) $\\mu = 13$, $\\tilde x = 11$ &nbsp; "], ["a) 13, 8, 6, 5, 18 &nbsp; b) 15, 2, 7, 8, 10, 24 &nbsp; ", "a) $\\mu = 10$, $\\tilde x = 8$ &nbsp; b) $\\mu = 11$, $\\tilde x = 9$ &nbsp; "], ["a) 18, 12, 11, 18, 1 &nbsp; b) 20, 16, 7, 14, 14, 1 &nbsp; ", "a) $\\mu = 12$, $\\tilde x = 12$ &nbsp; b) $\\mu = 12$, $\\tilde x = 14$ &nbsp; "], ["a) 15, 22, 13, 19, 6 &nbsp; b) 16, 10, 14, 12, 19, -5 &nbsp; ", "a) $\\mu = 15$, $\\tilde x = 15$ &nbsp; b) $\\mu = 11$, $\\tilde x = 13$ &nbsp; "], ["a) 21, 13, 10, 8, 23 &nbsp; b) 11, 18, 11, 17, 17, 34 &nbsp; ", "a) $\\mu = 15$, $\\tilde x = 13$ &nbsp; b) $\\mu = 18$, $\\tilde x = 17$ &nbsp; "], ["a) 18, 17, 22, 23, 0 &nbsp; b) 10, 12, 12, 14, 6, 18 &nbsp; ", "a) $\\mu = 16$, $\\tilde x = 18$ &nbsp; b) $\\mu = 12$, $\\tilde x = 12$ &nbsp; "], ["a) 25, 17, 13, 16, 19 &nbsp; b) 16, 14, 17, 16, 23, 16 &nbsp; ", "a) $\\mu = 18$, $\\tilde x = 17$ &nbsp; b) $\\mu = 17$, $\\tilde x = 16$ &nbsp; "], ["a) 11, 8, 9, 10, 12 &nbsp; b) 19, 13, 20, 18, 17, 15 &nbsp; ", "a) $\\mu = 10$, $\\tilde x = 10$ &nbsp; b) $\\mu = 17$, $\\tilde x = \\frac{35}{2} = 17.5$ &nbsp; "], ["a) 14, 1, 7, 14, 14 &nbsp; b) 10, 13, 13, 21, 16, 17 &nbsp; ", "a) $\\mu = 10$, $\\tilde x = 14$ &nbsp; b) $\\mu = 15$, $\\tilde x = \\frac{29}{2} = 14.5$ &nbsp; "]],
+" <hr> ", " <hr> ");
+</JS>
+<HTML>
+<div id="exomeanmed"></div>
+</HTML>
+<hidden Lösungen>
+<HTML>
+<div id="solmeanmed"></div>
+</HTML>
+</hidden>
+=== Donnerstag 20. Februar 2020 ===
+Berechnen Sie Standardabweichung der folgenden Wertereihe.
+<JS>miniAufgabe("#exostandardabweichung","#solstandardabweichung",
+[["20, 17, 13, 21, 9", "$\\mu = 16$, $\\sigma^2 = \\frac{1}{n-1} \\cdot \\sum_{i=1}^{n} (x_i-\\mu)^2 = \\frac{1}{4} \\cdot \\left(16+1+9+25+49\\right)  = \\frac{1}{4}\\cdot 100 = 25$ also $\\sigma = \\sqrt{25} = 5$"], ["18, 17, 20, 18, 12", "$\\mu = 17$, $\\sigma^2 = \\frac{1}{n-1} \\cdot \\sum_{i=1}^{n} (x_i-\\mu)^2 = \\frac{1}{4} \\cdot \\left(1+0+9+1+25\\right)  = \\frac{1}{4}\\cdot 36 = 9$ also $\\sigma = \\sqrt{9} = 3$"], ["7, 15, 13, 13, 12", "$\\mu = 12$, $\\sigma^2 = \\frac{1}{n-1} \\cdot \\sum_{i=1}^{n} (x_i-\\mu)^2 = \\frac{1}{4} \\cdot \\left(25+9+1+1+0\\right)  = \\frac{1}{4}\\cdot 36 = 9$ also $\\sigma = \\sqrt{9} = 3$"], ["21, 21, 18, 13, 7", "$\\mu = 16$, $\\sigma^2 = \\frac{1}{n-1} \\cdot \\sum_{i=1}^{n} (x_i-\\mu)^2 = \\frac{1}{4} \\cdot \\left(25+25+4+9+81\\right)  = \\frac{1}{4}\\cdot 144 = 36$ also $\\sigma = \\sqrt{36} = 6$"], ["15, 17, 14, 15, 9", "$\\mu = 14$, $\\sigma^2 = \\frac{1}{n-1} \\cdot \\sum_{i=1}^{n} (x_i-\\mu)^2 = \\frac{1}{4} \\cdot \\left(1+9+0+1+25\\right)  = \\frac{1}{4}\\cdot 36 = 9$ also $\\sigma = \\sqrt{9} = 3$"], ["14, 20, 14, 12, 15", "$\\mu = 15$, $\\sigma^2 = \\frac{1}{n-1} \\cdot \\sum_{i=1}^{n} (x_i-\\mu)^2 = \\frac{1}{4} \\cdot \\left(1+25+1+9+0\\right)  = \\frac{1}{4}\\cdot 36 = 9$ also $\\sigma = \\sqrt{9} = 3$"], ["11, 6, 14, 12, 12", "$\\mu = 11$, $\\sigma^2 = \\frac{1}{n-1} \\cdot \\sum_{i=1}^{n} (x_i-\\mu)^2 = \\frac{1}{4} \\cdot \\left(0+25+9+1+1\\right)  = \\frac{1}{4}\\cdot 36 = 9$ also $\\sigma = \\sqrt{9} = 3$"], ["13, 14, 9, 7, 17", "$\\mu = 12$, $\\sigma^2 = \\frac{1}{n-1} \\cdot \\sum_{i=1}^{n} (x_i-\\mu)^2 = \\frac{1}{4} \\cdot \\left(1+4+9+25+25\\right)  = \\frac{1}{4}\\cdot 64 = 16$ also $\\sigma = \\sqrt{16} = 4$"], ["21, 15, 15, 16, 13", "$\\mu = 16$, $\\sigma^2 = \\frac{1}{n-1} \\cdot \\sum_{i=1}^{n} (x_i-\\mu)^2 = \\frac{1}{4} \\cdot \\left(25+1+1+0+9\\right)  = \\frac{1}{4}\\cdot 36 = 9$ also $\\sigma = \\sqrt{9} = 3$"], ["15, 17, 11, 13, 4", "$\\mu = 12$, $\\sigma^2 = \\frac{1}{n-1} \\cdot \\sum_{i=1}^{n} (x_i-\\mu)^2 = \\frac{1}{4} \\cdot \\left(9+25+1+1+64\\right)  = \\frac{1}{4}\\cdot 100 = 25$ also $\\sigma = \\sqrt{25} = 5$"]],
+" <hr> ", " <hr> ");
+</JS>
+<HTML>
+<div id="exostandardabweichung"></div>
+</HTML>
+<hidden Lösungen>
+<HTML>
+<div id="solstandardabweichung"></div>
+</HTML>
+</hidden>