document.write( "Question 1203810: A random variable X has the pdf\r
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Algebra.Com's Answer #839942 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "That's the graph.  Now we'll shade the area under it:\r\n" );
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document.write( "Now we find the area between the graph and the x-axis, take half\r\n" );
document.write( "of it and find the value of x which divides the entire area into\r\n" );
document.write( "two parts.\r\n" );
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document.write( "First we find the area under the curved (parabola) part\r\n" );
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document.write( "\"int%28x%5E2%2Cdx%2C0%2C1%29\"\"%22%22=%22%22\"\"%22%22=%22%22\"\"expr%281%2F3%29%281%29%5E3-expr%281%2F3%29%280%29%5E3\"\"%22%22=%22%22\"\"1%2F3\"\r\n" );
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document.write( "The right part of the graph is just a rectangle with base 1 and height \"2%2F3\", \r\n" );
document.write( "so its area is just \"base%2Aheight=%281%29%282%2F3%29=2%2F3\".\r\n" );
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document.write( "The entire shaded area is \"1%2F3%2B2%2F3\"\"%22%22=%22%22\"\"1\"\r\n" );
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document.write( "So we need to find the value of x such that half the area \"1%2F2\" is left of\r\n" );
document.write( "it and the other half \"1%2F2\", is on the right of it.\r\n" );
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document.write( "The area under the parabola is \"1%2F3\" so that leaves \"1%2F2-1%2F3=3%2F6-2%2F6=1%2F6\" the we must take \r\n" );
document.write( "of the rectangle. To get the base of a rectangle with area \"1%2F6\" that has height \"2%2F3\", \r\n" );
document.write( "we divide the area by the height and get \"%281%2F6%29%2F%282%2F3%29\"\"%22%22=%22%22\"\"%281%2F6%29%283%2F2%29\"\"%22%22=%22%22\"\"3%2F12\"\"%22%22=%22%22\"\"1%2F4\".  So the\r\n" );
document.write( "median is \"1%2F4\" on a unit past 1, or \"1%261%2F4\".\r\n" );
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document.write( "So the median is \"1%261%2F4\". To indicate the median, we can draw a green\r\n" );
document.write( "vertical line through the graph at \"x=1%261%2F4\": \r\n" );
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document.write( "Edwin

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