document.write( "Question 349379: Please help me with this probability question.It is due tomorrow.
\n" ); document.write( "5. Consider the ‘symmetric’ distribution as shown in the table below, where a and b are any numbers between 0 and 1.
\n" ); document.write( "x=1 Pr(X)=b,x=2 Pr(X)=a,x=3 Pr(X)=a,x=4 Pr(X)=b
\n" ); document.write( " Prove that E(X)=2.5 \n" ); document.write( "

Algebra.Com's Answer #249751 by Edwin McCravy(20059) $\"\"$ $\"About$

You can put this solution on YOUR website!

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document.write( "        x   P(x)    x*P(x)\r\n" );
document.write( "        1    b       b\r\n" );
document.write( "        2    a      2a\r\n" );
document.write( "        3    a      3a \r\n" );
document.write( "        4    b      4b\r\n" );
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document.write( "    Totals:  1      E(X)\r\n" );
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document.write( "The totals of the individual probabilities must be 1. \r\n" );
document.write( "The totals of the individual expectations equals the\r\n" );
document.write( "total expectation E(X).\r\n" );
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document.write( "Therefore\r\n" );
document.write( "summing the middle column:\r\n" );
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document.write( "b + a + a + b = 1\r\n" );
document.write( "      2b + 2a = 1\r\n" );
document.write( "     2(b + a) = 1\r\n" );
document.write( "        b + a = \r\n" );
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document.write( "Summing the right-most column:\r\n" );
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document.write( "E(X) = b + 2a + 3a + 4b\r\n" );
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document.write( "E(X) = 5b + 5a\r\n" );
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document.write( "E(X) = 5(b + a)\r\n" );
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document.write( "Substituting  for b + a,\r\n" );
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document.write( "E(X) = 5() =  = 2.5\r\n" );
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document.write( "Edwin

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