document.write( "Question 1182516: If X represents a random variable coming from a normal distribution and P (X < 10.7) =0.6, then P (X > 10.7) =0.4.\r
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document.write( "True\r
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document.write( "False \n" );
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Algebra.Com's Answer #812620 by math_tutor2020(3820) ![]() You can put this solution on YOUR website! \n" ); document.write( "Answer: True\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Explanation: The two probabilities are complements, meaning that they add to 1. \n" ); document.write( "So 0.6+0.4 = 1\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The fact that P(A)+P(B) = 1 is being used here has nothing to do with the underlying probability distribution. So the \"normal distribution\" seems like a red herring in my opinion (ie it may be \"clue\" that's added intentionally to distract/mislead the reader).\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Either x is less than 10.7, or it is larger than 10.7 \n" ); document.write( "We consider the case x = 10.7 itself to happen with probability 0 since throwing a dart to land *exactly* at this location is pretty much 0 chance of it happening. There's always going to be some slight error involved. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "I guess to be more thorough, we could say \n" ); document.write( "Or we could say \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |