document.write( "Question 1068944: It is known that IQ scores form a normal distribution with a mu of 100 and a standard deviation of 15. Given this information, what is the probability of randomly selecting an individual with an IQ score less than 120?
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document.write( "a)9.72%
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document.write( "b)95%
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document.write( "c)Unable to answer this question with information given.
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document.write( "d)90.15%\r
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Algebra.Com's Answer #684188 by Theo(13342)![]() ![]() You can put this solution on YOUR website! first you find the z-score.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the z-score is equal to (x-m)/s\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "x is the x-score \n" ); document.write( "x is the raw score. \n" ); document.write( "m is the mean. \n" ); document.write( "s is the standard deviation.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the z-score in this case is z = (120-100)/15 = 20/15 = 4/3 = 1.33.....\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "use a z-score calculator to the highest degree of accuracy that it possesses and the probability becomes .9087887181 which is equal to roughly 90.88%.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "closest selection you have is 90.15%, even though it's not right on.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "a normal distribution is assumed.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |