document.write( "Question 1135014: A worldwide organization of academics claims that the mean IQ score of its members is 115
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document.write( ", with a standard deviation of 16
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document.write( ". A randomly selected group of 40
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document.write( " members of this organization is tested, and the results reveal that the mean IQ score in this sample is 115.8
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document.write( ". If the organization's claim is correct, what is the probability of having a sample mean of 115.8
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document.write( " or less for a random sample of this size? \n" );
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Algebra.Com's Answer #752546 by Theo(13342) You can put this solution on YOUR website! z-score formula is z = (x - m) / s\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "z is the z-score \n" ); document.write( "x is the raw score \n" ); document.write( "m is the mean \n" ); document.write( "s is the standard error.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the formula for standard error is s = sd / sqrt(n)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "sd is the standard deviation of the population. \n" ); document.write( "sqrt(n) is the square root of the sample size.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the claim is that the population has a mean of 115 with a standard deviation of 16.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the sample of 40 members has a mean of 115.8.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the standard error for the distribuion of sample means is equal to 16 / sqrt(40) which is equal to 2.529822128.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the z-score for the sample mean compared to the population mean would be equal to (115.8 - 115) / 2.529822128 which is equal to .316227766.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the probability of having a sample mean of 15.8 or lessfor a random sample of this size would be the area under the normal distribution curve to the left of a z-score of .316227766.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "using the z-score calculator on the TI-84 Plus, i get this area to be equal to .6240851183.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the probability of getting a sample of size 40 with a mean less than 115.8 is approximately 62.41%.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "this can be seen visually through the use of the following online normal distribution calculator.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "http://davidmlane.com/hyperstat/z_table.html\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "here's a picture of the results.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \r\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "calculation of the standard error correctly is critical to getting the right probability.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "standard error = standard deviation of the population divided by the square root of the sample size.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |