document.write( "Question 1196427: Finishing times in the annual cross country meet hosted by Altoona High School have a symmetrical distribution with a mean of 22.7 minutes and a standard deviation of 3.22 minutes.\r
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document.write( "What number would correctly complete the sentence below?\r
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document.write( "Approximately 16% of the runners in the Altoona High School cross country meet have finishing times below _____ minutes. Round your answer to one decimal place.
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Algebra.Com's Answer #829247 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "According to the Empirical rule, roughly 68% of the population from a normal distribution is between z = -1 and z = 1 \n" ); document.write( "I.e. about 68% of the population is within 1 standard deviation of the mean.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "P(-1 < z < 1) = 0.68 approximately \n" ); document.write( "This leaves 1-0.68 = 0.32 as the area of the two tails combined \n" ); document.write( "P(z < -1 or z > 1) = 0.32 approximately\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "One tail is 0.32/2 = 0.16 \n" ); document.write( "About 16% of the area is to the left of z = -1 \n" ); document.write( "P(z < -1) = 0.16 approximately\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "mu = 22.7 = mean \n" ); document.write( "sigma = 3.22 = standard deviation\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let's find the raw score x based on the z score z = -1 \n" ); document.write( "So, \n" ); document.write( "z = (x - mu)/sigma \n" ); document.write( "-1 = (x - 22.7)/(3.22) \n" ); document.write( "-1*3.22 = x - 22.7 \n" ); document.write( "-3.22 = x - 22.7 \n" ); document.write( "-3.22 + 22.7 = x \n" ); document.write( "19.48 = x \n" ); document.write( "x = 19.48 \n" ); document.write( "x = 19.5\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Answer: 19.5 \n" ); document.write( " \n" ); document.write( " |