document.write( "Question 1196804: The time to complete a standardized exam in the BYU-Idaho Testing Center is approximately normal with a mean of 70 minutes and a standard deviation of 10 minutes. Using the 68-95-99.7% Rule, approximately what percentage of students will complete the exam in under fifty minutes? Give your answer as a percentage accurate to one decimal place. \n" ); document.write( "
Algebra.Com's Answer #829804 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "Answer: 2.5%\r
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\n" ); document.write( "\n" ); document.write( "Explanation:\r
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\n" ); document.write( "\n" ); document.write( "Given info
\n" ); document.write( "mu = 70 and sigma = 10
\n" ); document.write( "this represents the mean and standard deviation respectively\r
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\n" ); document.write( "\n" ); document.write( "The 68-95-99.7% Rule is also known as the Empirical Rule.\r
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\n" ); document.write( "\n" ); document.write( "That rule lays out three basic properties
  • Approximately 68% of the normal distribution is within 1 standard deviation of the mean.
  • Approximately 95% of the normal distribution is within 2 standard deviations of the mean.
  • Approximately 99.7% of the normal distribution is within 3 standard deviations of the mean.
Compute the z score for x = 50
\n" ); document.write( "z = (x - mu)/sigma
\n" ); document.write( "z = (50 - 70)/10
\n" ); document.write( "z = -20/10
\n" ); document.write( "z = -2
\n" ); document.write( "We're exactly 2 standard deviations below the mean.\r
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\n" ); document.write( "\n" ); document.write( "We have roughly 95% of the values between z = -2 and z = 2
\n" ); document.write( "I.e. P(-2 < z < 2) = 0.95 approximately\r
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\n" ); document.write( "\n" ); document.write( "That leaves 100% - 95% = 5% of the area for the tails to split up
\n" ); document.write( "Each tail gets (5%)/2 = 2.5%\r
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\n" ); document.write( "\n" ); document.write( "Therefore, about 2.5% of the normally distributed population will have a z score smaller than -2
\n" ); document.write( "P(z < -2) = 0.025 approximately
\n" ); document.write( "P(x < 50) = 0.025 approximately when mu = 70 and sigma = 10
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