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The shape of the distribution of the time required to get an oil change at a 20 min. Oil facility is unknown. However, records indicate that the mean is 21.8 mins., and the standard deviation is 4.8 mins. What is the probability that a random sample of n= 40 oil changes results in a sample mean time less than 20 mins.. The probability is approximately .? ( round to four decimal places as needed). I have tried many solutions on my TI 83 and come up with nothing cloce to four decimals. Help!
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document.write( "Since 40 is a large sample, the central limit theorem tells us that\r\n" );
document.write( "the distribution is approximately normal even though the shape of \r\n" );
document.write( "the distribution is unknown. However this is a sample, not just a \r\n" );
document.write( "single value, so we must divide the standard deviation by √n.\r\n" );
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document.write( "Press ON\r\n" );
document.write( "Press CLEAR\r\n" );
document.write( "Press 2ND\r\n" );
document.write( "Press VARS\r\n" );
document.write( "Press 2\r\n" );
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document.write( "You should see this---> normalcdf(\r\n" );
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document.write( "Make it read this way:\r\n" );
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document.write( "normalcdf(-1E99,20,21.8,4.8/√(40))\r\n" );
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document.write( "[To get the little E, press ALPHA then , (the comma key)]\r\n" );
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document.write( "Press ENTER\r\n" );
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document.write( "Read .00088530196 \r\n" );
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document.write( "round to 0.0009\r\n" );
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document.write( "Edwin \n" );
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