document.write( "Question 867588: I need help...\r
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document.write( "The shelf life of a battery produced by one major company is known to be normally distributed, with a mean life of 3.9 years and a standard deviation of 0.3 years. What is the probability that a randomly chosen battery will last fewer than 3.2 years?\r
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document.write( "Thanks! \n" );
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Algebra.Com's Answer #523106 by jim_thompson5910(35256)![]() ![]() ![]() You can put this solution on YOUR website! First you need the standard z-score\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "z = (x - mu)/sigma\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "z = (3.2 - 3.9)/0.3\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "z = -2.33333333333333\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "z = -2.33 round to 2 decimal places\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Now we use a table like this one here to find the area under the standard normal distribution curve that is to the left of z = -2.33\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "According to the table, the area is roughly 0.0099 (look in the row that starts with -2.3 and then look in the column that starts with 0.03, where they intersect is at the cell with the number 0.0099 in it)\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So the approximate answer is 0.0099\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If you need a more accurate answer, use a calculator that can compute normal cdfs \n" ); document.write( " |