document.write( "Question 1151818: Stress at work: In a poll conducted by the General Social Survey, 81% of
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Algebra.Com's Answer #773729 by Edwin McCravy(20062) $\"\"$ $\"About$

You can put this solution on YOUR website!
Stress at work: In a poll conducted by the General Social Survey, 81% of
\n" ); document.write( "respondents said that their jobs were sometimes or always stressful. Two hundred
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document.write( "This is a binomial problem, number of trials = n = 200,\r\n" );
document.write( "probability of \"success\" in one trial = p = 81% = 0.81\r\n" );
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document.write( "m = n∙p = 200∙0.81 = 162\r\n" );
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document.write( "We use the normal to approximate the binomial.\r\n" );
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document.write( "Since we are including 160 (\"160 or less\") and since it is \"or less\",\r\n" );
document.write( "we consider the upper limit to be 0.5 more than 160, or 160.5.\r\n" );
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document.write( "Round to two decimals -0.36 \r\n" );
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document.write( "There are two kinds of tables in use today. \r\n" );
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document.write( "If your table is like this one, giving all the area to the left of z,\r\n" );
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document.write( "http://www.z-table.com/\r\n" );
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document.write( "scroll down to find -0.3 in the left-most z-column, then go over to the column headed 0.06, and read 0.3594\r\n" );
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document.write( "If your table is like this one, giving area to the right of 0,\r\n" );
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document.write( "https://www.mathsisfun.com/data/standard-normal-distribution-table.html\r\n" );
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document.write( "find 0.3 in the left-most z-column, then go over to the column headed\r\n" );
document.write( "0.06, and read 0.1406.  Then subtract from 0.5 to get 0.3594\r\n" );
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document.write( "Or on your TI-83 or 84,\r\n" );
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document.write( "Press 2ND VARS\r\n" );
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document.write( "normalcdf(-1E99,-0.3604920469,0,1)\r\n" );
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document.write( "Press ENTER\r\n" );
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document.write( "Read 0.3592396598, which differs in the 4th place because the calculator\r\n" );
document.write( "is more accurate than old-fashioned tables, which nobody but old-fashioned\r\n" );
document.write( "teachers use anymore. lol\r\n" );
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document.write( "Edwin

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