document.write( "Question 921798: I am trying to prepare for my test and I have absolutely no idea how to do this!\r
\n" ); document.write( "\n" ); document.write( "Each month, an American household generates an average of 28 pounds of newspaper for garbage or recycling. Assume the variable is approximately normally distributed and the standard deviation is 2 pounds. If a household is selected at random, find the probability of its generating:
\n" ); document.write( "a. Between 27 and 31 pounds per month
\n" ); document.write( "b. More than 30.2 pounds per month
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Algebra.Com's Answer #559214 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!
given: an average of 28 pounds\r
\n" ); document.write( "\n" ); document.write( "a.
\n" ); document.write( "Find \"z\" values corresponding to \"27\" and \"31\":\r
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\n" ); document.write( "\n" ); document.write( "\"z=%2827-28%29%2F2=-1%2F2=-0.5\"\r
\n" ); document.write( "\n" ); document.write( "\"z=%2831-28%29%2F2=3%2F2=1.5\"\r
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\n" ); document.write( "\n" ); document.write( "Find the area between \"z+=+-0.5\" and \"z=+1.5\"\r
\n" ); document.write( "\n" ); document.write( "Table E gives us an area of \"0.9332-0.30=0.6247\"=> The probability is \"62.47\"%\r
\n" ); document.write( "\n" ); document.write( "b.\r
\n" ); document.write( "\n" ); document.write( "Find \"z\" values corresponding to \"30.2\"\r
\n" ); document.write( "\n" ); document.write( "\"z=%2830.2-28%29%2F2=2.2%2F2=1.1\" Find the area to the right of \"z+=1.1\"\r
\n" ); document.write( "\n" ); document.write( "it gives you an area of \"0.1357\"=> The probability is \"13.57\"%\r
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