document.write( "Question 150420: This is a multiple choice proble that I am not getting any of the answers to.
\n" ); document.write( "The average amount customers at a certain grocery store spend yearly is $636.55. Assume the variable is normally distributed. If the standard deviation is $89.46, find the probability that a randomly selected customer spends between $550.67 and $836.94.
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\n" ); document.write( " 0.144 = 14.4%
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\n" ); document.write( " 0.820 = 82.0%
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\n" ); document.write( " 0.156 = 15.6%
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\n" ); document.write( " 0.943 = 94.3%
\n" ); document.write( "Please help,
\n" ); document.write( " Ellen \n" ); document.write( "

Algebra.Com's Answer #110372 by scott8148(6628) $\"\"$ $\"About$

You can put this solution on YOUR website!
find the z values for the upper and bounds of the range and then find the portion of the distribution represented\r
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\n" ); document.write( "\n" ); document.write( "lower __ z=(550.67-636.55)/89.46 __ z=-.96 (approx)\r
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\n" ); document.write( "\n" ); document.write( "upper __ z=(836.94-636.55)/89.46 __ z=2.24 (approx)\r
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\n" ); document.write( "\n" ); document.write( "this range represents about 82% of the distribution \n" ); document.write( "

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