document.write( "Question 807050: A study of a company's practice regarding the payment of invoices revealed that on the average an invoice was paid 20 days after it was received. The standard deviation equalled five days. A sample of 25 invoices is selected. Assuming that the distribution is normal, what percent of the sampled invoices were paid within 15 days of receipt?\r
\n" ); document.write( "\n" ); document.write( "so here is what I thought it was written as:\r
\n" ); document.write( "\n" ); document.write( "(15-20)/(5/sqrt of 25)= 5 but how do you look that up on the z value
\n" ); document.write( "or is it 0.5 you look up or 0.05? \n" ); document.write( "
Algebra.Com's Answer #486175 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "Your arithmetic is off a bit, as your calculation should have come up with -5. But you are using the wrong Z-score formula anyway. You get a Z-score of -5 against the proposition that you have a sample mean of 15 across your sample of 25. This is a much smaller probability than what is actually being asked for. You just want the probability that a given invoice would be paid in 15 days, which is just the percentage of the general population that is 1 standard deviation below the mean, or Z = -1.000. Just look it up.\r
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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "Egw to Beta kai to Sigma
\n" ); document.write( "My calculator said it, I believe it, that settles it
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