document.write( "Question 1053922:  The owner of a computer repair shop has determined that their daily revenue has mean $7200 and standard deviation $1200. The daily revenue totals for the next 30 days will be monitored. What is the probability that the mean daily revenue for the next 30 days will exceed $7500?\r
\n" );
document.write( "\n" );
document.write( "How can I solve this with a calculator? \n" );
document.write( "
| Algebra.Com's Answer #669146 by ewatrrr(24785)     You can put this solution on YOUR website! mean $7200 and standard deviation $1200. \n" ); document.write( "n = 30, x̄ = mean daily revenue for the next 30 days \n" ); document.write( "Need to Know: \n" ); document.write( "P(x̄ >7500) = 1- P(x̄ <= 7500) = 1- P(z <= \n" ); document.write( "Use Calculator to find value of z \n" ); document.write( "then use Calculator to find P( z <= that value) = normalcdf(-100, z-value) \n" ); document.write( "-100 a place value \n" ); document.write( " \n" ); document.write( " |