document.write( "Question 1156965: 1) Suppose it is known that the distribution of purchase amounts by customers entering a
\n" );
document.write( "popular retail store is approximately normal with mean $75 and standard deviation
\n" );
document.write( "$20.\r
\n" );
document.write( "\n" );
document.write( "a. What is the probability that a randomly selected customer spends less than $85
\n" );
document.write( "at this store?
\n" );
document.write( "b. What is the probability that a randomly selected customer spends between $65
\n" );
document.write( "and $85 at this store?
\n" );
document.write( "c. What is the probability that a randomly selected customer spends more than
\n" );
document.write( "$45 at this store?
\n" );
document.write( "d. Find the dollar amount such that 80% of all customers spend at least this
\n" );
document.write( "amount.
\n" );
document.write( "e. Find two dollar amounts, equidistant from the mean, such that 90% of all
\n" );
document.write( "customer purchases are between these values. \n" );
document.write( "
Algebra.Com's Answer #780121 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! z=(x-mean)/sd \n" ); document.write( "a. this is (85-75)/20=z and want z<0.5. prob is 0.6915 \n" ); document.write( "b. This is a z between -0.5 and +0.5. probability is 0.3829 \n" ); document.write( "c. this is a z> -1.5 with probability 0.9332 \n" ); document.write( "d.80% is the area between z=-1.282 and +1.282 \n" ); document.write( "1.282=(x-mean)sd and -1.282=(x-mean)/sd \n" ); document.write( "25.64=x-mean or $100.64 top end \n" ); document.write( "and $49.36 bottom end. ($49.36, $100.64)\r \n" ); document.write( "\n" ); document.write( "1.645 =(x-mean)/sd \n" ); document.write( "32.90=x-mean and -32.90=(x-mean) \n" ); document.write( "($42.10, $107.90) \n" ); document.write( " |