document.write( "Question 1150147: A gym knows that each member, on average, spends 70 minutes at the gym per week, with a standard deviation of 20 minutes. Assume the amount of time each customer spends at the gym is normally distributed.\r
\n" );
document.write( "\n" );
document.write( "a. What is the probability that a randomly selected customer spends between 65 minutes to 75 minutes at the gym? \n" );
document.write( "
Algebra.Com's Answer #771550 by VFBundy(438)![]() ![]() You can put this solution on YOUR website! Probability a gym member spends less than 75 minutes at the gym: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "Look up 0.25 on a z-table. The result is 0.5987. \n" ); document.write( " \n" ); document.write( "This means there is a 0.5987 probability a gym member spends less than 75 minutes at the gym. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "Probability a gym member spends less than 65 minutes at the gym: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "Look up -0.25 on a z-table. The result is 0.4013. \n" ); document.write( " \n" ); document.write( "This means there is a 0.4013 probability a gym member spends less than 65 minutes at the gym. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "So, the probability that a gym member spends between 65 and 75 minutes at the gym: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |