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)\"\" \"About 
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( "\"%28x-+mean%29%2FSD\" = \"%2875+-+70%29%2F20\" = \"5%2F20\" = 0.25
\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( "\"%28x-+mean%29%2FSD\" = \"%2865+-+70%29%2F20\" = \"%28-5%29%2F20\" = -0.25
\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( "\"0.5987+-+0.4013\" = 0.1974
\n" ); document.write( "
\n" ); document.write( "
\n" );