SOLUTION: 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

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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.
a. What is the probability that a randomly selected customer spends between 65 minutes to 75 minutes at the gym?
Answer by VFBundy(438) About Me  (Show Source):
You can put this solution on YOUR website!
Probability a gym member spends less than 75 minutes at the gym:

%28x-+mean%29%2FSD = %2875+-+70%29%2F20 = 5%2F20 = 0.25

Look up 0.25 on a z-table. The result is 0.5987.

This means there is a 0.5987 probability a gym member spends less than 75 minutes at the gym.


Probability a gym member spends less than 65 minutes at the gym:

%28x-+mean%29%2FSD = %2865+-+70%29%2F20 = %28-5%29%2F20 = -0.25

Look up -0.25 on a z-table. The result is 0.4013.

This means there is a 0.4013 probability a gym member spends less than 65 minutes at the gym.


So, the probability that a gym member spends between 65 and 75 minutes at the gym:

0.5987+-+0.4013 = 0.1974