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

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) (Show Source): You can put this solution on YOUR website!
Probability a gym member spends less than 75 minutes at the gym:

= = = 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:

= = = -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:

= 0.1974

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