SOLUTION: A gym knows that each member, on average, spends 50 minutes at the gym per week, with a standard deviation of 10 minutes. Assume the amount of time each customer spends at the gym
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Question 1161577: A gym knows that each member, on average, spends 50 minutes at the gym per week, with a standard deviation of 10 minutes. Assume the amount of time each customer spends at the gym is normally distributed. If 30 members are randomly selected, what is the probability that a randomly selected customer spends between 45 minutes to 55 minutes at the gym? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
z is between -.5 and +.5, since the difference between the mean is -5 and 5, and the sd is 10
That probability is 0.3829
