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

Question 1160812: 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.
a. Suppose the gym surveys a random sample of 30 members about the amount of time they spend at the gym each week. What are the expected value and standard deviation (standard error) of the sample mean of the time spent at the gym?

b. If 30 members are randomly selected, what is the probability that the average time spent at the gym exceeds 55 minutes?

c. 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!
a. E(X)=50 minutes, same as individual average
SEM=10/sqrt(30)=1.826 minutes
z=(x-mean)/sd/sqrt(n)
> 5*sqrt(30)10=2.74
prob. z>2.74 is 0.0031
this would be z between -.5 and +.5 which is 0.3829
Much more difficult for the average of 30 to differ greatly from the mean than a single person.

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