Answered

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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. Suppose the gym surveys a random sample of 49 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

Sagot :

Answer:

the expected value and the standard deviation of the sample mean is 70 and 2.86 respectively

Step-by-step explanation:

The computation of the expected value and the standard deviation of the sample mean is shown below:

The mean is 70

And, the standard deviation is

= 20 ÷ √49

= 20 ÷ 7

= 2.86

Hence, the expected value and the standard deviation of the sample mean is 70 and 2.86 respectively

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