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    Normal Distrubtion question




    Hi, I have this question that I can not figure out:

    Suppose that in an apartment building run by the Probandstat rental company, the inhabitants's weights follow a normal distribution with mean 75.0 kilos and standard deviation 13.8 kilos. A new elevator must be installed. The elevator company rates the safe carrying capacity of the new elevator to be 900 kilos, but publishes the rated capacity as only 750 kilos. A manager at the apartment reasoned as follows, “Since the average resident weighs 75 kilos and the elevator safely carries 750 kilos, the safe capacity for the elevator in our building must be 10 passengers.”

    How do I show that this is incorrect by showing that the probability that ten randomly selected inhabitants would have a total weight exceeding the rated weight capacity is one-half ?

    I tried to use u = 75 std = 13.8 and P(x>750) = (750 - 75) / 13.8 to get the Z value then look it up on the Z table, but it is obviously not correct, what is another way to do this?

    Then I have to find the probability that ten randomly selected inhabitants would have a total weight exceeding the actual carrying capacity. What should I use?

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    Re: Normal Distrubtion question


    Note that if 10 people have a total weight of 750 that the average value of their weight is 75. Note that the standard error of a sample average is \frac{\sigma}{\sqrt{n}}

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