<PRE><B>Question 314788</B>
Suppose that the weight of apples eaten by individual Americans can be described by a normal probability distribution with mean &amp;#956; = 15 pounds and standard deviation &amp;#963; = 5 pounds per &lt; year.
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For a random sample of n = 16 people, what is the standard deviation of the sampling distribution of the sample mean &amp;#772;x?
Ans: 5/Sqrt(16) = 5/4
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For a random sample of n = 16 people, what is the mean of the sampling distribution of the sample mean &amp;#772;x?
Ans: x-bar = 15
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For a random sample of n = 16 people, what is the z-score for a sample mean &amp;#772;x = 10 pounds?
z(10) = (10-15)/[5/sqrt(16)] = -5/(5/4) = -4
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For a random sample of n = 9 people, what is the z-score for a sample mean &amp;#772;x = 18 pounds?
z(18) = (18-15)/[5/sqrt(3)] = 3/(5/sqrt(3)) = 3*sqrt(3)/5 = 1.0392
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Cheers,
Stan H.
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