<PRE><B>Question 344868</B>
Suppose that the weight of apples eaten by individual Americans can be described by a normal probability distribution with mean u=15 pounds and standard deviation o= 5 pounds per 
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1. for a random sample of n= 16 people, what is the standard deviation of the sampling distribution of the sample mean x?
Ans: 5/sqrt(16) = 5/4
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2. for a random sample of n=16 people, what is the mean of the sampling distribution of the sample mean x?
Ans: 15
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3. for a random sample of n=16 people, what is the z-score for a sample means x=10 pounds?
Ans: z(10) = (10-15)/(5/4) = 5*4/5 = 4
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4. for a random sample of n=9 people, what is the z-score for a sample means x=18 pounds
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z(18) = (18-15)/sqrt(9) = -3/3 = -1
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Cheers,
Stan H.
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