document.write( "Question 344868: 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 \n" ); document.write( "1. for a random sample of n= 16 people, what is the standard deviation of the sampling distribution of the sample mean x?
\n" ); document.write( "2. for a random sample of n=16 people, what is the mean of the sampling distribution of the sample mean x?
\n" ); document.write( "3. for a random sample of n=16 people, what is the z-score for a sample means x=10 pounds?
\n" ); document.write( "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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Algebra.Com's Answer #246772 by stanbon(75887)\"\" \"About 
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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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\n" ); document.write( "1. for a random sample of n= 16 people, what is the standard deviation of the sampling distribution of the sample mean x?
\n" ); document.write( "Ans: 5/sqrt(16) = 5/4
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\n" ); document.write( "2. for a random sample of n=16 people, what is the mean of the sampling distribution of the sample mean x?
\n" ); document.write( "Ans: 15
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\n" ); document.write( "3. for a random sample of n=16 people, what is the z-score for a sample means x=10 pounds?
\n" ); document.write( "Ans: z(10) = (10-15)/(5/4) = 5*4/5 = 4
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\n" ); document.write( "\n" ); document.write( "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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\n" ); document.write( "z(18) = (18-15)/sqrt(9) = -3/3 = -1
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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