document.write( "Question 219059: The mean amount of life insurance per household is 110,000 normal distribution a) with a standarf deviation of $40,000 what is the likelihood of selecting a sample with a mean of at least $112,000?
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document.write( "b) What is the expected shape of the distribution of the sample mean? \n" );
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Algebra.Com's Answer #164811 by stanbon(75887) ![]() You can put this solution on YOUR website! The mean amount of life insurance per household is 110,000 normal distribution a) with a standard deviation of $40,000 what is the likelihood of selecting a sample with a mean of at least $112,000? \n" ); document.write( "--- \n" ); document.write( "z(112,000) = (112,000-110,000)/[40,000] = 2/40 \n" ); document.write( "--- \n" ); document.write( "P(z>(2/40) = normalcdf((2/40),10) = 0.48 \n" ); document.write( "------------------------------------------------------ \n" ); document.write( "b) What is the expected shape of the distribution of the sample mean? \n" ); document.write( "The sample means will have a normal distribution \n" ); document.write( "as guaranteed by the Central Limit Theorem. \n" ); document.write( "============================================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " \n" ); document.write( " |