document.write( "Question 1171191: Suppose the mean amount of life insurance purchased by Canadian households is normally distributed with mean of $110,000 and standard deviation of $40,000.\r
\n" ); document.write( "\n" ); document.write( "a)What is the probability that a randomly selected Canadian household purchases at least $112,000 of life insurance? \r
\n" ); document.write( "\n" ); document.write( "b)If a sample of 50 Canadian households was selected at random, what is the probability that the mean amount of life insurance purchased is at least $112,000?
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Algebra.Com's Answer #796137 by Boreal(15235)\"\" \"About 
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This is z>(112000-110000)/40000 or >2/40 or 0.05. That probability is 0.4801.
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\n" ); document.write( "This is z > (112-110)/40/sqrt(50)= 2* sqrt(50)/40=0.3536. That probability is 0.3618.
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