SOLUTION: 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. a)What is the probabi

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.
a)What is the probability that a randomly selected Canadian household purchases at least $112,000 of life insurance?
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?

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
This is z>(112000-110000)/40000 or >2/40 or 0.05. That probability is 0.4801.
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This is z > (112-110)/40/sqrt(50)= 2* sqrt(50)/40=0.3536. That probability is 0.3618.
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