Question 1165213: Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $136,000. This distribution follows the normal distribution with a standard deviation of $36,000.
If we select a random sample of 78 households, what is the standard error of the mean? (Round your answer to the nearest whole number.)
What is the expected shape of the distribution of the sample mean?
What is the likelihood of selecting a sample with a mean of at least $141,000? (Round your z value to 2 decimal places and final answer to 4 decimal places.)
What is the likelihood of selecting a sample with a mean of more than $131,000? (Round your z value to 2 decimal places and final answer to 4 decimal places.)
Find the likelihood of selecting a sample with a mean of more than $131,000 but less than $141,000. (Round your z value to 2 decimal places and final answer to 4 decimal places.)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! a. SEM is 36000/sqrt(78)=$4076.20
b. A sample from a normal distribution may be considered normal as well. The curve is narrower and steeper, as the variance is much less than for an individual.
c.the z would be 5000/4076.20 or 1.23. That would be 0.1093
d. c would also be the answer for < 131000, so the answer to d is 0.8997, the complement
e. Here, the z value is between -1.23 and +1.23, and that probability is 0.7813
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