Question 862717: (Please help me with this question?)
Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $120,000. This distribution follows the normal distribution with a standard deviation of $38,000.
(a)
If we select a random sample of 58 households, what is the standard error of the mean? (Round your answer to the nearest whole number.)
Standard error of the mean=?
(b) What is the expected shape of the distribution of the sample mean?
Not normal, the standard deviation is unknown.
Unknown.
Uniform
Normal.
(c)
What is the likelihood of selecting a sample with a mean of at least $124,000? (Round z value to 2 decimal places and final answer to 4 decimal places.)
Probability=?
(d)
What is the likelihood of selecting a sample with a mean of more than $112,000? (Round z value to 2 decimal places and final answer to 4 decimal places.)
Probability=?
(e)
Find the likelihood of selecting a sample with a mean of more than $112,000 but less than $124,000. (Round z value to 2 decimal places and final answer to 4 decimal places.)
Probability=?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! a) standard error of the mean = 38000 / square root(56) = 5078
b) normal
c) P(x > or = 124000) = 1 - P(x<124000)
we calculate the z score for P(x<124000)
124000 - 120000 / 5078 = 0.78771169751870815282 = 0.79
consult z table for probability assocated with z score = .79
P(x > or = 124000) = 1 - .79 = 0.21
d) P(x > or = 112000) = 1 - P(x<112000)
we calculate the z score for P(x<112000)
112000 - 120000 / 5078 = -1.58
consult -z table for probability assocated with z score = -1.58
P(x > 112000) = 1 - P(x<112000) = 1 - 0.06 = 0.94
e) P(x > 112000 and x < 124000) = P(x<124000) - P(x<112000) = 0.79 - 0.06 = 0.73
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