Question 862746: (I'm stuck on this problem. Can someone please help?)
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 jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! (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 = 4,990
Work Shown:
SEM = sigma/sqrt(n) SEM stands for the Standard Error of the Mean
SEM = 38000/sqrt(58)
SEM = 4,989.64444866946 This is approximate
SEM = 4,990 Rounding to the nearest whole number.
Note: you'll often see "SE" used instead of "SEM", so either one works
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(B) What is the expected shape of the distribution of the sample mean?
Not normal, the standard deviation is unknown.
Unknown.
Uniform
Normal.
This is because the parent distribution is normally distributed. All sampling distributions (regardless of size) drawn from a normal distribution will be normally distributed as well.
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(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 = 0.2119
Work Shown:
First convert to a z-score:
z = (x-mu)/(sigma/sqrt(n))
z = (124000-120000)/(38000/sqrt(58))
z = 0.80166032693304
z = 0.80
Then find the area to the left of z = 0.80 using the table found on this link: http://www.had2know.com/images/z-score-normal-table.png
Note: you can use the table given to you by your book or your instructor if you want
According to the table, the area to the left of z = 0.80 is 0.7881
So the area to the right of z = 0.80 is 1-0.7881 = 0.2119
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(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 = 0.9452
Work Shown:
First convert to a z-score:
z = (x-mu)/(sigma/sqrt(n))
z = (112000-120000)/(38000/sqrt(58))
z = -1.60332065386609
z = -1.60
Then find the area to the left of z = -1.60 using the table found on this link: http://www.had2know.com/images/z-score-normal-table.png.
Since the distribution is symmetric about 0, we can find the area to the right of z = 1.60, which is 1 - P(Z < 1.60) = 1 - 0.9452 = 0.0548
The area to the right of 1.60 is 0.0548
Due to symmetry, this means the area to the left of z = -1.60 is also 0.0548
So the area to the right of z = -1.60 is 1 - 0.0548 = 0.9452
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(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 = 0.7333
Work Shown:
From parts (D) and (C), the z-scores for x = 112000 and x = 124000 are z = -1.60 and z = 0.80 respectively
The area to the left of z = -1.60 is 0.0548 (shown in part D)
The area to the left of z = 0.80 is 0.7881 (shown in part C)
So the area between z = -1.60 and z = 0.80 is
Area to left of z = 0.80 - Area to left of z = -1.60 = 0.7881 - 0.0548 = 0.7333
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