Question 862746
(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 = <font color="red">4,990</font>



Work Shown:


SEM = sigma/sqrt(n) <font color="blue">SEM stands for the Standard Error of the Mean</font>


SEM = 38000/sqrt(58)


SEM = 4,989.64444866946 <font color="blue">This is approximate</font>


SEM = 4,990 <font color="blue">Rounding to the nearest whole number.</font>


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
<font color="red">Normal.</font>



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 = <font color="red">0.2119</font>



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: <a href="http://www.had2know.com/images/z-score-normal-table.png">http://www.had2know.com/images/z-score-normal-table.png</a>


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 = <font color="red">0.9452</font>



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: <a href="http://www.had2know.com/images/z-score-normal-table.png">http://www.had2know.com/images/z-score-normal-table.png</a>. 



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 = <font color="red">0.7333</font>


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 <font color="blue">0.0548</font> (shown in part D)


The area to the left of z = 0.80 is <font color="green">0.7881</font> (shown in part C)


So the area between z = -1.60 and z = 0.80 is 


<font color="green">Area to left of z = 0.80</font> - <font color="blue">Area to left of z = -1.60</font> = <font color="green">0.7881</font> - <font color="blue">0.0548</font> = 0.7333