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Re: Susan read in her local paper that the average (arithmetic mean) [#permalink]
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­Let's evaluate each option and see which one most directly calls into question Susan's belief:

A) The standard deviation of earnings in her community was greater than $2,000.
   - If the standard deviation of earnings in Susan's community is large, it indicates that there is considerable variability or spread in income levels. This means that even though Susan's income is below the mean income of $38,000, there could still be a significant portion of earners who make less than $38,000. Therefore, option A suggests that Susan's belief may be questionable because the large standard deviation implies a wide range of incomes, potentially including many below the mean.

B) The mean earnings in the community were $3,000 greater than the median earnings.
   - This option provides information about the difference between the mean and median earnings. However, it doesn't directly address Susan's belief that she earned less than the majority of earners in her community.

C) The mean earnings in the community were $1,000 greater than the median earnings.
   - Similar to option B, this option provides information about the difference between the mean and median earnings but doesn't directly address Susan's belief.

D) Doctors in Susan's community had an average income over $200,000.
   - This option provides information about the income of doctors specifically and doesn't directly address Susan's belief about her relative income compared to the majority of earners in her community.

E) The average income of $38,000 is calculated from the incomes of both full- and part-time workers.
   - This option provides information about the calculation of the average income but doesn't directly address Susan's belief.

Among the options, option A is the most relevant because it directly suggests that there is significant variability in earnings within Susan's community. This variability implies that even though Susan's income is below the mean income of $38,000, there could still be a substantial portion of earners who make less than $38,000, thus calling into question Susan's belief that she earned less than the majority of earners in her community. Therefore, option A most directly challenges Susan's belief.
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Re: Susan read in her local paper that the average (arithmetic mean) [#permalink]
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Bunuel wrote:
­Susan read in her local paper that the average (arithmetic mean) income in her community last year was $38,000. She worked very hard to make $36,000 last year and was upset that she earned less than the majority of the earners in her community.

Which of the following, if true, most calls into question Susan’s belief that she earned less than the majority of the earners in her community?

(A) The standard deviation of earnings in her community was greater than $2,000.

(B) The mean earnings in the community were $3,000 greater than the median earnings.

(C) The mean earnings in the community were $1,000 greater than the median earnings.

(D) Doctors in Susan's community had an average income over $200,000.

(E) The average income of $38,000 is calculated from the incomes of both full- and part-time workers.

­

­
Summary: Susan's concern comes from misunderstanding that the mean income necessarily reflects where most people's incomes lie, which isn't always true, especially in skewed distributions.

Analyzing the Options:

A) The standard deviation of earnings in her community was greater than $2,000.
- A large standard deviation indicates wide variance in income but doesn't directly address whether Susan earned less than the majority.

B) The mean earnings in the community were $3,000 greater than the median earnings.
- This is critical because if the mean is higher than the median, it suggests that the income distribution is skewed to the right, meaning there are outliers on the higher end that are pulling the average up. This implies that the majority of people actually earn less than the mean, directly challenging Susan’s belief. This could mean Susan may very well be earning more than the majority of her community.

C) The mean earnings in the community were $1,000 greater than the median earnings.
- Similar to B, but the smaller difference between the mean and median suggests a less pronounced skew. This still suggests that the distribution is skewed and that Susan might be closer to or even above the majority's earnings, but B offers a stronger argument.

D) Doctors in Susan's community had an average income over $200,000.
- While this indicates the presence of high earners that could skew the average, it doesn't directly address the distribution of the entire community's earnings in relation to Susan's income.

E) The average income of $38,000 is calculated from the incomes of both full- and part-time workers.
- This adds context but doesn’t directly challenge Susan's belief about her income relative to the majority since it doesn't address the distribution of earnings.

Conclusion:
B) The mean earnings in the community were $3,000 greater than the median earnings.
This option most directly calls into question Susan's belief by indicating that the distribution of incomes is skewed, with a few high earners likely pushing the mean above the median. Therefore, most people (the majority) in her community could be earning less than the mean, possibly making Susan's income closer to or even above what the majority earns.­
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Re: Susan read in her local paper that the average (arithmetic mean) [#permalink]
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Bunuel wrote:
­Susan read in her local paper that the average (arithmetic mean) income in her community last year was $38,000. She worked very hard to make $36,000 last year and was upset that she earned less than the majority of the earners in her community.

Which of the following, if true, most calls into question Susan’s belief that she earned less than the majority of the earners in her community?

(A) The standard deviation of earnings in her community was greater than $2,000.

(B) The mean earnings in the community were $3,000 greater than the median earnings.

(C) The mean earnings in the community were $1,000 greater than the median earnings.

(D) Doctors in Susan's community had an average income over $200,000.

(E) The average income of $38,000 is calculated from the incomes of both full- and part-time workers.


­
This is a CR Butler Question

    Check the links to other Butler Projects:
 
­

­

KAPLAN OFFICIAL EXPLANATION:



Step 1: The phrase “calls into question” signals that this is a Weaken question.

Step 2: Susan concludes that, since her income was less than the average in her community, her earnings were below the 50th percentile. She has conflated the meanings of median and mean in assuming earnings below the arithmetic mean put her in the bottom half of wage earners.

Step 3: The average income is computed by totaling all the earnings and dividing by the number of individuals counted. Susan's conclusion that she earned less than the majority of the earners in her community actually relates to the median earnings level rather than the mean. Thus, you could predict that a statement that provides information about the relationship between the median and mean earnings will be the correct answer.

Step 4: Choice (B) stipulates earnings $3,000 greater than the median level. Therefore, the median earnings would have been $38,000 – $3,000 = $35,000. Since Susan earned $36,000 her income would have been greater than the median amount. The median is the middle value for all the members of a set, in this case the set of wage earners in Susan’s community. Therefore, Susan actually earned more than half the earners in her community.

If choice (C) were applied, the median level would be $37,000 and Susan’s conclusion would actually be correct.

Standard deviation, in choice (A), refers to how widely dispersed the earnings levels are in the community, which has no bearing on Susan's reasoning.

While choice (D) might make you think that the mean earnings in Susan’s community must have been well above the median, this isn’t necessarily the case; it would depend on how many doctors there are relative to lower earners, and more generally on how earnings are distributed.

Finally, whether workers were part- or full-time, choice (E), has no bearing on the argument. Again, don't over-think the question by trying to figure out what the effect of part-time workers might be on the relationship between the mean and the median levels.­
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Re: Susan read in her local paper that the average (arithmetic mean) [#permalink]
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