Susan read in her local paper that the average (arithmetic mean)

Question

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.

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Reva22 · Answer

B, it's pretty obvious. If Mean > Median implies distribution is right skewed. Median is 50th %le, so median = 38000-3000 = 35000, which is less that what Susan is earning.

abhishekb1352 · Answer

I think option B is right as it says the Median is 3000 dollars lesser than the mean, suggesting Susan is earning decently and above the 50th Percentile in her community.

Vivek1707 · Answer

A. Whether the standard deviation is above $2,000 doesn't tell us whether most people earned more or less than Susan.
B. Correct If the mean ($38k) is $3,000 higher than the median ($35k), Susan's income of $36k is actually above the median. That means she earned more than at least half the people in her community, so her belief is clearly mistaken.
C. If the median is $37k, Susan's $36k is still below the median, so this doesn't challenge her belief much.
D. A few doctors earning very high incomes could raise the mean, but this alone doesn't tell us where Susan stands relative to everyone else.
E. Including part-time workers affects the average, but it doesn't tell us whether Susan earned less than the majority.

Bunuel · Answer

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.

Susan read in her local paper that the average (arithmetic mean)

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­Susan read in her local paper that the average (arithmetic mean)

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My Rewards

Susan read in her local paper that the average (arithmetic mean)