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In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?

(1) In City X last April, the sum of the 30 daily high temperatures was 2,160°. (2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature.

Diagnostic Test Question: 43 Page: 26 Difficulty: 650

In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?

Let's consider the temperatures in April in ascending order: \(t_1\leq{t_2}...\leq{t_{15}}\leq{t_{16}}\leq...\leq{t_{29}}\leq{t_{30}}\). The median temperature would be the average of 15th and 16th greatest temperatures. The question asks: is this value less than the average temperature, (which would be the sum of temperatures divided by 30).

(1) In City X last April, the sum of the 30 daily high temperatures was 2,160° --> gives the average temperature, but not info about the median. Not sufficient.

(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature --> each temperature (term) from the 18 lowest ones (from \(t_1\) to \(t_{18}\)) given to be less than average, hence the median (average of 15th and 16th terms) would also be less than the average. Sufficient.

Re: In City X last April, was the average (arithmetic mean) [#permalink]

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09 Jul 2012, 05:57

4

This post received KUDOS

Bunuel wrote:

In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?

(1) In City X last April, the sum of the 30 daily high temperatures was 2,160°. (2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature.

Hi,

Difficulty level: 600

Is AM > Median? Median = Middle value of a sequence. Arthimetic meam (AM) = Central tendency of a sequence.

Using (1), Sum of 30 daily high temperature = 2160. We can find the mean, but no median. Insufficient.

Using (2), 60% of the daily high temperatures were less than the average daily high temperature. xxxxxxxxxxxxxxxxxx, the median will be in the green region and on the left side of mean. Thus, Median < AM. Sufficient.

Re: In City X last April, was the average (arithmetic mean) [#permalink]

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11 Jul 2012, 11:50

I think its B.. bunuel plz post the ans as soon as possible because i have exam in few days ..so i have to make sure m doing rite or wrong..
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Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?

Let's consider the temperatures in April in ascending order: \(t_1\leq{t_2}...\leq{t_{15}}\leq{t_{16}}\leq...\leq{t_{29}}\leq{t_{30}}\). The median temperature would be the average of 15th and 16th greatest temperatures. The question asks: is this value less than the average temperature, (which would be the sum of temperatures divided by 30).

(1) In City X last April, the sum of the 30 daily high temperatures was 2,160° --> gives the average temperature, but not info about the median. Not sufficient.

(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature --> each temperature (term) from the 18 lowest ones (from \(t_1\) to \(t_{18}\)) given to be less than average, hence the median (average of 15th and 16th terms) would also be less than the average. Sufficient.

In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?

(1) In City X last April, the sum of the 30 daily high temperatures was 2,160°. (2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature.

Diagnostic Test Question: 43 Page: 26 Difficulty: 650

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I solved this in a rather unconventional way, and I do not know if it works generally:

1) is clearly insufficient. But 2) basically is telling us that "there are outliers to the right that jack up the average", and a key rule of thumb in statistics is that if there are observations in the sample that jack up the arithmetic mean, then the median is a better representation of the "real" population value. The median is in that case always higher/lowr than the average (in our case, lower because 60% of the observations are lower than the average value) and thus through inference and common sense I concluded that the arithmetic mean is higher than the median. B is therefore sufficient.

Re: In City X last April, was the average (arithmetic mean) [#permalink]

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08 Apr 2014, 00:50

Bunuel wrote:

SOLUTION

In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?

Let's consider the temperatures in April in ascending order: \(t_1\leq{t_2}...\leq{t_{15}}\leq{t_{16}}\leq...\leq{t_{29}}\leq{t_{30}}\). The median temperature would be the average of 15th and 16th greatest temperatures. The question asks: is this value less than the average temperature, (which would be the sum of temperatures divided by 30).

(1) In City X last April, the sum of the 30 daily high temperatures was 2,160° --> gives the average temperature, but not info about the median. Not sufficient.

(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature --> each temperature (term) from the 18 lowest ones (from \(t_1\) to \(t_{18}\)) given to be less than average, hence the median (average of 15th and 16th terms) would also be less than the average. Sufficient.

Answer: B.

Hi Bunuel,

Cant 60% of daily temperature less than average daily temperature range from t1 to t9 and t21 to t30 then these wont accomodate the median(t15 and t16)

In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?

Let's consider the temperatures in April in ascending order: \(t_1\leq{t_2}...\leq{t_{15}}\leq{t_{16}}\leq...\leq{t_{29}}\leq{t_{30}}\). The median temperature would be the average of 15th and 16th greatest temperatures. The question asks: is this value less than the average temperature, (which would be the sum of temperatures divided by 30).

(1) In City X last April, the sum of the 30 daily high temperatures was 2,160° --> gives the average temperature, but not info about the median. Not sufficient.

(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature --> each temperature (term) from the 18 lowest ones (from \(t_1\) to \(t_{18}\)) given to be less than average, hence the median (average of 15th and 16th terms) would also be less than the average. Sufficient.

Answer: B.

Hi Bunuel,

Cant 60% of daily temperature less than average daily temperature range from t1 to t9 and t21 to t30 then these wont accomodate the median(t15 and t16)

No, since \(t_1\leq{t_2}...\leq{t_{15}}\leq{t_{16}}\leq...\leq{t_{29}}\leq{t_{30}}\).
_________________

Re: In City X last April, was the average (arithmetic mean) [#permalink]

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24 Apr 2014, 04:10

Bunuel wrote:

abid1986 wrote:

Bunuel wrote:

SOLUTION

In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?

Let's consider the temperatures in April in ascending order: \(t_1\leq{t_2}...\leq{t_{15}}\leq{t_{16}}\leq...\leq{t_{29}}\leq{t_{30}}\). The median temperature would be the average of 15th and 16th greatest temperatures. The question asks: is this value less than the average temperature, (which would be the sum of temperatures divided by 30).

(1) In City X last April, the sum of the 30 daily high temperatures was 2,160° --> gives the average temperature, but not info about the median. Not sufficient.

(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature --> each temperature (term) from the 18 lowest ones (from \(t_1\) to \(t_{18}\)) given to be less than average, hence the median (average of 15th and 16th terms) would also be less than the average. Sufficient.

Answer: B.

Hi Bunuel,

Cant 60% of daily temperature less than average daily temperature range from t1 to t9 and t21 to t30 then these wont accomodate the median(t15 and t16)

No, since \(t_1\leq{t_2}...\leq{t_{15}}\leq{t_{16}}\leq...\leq{t_{29}}\leq{t_{30}}\).

are u assuming this because we are dealing with median temperature? because for median, set is ordered from the lowest to highest.

Re: In City X last April, was the average (arithmetic mean) [#permalink]

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14 Dec 2015, 02:29

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