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# For the 5 days shown in the graph, how many kilowatt-hours greater was

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For the 5 days shown in the graph, how many kilowatt-hours greater was  [#permalink]

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13 Jan 2017, 10:50
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Difficulty:

15% (low)

Question Stats:

79% (01:20) correct 21% (01:24) wrong based on 217 sessions

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For the 5 days shown in the graph, how many kilowatt-hours greater was the median daily electricity use than the average (arithmetic mean) daily electricity use?

A) 1
B) 2
C) 3
D) 4
E) 5

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Re: For the 5 days shown in the graph, how many kilowatt-hours greater was  [#permalink]

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13 Jan 2017, 20:23
felippemed wrote:
For the 5 days shown in the graph, how many kilowatt-hours greater was the median daily electricity use than the average (arithmetic mean) daily electricity use?

A) 1
B) 2
C) 3
D) 4
E) 5

well the median is =27 from set (19,24,27,29,31)
and mean is (31+29+27+24+19)/5 =26
thus difference =27-26 =1
Ans A
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Re: For the 5 days shown in the graph, how many kilowatt-hours greater was  [#permalink]

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14 Jan 2017, 00:20
felippemed wrote:
For the 5 days shown in the graph, how many kilowatt-hours greater was the median daily electricity use than the average (arithmetic mean) daily electricity use?

A) 1
B) 2
C) 3
D) 4
E) 5

Median Data $$= { 19 , 24 , 27 , 29 , 31 }$$ & $$Median = 27$$

Mean= $$\frac{( 31 + 27 + 29 + 24 + 19 )}{5} = 26$$

Difference between Median and Mean = 27 - 26 => 1

Hence, answer will be (A) 1

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Re: For the 5 days shown in the graph, how many kilowatt-hours greater was  [#permalink]

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08 May 2017, 02:07
Median of 19, 24, 27, 29 and 31 is 27 (middle value in the list)

Mean = (19+24+27+29+31) / 5 = 130/5 = 26

Difference = 27-26 = 1
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Re: For the 5 days shown in the graph, how many kilowatt-hours greater was  [#permalink]

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09 May 2017, 00:07
To find the Median, first of all arrange the data in ascending or descending order

(19,24,27,29,31) Median value is the middle value of the series

Median = 27

Mean= $$\frac{(19+24+27+29+31)}{5}$$= 26

Difference between Median and Mean = 27 - 26= 1

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Re: For the 5 days shown in the graph, how many kilowatt-hours greater was  [#permalink]

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30 Sep 2017, 10:12
1
Abhishek009 VeritasPrepKarishma Bunuel Engr2012

Quote:
Mean= $$\frac{( 31 + 27 + 29 + 24 + 19 )}{5} = 26$$

There is no shortcut to this? for eg: for AP, ,mean = $$\frac{(first term + last term)}{2}$$
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Re: For the 5 days shown in the graph, how many kilowatt-hours greater was  [#permalink]

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08 Mar 2019, 08:06
felippemed wrote:
For the 5 days shown in the graph, how many kilowatt-hours greater was the median daily electricity use than the average (arithmetic mean) daily electricity use?

A) 1
B) 2
C) 3
D) 4
E) 5

Listing the numbers from smallest to largest, we have: 19, 24, 27, 29 and 31. So 27 is the median. The average is (19 + 24 + 27 + 29 + 31)/5 = 130/5 = 26. Therefore, the median exceeds the average by 1.

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Re: For the 5 days shown in the graph, how many kilowatt-hours greater was   [#permalink] 08 Mar 2019, 08:06
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