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# M02-28

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Math Expert
Joined: 02 Sep 2009
Posts: 52971

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15 Sep 2014, 23:18
11
00:00

Difficulty:

65% (hard)

Question Stats:

59% (01:16) correct 41% (01:25) wrong based on 191 sessions

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The following table shows results of a quality inspection of a lot of 15 mirrors.

The difference between the median number of defects and the average number of defects in the sample checked is between:

A. -1 and 0
B. 0 and 0.5
C. 0.5 and 1
D. 1 and 1.5
E. 1.5 and 2

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 52971

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15 Sep 2014, 23:18
2
1
Official Solution:

The following table shows results of a quality inspection of a lot of 15 mirrors.

The difference between the median number of defects and the average number of defects in the sample checked is between:

A. -1 and 0
B. 0 and 0.5
C. 0.5 and 1
D. 1 and 1.5
E. 1.5 and 2

Sort the observations of the number of defects: 0 0 0 0 0 0 1 2 2 2 2 3 3 3 4

The median is the middle term, which is 2.

The average can be calculated as follows: $$\frac{0*6 + 1*1 + 2*4 + 3*3 + 4*1}{15} = \frac{22}{15}$$

The difference between the median and the average is $$2 -\frac{22}{15} = \frac{8}{15} = 0.53$$.

_________________
Manager
Joined: 05 Jul 2015
Posts: 100
GMAT 1: 600 Q33 V40
GPA: 3.3

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11 Jan 2016, 13:23
I think this is a poor-quality question and I agree with explanation. This does not seem like a 700 level question.
Manager
Joined: 09 Jun 2014
Posts: 240
Location: India
Concentration: General Management, Operations
Schools: Tuck '19

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20 Mar 2018, 04:56
Bunuel wrote:
Official Solution:

The following table shows results of a quality inspection of a lot of 15 mirrors.

The difference between the median number of defects and the average number of defects in the sample checked is between:

A. -1 and 0
B. 0 and 0.5
C. 0.5 and 1
D. 1 and 1.5
E. 1.5 and 2

Sort the observations of the number of defects: 0 0 0 0 0 0 1 2 2 2 2 3 3 3 4

The median is the middle term, which is 2.

The average can be calculated as follows: $$\frac{0*6 + 1*1 + 2*4 + 3*3 + 4*1}{15} = \frac{22}{15}$$

The difference between the median and the average is $$2 -\frac{22}{15} = \frac{8}{15} = 0.53$$.

Hi Bunnel,

The question asks the difference between Median and average number of defects ..I believe it should be just average or average of defects ..(No average number number of defects--gets me into thinking its the average of sum of frequency..)
Intern
Joined: 25 Jan 2018
Posts: 5

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27 Apr 2018, 18:27
I think this is a poor-quality question and I agree with explanation. The language of question is such that I subtracted median from average which resulted in choice A, I think they should specify it more clearly.
Manager
Joined: 17 Jan 2017
Posts: 61

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07 May 2018, 10:25
There is nothing wrong with that question. Clear explanation.
Manager
Joined: 20 Jul 2018
Posts: 75
WE: Corporate Finance (Investment Banking)

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27 Oct 2018, 04:05
It is indeed a very good question; it requires that you think before you jump into solving for arithmetic mean.
Re: M02-28   [#permalink] 27 Oct 2018, 04:05
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# M02-28

Moderators: chetan2u, Bunuel

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