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Mean and median

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Manager
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Mean and median [#permalink] New post 29 Dec 2009, 04:51
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25% (00:00) correct 75% (00:32) wrong based on 5 sessions
The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the sixth car through the production line has either 3, 7, or 12 defects, for which of theses values does the mean number of defects per car for the first six cars equal the median?

I. 3
II. 7
III. 12



A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III
Senior Manager
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Location: India
Concentration: General Management
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Re: Mean and median [#permalink] New post 29 Dec 2009, 05:05
kirankp wrote:
The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the sixth car through the production line has either 3, 7, or 12 defects, for which of theses values does the mean number of defects per car for the first six cars equal the median?

I. 3
II. 7
III. 12



A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III



D - I and III only
for 7 we will have mean as 7.16(approx) and median as 7 but for 3 we will have mean and median = 6.5 and for 12 we will have median and mean =8
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Re: Mean and median [#permalink] New post 01 Jan 2010, 20:41
I disagree with the above response. The answer should be B) II only. 9, 7, 10, 4, and 6 added to 7 = 42. The mean of the set is therefore 7. The Median of the set is also 7. 3 doesn't work and there is only on option that includes 7 and excludes 3. B. Please post OA.

BTW to the response above: I believe that the median must be a number in the set.
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Re: Mean and median [#permalink] New post 01 Jan 2010, 23:03
9 + 7 + 10 + 4 + 6 + 7 = 43

As for the median, when the series is even-numbered then the median is the average of the two middle numbers. Answer choice D (I and III) is correct.
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Re: Mean and median [#permalink] New post 01 Jan 2010, 23:07
craigmcdermott wrote:
I disagree with the above response. The answer should be B) II only. 9, 7, 10, 4, and 6 added to 7 = 42. The mean of the set is therefore 7. The Median of the set is also 7. 3 doesn't work and there is only on option that includes 7 and excludes 3. B. Please post OA.

BTW to the response above: I believe that the median must be a number in the set.


we have set 9, 7, 10, 4, and 6 which is arranged as 4 6 7 9 10 [ total sum = 36] so adding 7 gives us 43 and not 42. Also mean will be 7.16 and median here will be 7
if we have 3 then the set which we get is 3 4 6 7 9 10 [ total sum = 39] median and mean are both 6.5 here
if we have 12 then the set will be 4 6 7 9 10 12 [ total sum = 48] median and mean both will be 8

Also Median necessarily doesnt need to be a number in the set
Re: Mean and median   [#permalink] 01 Jan 2010, 23:07
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