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Joined: 05 Jan 2019
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Five batches of 100 nails each are taken from a production line. The
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01 Mar 2019, 15:52
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65% (01:49) correct 35% (02:04) wrong based on 17 sessions
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Five batches of 100 nails each are taken from a production line. The numbers of defective nails in the first four batches are 2, 4, 3, and 5, respectively. If the fifth batch has either 1, 2, or 6 defective nails, for which of these values does the average (arithmetic mean) number of defective nails per batch for the five batches equal the median number of defective nails for the five batches? I. 1 II. 2 III. 6 A. I only B. II only C. III only D. I and III only E. I, II, and III
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Joined: 17 Jul 2016
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Five batches of 100 nails each are taken from a production line. The
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Updated on: 02 Mar 2019, 09:19
Consider the cases.
The first four batches have a total of 2+3+4+5=14 defective nails.
If the fifth batch has 1 defective nail, then the total number of defective nails is 14+1=15. The average is 15/5=3 If we order them (1,2,3,4,5) we see the median IS 3. Since the mean and median are both 3, having 1 defective nail in the fifth batch satisfies the condition.
If the fifth batch has 2 defective nails, the average would be 16/5=3.2 If we order them (2,2,3,4,5) we see the median is 3. The median doesn't equal the mean in this case.
If the fifth batch has 6 defective nails, the total would be 14+6=20. The average would be 20/5=4 In order (2,3,4,5,6) we see the median is 4. The mean DOES equal the median.
Having 1 or 6 nails in the fifth batch satisfies the condition.



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Re: Five batches of 100 nails each are taken from a production line. The
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01 Mar 2019, 23:45
In any 'equally spaced' list of numbers, the mean and median are always equal. So in this list, say: 1, 2, 3, 4, 5 (where the list increases by 1 each time), or in this list: 10, 17, 24, 31, 38, 45, 52 (where the list increases by 7 each time), the mean and the median are definitely equal. If you know that, you can instantly see in this question that 1, 2, 3, 4, 5 and 2, 3, 4, 5, 6 will both be lists with equal mean and median. And since 2, 2, 3, 4, 5 has the same median as, but a larger mean than 1, 2, 3, 4, 5, then it can't be true that the mean and median are the same for this list. So the answer is I and III only.
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Re: Five batches of 100 nails each are taken from a production line. The
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02 Mar 2019, 09:01
new2on wrote: Five batches of 100 nails each are taken from a production line. The numbers of defective nails in the first four batches are 2, 4, 3, and 5, respectively. If the fifth batch has either 1, 2, or 6 defective nails, for which of these values does the average (arithmetic mean) number of defective nails per batch for the five batches equal the median number of defective nails for the five batches?
I. 1 II. 2 III. 6 A. I only B. II only C. III only D. I and III only E. I, II, and III Total number of nails \(= 100\) Defective nails in first \(4\) batches \(= 2, 4, 3,\) and \(5\) respectively. \(5th\) batch could have either \(1, 2,\) or \(6\) defective nails. Median of defective nails respectively for \(1, 2\) or \(6\) nails could be the following\(;\) Median when \(6th\) batch has \("1"\) defective nail \(=> 1,2,3,4,5 = 3\) Median when \(6th\) batch has \("2"\) defective nails \(=> 2,2,3,4,5 = 3\) Median when \(6th\) batch has \("6"\) defective nails \(=> 2,3,4,5,6 = 4\) Total number of defective nails in \(4\) batches \(= 2+3+4+5 = 14\) Therefore Average for defective nails in \(5\) batches could be the following \(;\) Average when \(6th\) batch has \("1"\) defective nail \(=> \frac{14 + 1}{5} = \frac{15}{5} = 3\) Average when \(6th\) batch has \("2"\) defective nails \(=>\frac{14+2}{5} = \frac{16}{5} = 3.2\) Average when \(6th\) batch has \("6"\) defective nails \(=>\frac{14+6}{5} = \frac{20}{5} = 4\) Therefore defective nails for \(6th\) batch where average equals median is \(= 1\) and \(6\) \(I\) and \(III\) only. Answer D




Re: Five batches of 100 nails each are taken from a production line. The
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