new2on
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