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In four production runs, Runs 1 to 4, a quality inspector found \(w\), \(x\), \(y\), and \(z\) defective items, respectively. If the standard deviation of these four defect counts was \(d\), was the standard deviation of the defect counts for the next four production runs, Runs 5 to 8, greater than \(d\)?
(1) In each of the next four runs, Runs 5, 6, 7, and 8, the number of defective items found was 3.
(2) In each of the next four runs, Runs 5, 6, 7, and 8, the number of defective items found was n greater than in Runs 1, 2, 3, and 4, respectively.
(1) In each of the next four runs, Runs 5, 6, 7, and 8, the number of defective items found was 3.
(2) In each of the next four runs, Runs 5, 6, 7, and 8, the number of defective items found was n greater than in Runs 1, 2, 3, and 4, respectively.
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Official Solution:
In four production runs, Runs 1 to 4, a quality inspector found \(w\), \(x\), \(y\), and \(z\) defective items, respectively. If the standard deviation of these four defect counts was \(d\), was the standard deviation of the defect counts for the next four production runs, Runs 5 to 8, greater than \(d\)?
(1) In each of the next four runs, Runs 5, 6, 7, and 8, the number of defective items found was 3.
If Runs 5 to 8 each have 3 defects, then all four values are the same, so the standard deviation for Runs 5 to 8 is 0. Since a standard deviation is always 0 or positive, the standard deviation for Runs 5 to 8, which is 0, cannot be more than \(d\), whatever it might be. Sufficient.
(2) In each of the next four runs, Runs 5, 6, 7, and 8, the number of defective items found was \(n\) greater than in Runs 1, 2, 3, and 4, respectively.
Each of Runs 5 to 8 is obtained by adding the same amount \(n\) to the corresponding earlier value, \(w\), \(x\), \(y\), and \(z\). Adding the same constant to every value does not change the spread, so the standard deviation for Runs 5 to 8 is still \(d\). We are asked whether \(d\) is greater than \(d\), so the answer is No. Sufficient.
Answer: D
In four production runs, Runs 1 to 4, a quality inspector found \(w\), \(x\), \(y\), and \(z\) defective items, respectively. If the standard deviation of these four defect counts was \(d\), was the standard deviation of the defect counts for the next four production runs, Runs 5 to 8, greater than \(d\)?
(1) In each of the next four runs, Runs 5, 6, 7, and 8, the number of defective items found was 3.
If Runs 5 to 8 each have 3 defects, then all four values are the same, so the standard deviation for Runs 5 to 8 is 0. Since a standard deviation is always 0 or positive, the standard deviation for Runs 5 to 8, which is 0, cannot be more than \(d\), whatever it might be. Sufficient.
(2) In each of the next four runs, Runs 5, 6, 7, and 8, the number of defective items found was \(n\) greater than in Runs 1, 2, 3, and 4, respectively.
Each of Runs 5 to 8 is obtained by adding the same amount \(n\) to the corresponding earlier value, \(w\), \(x\), \(y\), and \(z\). Adding the same constant to every value does not change the spread, so the standard deviation for Runs 5 to 8 is still \(d\). We are asked whether \(d\) is greater than \(d\), so the answer is No. Sufficient.
Answer: D
