This question's explanation and your answers so far.
ソフトウェアの品質を判定する指標として，機能単位の不良件数をその開発規模で割った値を“不良密度”と定義する。不良密度の下限値と上限値を設定し，実績値がその範囲を逸脱した場合に問題ありと判定するとき，A工程では問題がなく，B工程で問題があると判定される機能はどれか。ここで，不良密度の下限値は0.25件／KS，上限値は0.65件／KS とする。また，不良密度の下限値，上限値及び開発規模は，両工程とも同じとする。
| 機能 | 開発規模(KS) | A工程の不良件数(件) | B工程の不良件数(件) |
|---|---|---|---|
| 機能1 | 10 | 6 | 3 |
| 機能2 | 20 | 14 | 10 |
| 機能3 | 50 | 10 | 40 |
| 機能4 | 80 | 32 | 8 |
Function 1 has a development size of 10 KS. In process A, 6 ÷ 10 = 0.60 defects/KS, within the range (no problem); in process B as well, 3 ÷ 10 = 0.30 defects/KS, within the range (no problem). Since process B has no problem, it does not satisfy the condition «a problem in process B», so it is wrong.
Function 2 has a development size of 20 KS. In process A, 14 ÷ 20 = 0.70 defects/KS, which exceeds the upper limit of 0.65, so there is already a problem at process A. This violates the condition «no problem in process A», so it is wrong.
Function 3 has a development size of 50 KS. In process A, 10 ÷ 50 = 0.20 defects/KS, which is below the lower limit of 0.25, so process A has a problem (process B is 40 ÷ 50 = 0.80, also a problem, but process A is out of range to begin with). This violates the condition «no problem in process A», so it is wrong.
正解 The correct answer is エ (Function 4). Defect density = number of defects ÷ development size (KS); the judgment range is a lower limit of 0.25 to an upper limit of 0.65 defects/KS, and falling outside this range is judged as «having a problem». Function 4 has a development size of 80 KS. In process A, 32 ÷ 80 = 0.40 defects/KS, which is within the range, so there is no problem; in process B, 8 ÷ 80 = 0.10 defects/KS, which is below the lower limit of 0.25, so there is a problem. Therefore, the only function that satisfies the condition «no problem in process A and a problem in process B» is Function 4, so エ is correct.