The Ministry of Justice asks you to compare the amount of noise in several courts. To this purpose, you run the following study. Three judges are presented with evidence concerning three cases. They all evaluate the same cases and they all agree in finding the suspects guilty. However, they differ in the number of years in jail that they sentence. For instance, Judge A sentences Case 1 with 7.5 years, while Judge B sentences the same case with 5 years. What is the amount of system noise in this court? [table]
| Case 1 | Case 2 | Case 3 | MEAN Judge A | 7.5 | 6 | 6.5 | 6.67 Judge B | 5 | 6 | 6 | 5.67 Judge C | 4 | 8.5 | 7 | 6.50 MEAN | 5.50 | 6.83 | 6.50 | 6.28
[/table]

Question

BlackTom AI · Accepted Answer

System noise here is the average dispersion of judges’ sentences for the same case — i.e., compute the standard deviation of the three judges’ sentences for each case and then average those case-level SDs.

1. Relevant concepts
- For each case, variability (noise) among judges = standard deviation of the three sentences.
- We use the population standard deviation for each case (divide squared deviations by 3), because we are summarizing the actual set of three judges for these cases.
- System noise = mean of the three case-level SDs.

2. Step-by-step calculations

Case 1:
- Values: 7.5, 5, 4. Mean = (7.5 + 5 + 4) / 3 = 16.5 / 3 = 5.50
- Deviations: 7.5−5.5 = 2.0; 5−5.5 = −0.5; 4−5.5 = −1.5
- Squared deviations: 4.00, 0.25, 2.25 → sum = 6.50
- Variance = 6.50 / 3 = 2.1667 → SD = sqrt(2.1667) = 1.472

Case 2:
- Values: 6, 6, 8.5. Mean = (6 + 6 + 8.5) / 3 = 20.5 / 3 = 6.833...
- Deviations: −0.8333, −0.8333, +1.6667
- Squared deviations: 0.6944, 0.6944, 2.7778 → sum ≈ 4.1667
- Variance = 4.1667 / 3 = 1.3889 → SD = sqrt(1.3889) = 1.179

Case 3:
- Values: 6.5, 6, 7. Mean = (6.5 + 6 + 7) / 3 = 19.5 / 3 = 6.50
- Deviations: 0, −0.5, +0.5
- Squared deviations: 0, 0.25, 0.25 → sum = 0.50
- Variance = 0.50 / 3 = 0.1667 → SD = sqrt(0.1667) = 0.408

3. Combine to get system noise
- Average SD = (1.472 + 1.179 + 0.408) / 3 = 3.059 / 3 = 1.020 ≈ 1.02

4. Why the correct answer is correct
- The average case-level SD is ≈1.02, which lies between 1.0 and 1.2.

5. Why other options are incorrect
- (a) Below 1: contradicted by computed average SD ≈1.02 (greater than 1).
- (c) Between 1.2 and 1.4, (d) Between 1.4 and 1.6, (e) Above 1.6: all are higher ranges than the computed 1.02.

Conclusion: The amount of system noise is between 1 and 1.2, so option b is correct.

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