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FRM Part II · FRM Exam Part II · Risk Governance

A bank's operational risk appetite sets a tolerance that fraud losses stay below USD 10 million per year. The risk committee reports that expected annual fraud loss is USD 6 million with a standard deviation of USD 2 million, and losses are approximated as normal. The board wants no more than a 2.5% chance of breaching the tolerance (z of 1.96 at 97.5%). What is the best assessment?

The 97.5th percentile annual fraud loss is 6 + 1.96 x 2 = USD 9.92 million, which is below the USD 10 million tolerance. The probability of breach is therefore slightly under 2.5%, so the board's confidence requirement is met, though headroom is thin.

  1. AThe tolerance is met, because 6 + 1.96 x 2 = 9.92 is below 10Correct
  2. BThe tolerance is breached, because 6 + 1.96 x 2 = 11.92 exceeds 10
  3. CThe tolerance is met, because expected loss of 6 is below 10 regardless of variability
  4. DThe tolerance is breached, because 6 + 2 x 2 = 10 equals the limit

Explanation

The 97.5th percentile loss is 6 + 1.96 x 2 = 9.92, below USD 10 million, so breach probability is just under 2.5%. Option C ignores variability, which the board's confidence requirement explicitly includes.

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