Exercise 2: A Feature-Flag Rollout's Risk Profile — Possible Solution
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GIVEN
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P(X=0) = 0.85         (no issues)
P(X=2000) = 0.12       (minor rollback)
P(X=80000) = 0.03      (major outage)

STEP 1: E[X]
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E[X] = (0)(0.85) + (2000)(0.12) + (80000)(0.03)
     = 0 + 240 + 2400
     = 2640

The expected cost of the rollout is $2,640.

STEP 2: E[X^2]
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E[X^2] = (0^2)(0.85) + (2000^2)(0.12) + (80000^2)(0.03)
       = 0 + (4,000,000)(0.12) + (6,400,000,000)(0.03)
       = 0 + 480,000 + 192,000,000
       = 192,480,000

STEP 3: Var(X) AND SD(X)
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Var(X) = E[X^2] - (E[X])^2
       = 192,480,000 - (2640)^2
       = 192,480,000 - 6,969,600
       = 185,510,400

SD(X) = sqrt(185,510,400) ~= $13,620

WHAT THIS MEANS FOR "IS IT SAFE ENOUGH"
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The standard deviation (~$13,620) is more than five times larger than
the expected value ($2,640) - an even more extreme ratio than this
chapter's own incident-cost example. Per this chapter's own finding,
this signals a highly skewed, risky distribution: the expected value
alone makes the rollout look modest and manageable, but that number is
almost entirely an artifact of averaging in a rare, very expensive
outage scenario. Relying on E[X] alone to judge "is this safe" would
be misleading - the real question is whether the organization can
absorb an $80,000 major-outage cost if the unlucky 3% outcome actually
occurs, not just whether the average cost looks small.

WHY THIS WORKS AS AN ANSWER
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Both E[X] and Var(X)/SD(X) are computed directly from this chapter's
own formulas, and the risk interpretation is grounded in this
chapter's own finding about a large standard deviation relative to the
mean signalling understated risk, rather than treating the expected
value figure alone as a complete answer to whether the rollout is
safe.