Exercise 3: Why Averaging Many Skewed Weeks Becomes Predictable — Possible Solution
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THE SETUP
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Chapter 5's own weekly-incident-cost example has mean $1,500 but a
standard deviation of about $7,057 - more than four times the mean.
That distribution is nothing like a normal bell curve: it's dominated
by a 70% chance of $0 and a rare 2% chance of a $50,000 catastrophe,
making any single week's actual cost wildly unpredictable.

WHY THE AVERAGE ACROSS MANY WEEKS BEHAVES DIFFERENTLY
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Per this chapter's own Central Limit Theorem, the distribution of a
SAMPLE MEAN approaches a normal distribution as the number of samples
grows, regardless of how skewed or unusual the original underlying
distribution is. Chapter 5's own weekly cost distribution never has to
be normal itself for this to apply - the CLT specifically concerns the
average of many independent draws from it, not the distribution of any
single draw.

WHY THE AVERAGE IS MORE TIGHTLY CLUSTERED
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Per this chapter's own standard error formula, the standard deviation
of the sample mean shrinks as sigma/sqrt(n) - so averaging over many
independent weeks doesn't just make the result look more bell-shaped,
it also makes it genuinely less variable than any single week's own
cost. A single week might land anywhere from $0 to $50,000; the
average over, say, 100 weeks, will cluster in a much narrower band
around the true $1,500 mean, because the rare catastrophic weeks get
diluted across many ordinary weeks in the average rather than
dominating a single data point the way they would for just one week.

WHY THIS WORKS AS AN ANSWER
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The explanation applies this chapter's own Central Limit Theorem
directly to Chapter 5's own specific, already-established skewed
distribution, and separately invokes this chapter's own standard-
error-shrinks-with-n property to explain not just that the average
becomes more normal-shaped, but specifically why it also becomes
tighter/less variable than any individual week's cost.