STAT5002 Chap.8 The Law of Averages and the Central Limit Theorem
The Law of Averages and the Central Limit Theorem
Once you can find the expected value and standard error of a sum or an average, the next question is what happens as the number of draws grows. Two results answer it. The first is the law of averages: because the standard error of an average is σ/√n, the average of many draws settles ever more tightly around the mean of the box.
The second is easy to overlook: the standard error of the sum, σ√n, keeps growing, so the count of heads in a long run of coin tosses drifts further from exactly half even while the proportion of heads closes in on one half. A famous wartime coin experiment by John Kerrich recorded 5,067 heads in 10,000 tosses, a chance error that looks large as a count yet is ordinary once measured in standard errors.
You will then see the shape of the box of all possible averages change as n increases, drawn from exact chances for a lopsided 0-1 box, and why skewed boxes need far more draws than symmetric ones before that shape looks normal. The final part is the Central Limit Theorem: for large n, the chance that a sum is at most s is approximated by converting s to standard units and reading the area from pnorm.
A worked game example compares the approximation with the exact answer at several sample sizes, so you can see how close it gets and when to trust it. Expect to use these ideas constantly afterwards, since prediction intervals, confidence intervals and tests for proportions and means all rest on a normal approximation to a sum or an average.
What this chapter covers
- 01
Kerrich's coin experiment as a box model
- 02
The 0 1 box with mean and SD of one half
- 03
Chance error of a count grows like the square root of n
- 04
Chance error of a proportion shrinks like one over root n
- 05
Law of averages and the law of large numbers
- 06
Shape of the box of all possible averages
- 07
Symmetric boxes versus skewed boxes
- 08
Central Limit Theorem for sums of draws
- 09
Standard units and pnorm for sums and averages
- 10
How large n needs to be for the normal approximation
Worked example · free
Bonus items from a vending machine
- +1Model each spin as a draw from a 0-1 box with p = 1/8 = 0.125, so μ = 0.125 and σ = √(0.125 − 0.125²) = 0.3307.
- +1Number of bonus items S is the sum of 320 draws: E(S) = 320 × 0.125 = 40.
- +1SE(S) = σ√n = 0.3307 × √320 = 5.916.
- +1Standard units: z = (50 − 40)/5.916 = 1.690.
- +1P(S > 50) = 1 − P(S ≤ 50) ≈ 1 − pnorm(50, 40, 5.916) = 0.0455.
Key terms
- Law of averages
- The law of averages is the result that a sample average concentrates around the box mean as n grows, because its SE, σ/√n, tends to zero.
- Central Limit Theorem
- The Central Limit Theorem is the result that the box of all possible sums of n draws from a fixed box becomes normal-shaped for large n.
- Standard units
- Standard units are a value's distance from its expected value measured in standard errors, (s − nμ)/(σ√n) for a sum.
- Box of all possible sums
- The box of all possible sums is the collection of every equally likely sum of n draws, with mean nμ and SD σ√n.
- Skewed box
- A skewed box is a box whose ticket histogram is lopsided, such as one ticket in six marked 1, so sums need many draws to look normal.
- Normal approximation
- A normal approximation is an estimate of a chance from the area under a normal curve with matching expected value and standard error, found in R with pnorm.
The Law of Averages and the Central Limit Theorem FAQ
Does the law of averages mean the number of heads catches up with half the tosses?
Not at all. The standard error of the number of heads is σ√n, which grows with more tosses, so the gap between heads and half the tosses tends to widen. Only the proportion of heads settles, because its standard error σ/√n shrinks.
How large does n have to be for the Central Limit Theorem to work?
It depends on the shape of the box. A fairly symmetric box can give a good normal approximation with about 5 or 10 draws, while a very skewed box may need more than 100 draws before the sums look normal.
How do I use pnorm for a sum of draws?
Find E(S) = nμ and SE(S) = σ√n first, then call pnorm(s, nμ, σ√n) for the chance that the sum is at most s. Subtract from one for an upper tail, and use μ and σ/√n instead when the question is about an average.
Does the Central Limit Theorem make the box itself normal?
No. The theorem describes the box of all possible sums or averages, not the tickets you draw from. Tickets in a skewed box stay skewed; what becomes normal-shaped is the distribution of the total or average of many draws.
Why does the proportion of heads get closer to one half if the count does not?
The proportion is the count divided by n. The gap in the count grows only like the square root of n, so dividing by n makes the gap in the proportion shrink like one over the square root of n.
Exam move
Make a two-column table for n = 10, 100, 1,000 and 10,000 tosses of a fair coin and fill in the SE of the count and the SE of the proportion; seeing one column grow while the other shrinks is the core of the law of averages. Then practise turning any word problem into a box: write down the tickets, compute μ and σ with mean square minus squared mean, and decide whether the question is about a sum or an average.
For each Central Limit Theorem question, write the three numbers E, SE and z before calling pnorm, and sketch a normal curve with the shaded side so you choose the correct tail. Finally, simulate one case in R with sample() and compare your pnorm answer with the simulated proportion, noting whether the box is symmetric or skewed.
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