Averaging steadies the wobble: A coffee machine pours cups whose volume varies: mean μ = 250 mL, standard deviation σ = 12 mL. One cup might be 231 mL or 268 mL — quite a spread.
Now pour n = 9 cups and average them. That average X̄ also varies from one batch of nine to the next — but it wobbles less than a single cup, because a few low cups get cancelled by a few high ones.
How much less? Its centre is still μ = 250, but its standard deviation shrinks to the standard error:
Why √n, not n: Averaging n values divides the spread by √n, not n. So to halve the standard error you need four times as much data (because √4 = 2). That is why bigger samples give steadier averages — but with diminishing returns.
IB-style question — standard error
The coffee machine pours cups with mean μ = 250 mL and standard deviation σ = 12 mL. A batch of n = 9 cups is poured and the mean volume X̄ is recorded.
Find the mean and standard deviation of X̄.
Step by step
- The mean of the sample mean is unchanged — it is still the population mean.
- The standard deviation of X̄ is the standard error: σ over √n.
Final answer
X̄ has mean 250 mL and standard deviation 4 mL. The average of 9 cups varies three times less than a single cup (12 → 4).
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The shape becomes a bell — even when the population isn't: Single delivery times at a depot are skewed — most are quick, a few are very long. That population is not Normal.
But the Central Limit Theorem says: average enough of them and the average behaves Normally anyway. For a large sample size n, the distribution of the sample mean X̄ is approximately Normal:
When does 'large enough' kick in?: If the population is already Normal, X̄ is exactly Normal for any n — even n = 2.
If the population is not Normal, you need a reasonably large n (a common rule of thumb is n ≥ 30) before the bell shape is a good approximation. Smaller-but-still-skewed → the approximation is rough.
IB-style question — probability for a sample mean
Delivery times at a depot have mean μ = 40 min and standard deviation σ = 15 min (the distribution is skewed). A sample of n = 36 deliveries is taken.
Using the Central Limit Theorem, find the probability that the mean delivery time of the sample exceeds 44 minutes.
Step by step
- By the CLT, with n = 36 large, the sample mean is approximately Normal. Find its standard error.
- So X̄ is approximately Normal with this mean and standard deviation.
- Find P(X̄ > 44) on the GDC normal-cdf with mean 40, standard deviation 2.5.
Final answer
P(X̄ > 44) ≈ 0.0548. Even though single deliveries are skewed, the mean of 36 is close to Normal, so this calculation is valid.
Single value vs sample mean — don't mix them up: P(one delivery > 44) would use σ = 15. P(mean of 36 deliveries > 44) uses the standard error σ/√n = 2.5. The mean is far less likely to be extreme — that is why the second probability is so much smaller.