SME
Grade 12 Math · Problem type
The central limit theorem, simulated
Use this for “Wait times are skewed right with mean 4 minutes. How are the means of samples of 30 spread?”
Chart
Population: wait times skewed right, mean μ = 4, σ = 4
1,000
samples
of
n
=
30:
their
means
pile
up
around
μ,
spread
σ/√n
=
4
÷
√
30
=
0.73
The simulated means average 4.04, spread 0.71; n ≥ 30, so close to the normal curve
This is an example. Type your own values to start a new problem.
μ
Population mean wait
example · min
n
Sample size
example
m
Samples taken
example
σ/√n
Spread of the sample means
worked out (not typed) · min
N
Waits drawn in all
worked out (not typed)
Formulas
N
=
n
×
m
30,000 = 30 × 1,000
σ/√
n
=
μ
÷
√
n
0.7303
=
4
÷
√
30
Assumptions
•
The waits are exponential, so σ = μ: skewed right, most short and a few long.
•
The sample means center on μ and spread by σ ÷ √n, whatever the population’s shape.
•
For n of about 30 or more the sample means are close to normal, though the waits are not.
Step-by-step
WE KNOW
μ = 4 min, n = 30, m = 1,000
FIND
waits drawn in all (N), spread of the sample means (σ/√n)
Step 1 · Find waits drawn in all (N)
Use N = n × m
Each
of
the
m
samples
draws
n
waits
from
the
population.
N
=
30
×
1,000
N
=
30,000
Step 2 · Find spread of the sample means (σ/√n)
Use σ/√n = μ ÷ √n
Here
σ
=
μ
:
divide
it
by
√
n
,
the
square
root
of
the
sample
size.
σ/√
n
=
4
÷
√
30
σ/√
n
=
4
5.477
σ/√
n
=
0.7303
min
Check
Put the numbers back in. Both sides should match.
30,000 = 30 × 1,000 ✓
0.7303
≈
4
÷
√
30
✓
Topic check
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