SME
Grade 12 Math · Statistics
Sampling distributions and the central limit theorem
Use this for “Heights have μ = 170 cm and σ = 10 cm. What is the chance a sample of 25 averages under 173 cm?”
Chart
The population has μ = 170 cm and σ = 10 cm.
Means
of
samples
of
n
=
25:
σ/√n
=
10/
√
25
=
2.
P(X̄ ≤ 173) = 0.9332.
This is an example. Type your own values to start a new problem.
SE (cm)
=
σ (cm)
n
z
=
x̄ (cm)
−
μ (cm)
SE (cm)
P
Chance the sample mean is at most x̄
calculated
Formulas
SE
=
σ
÷
√
n
2
=
10
÷
√
25
z
=
x̄
−
μ
SE
1.5
=
173 − 170
2
P
=
Φ(
z
)
0.9332 = Φ(1.5)
Assumptions
•
The sample means x̄ center on the population mean μ and spread by σ ÷ √n.
•
x̄ is close to normal when the population is normal or n ≥ 30 (the central limit theorem).
•
Samples are random and less than 10% of the population.
Step-by-step
WE KNOW
μ = 170 cm, σ = 10 cm, n = 25, x̄ = 173 cm
FIND
standard error of x̄ (SE), z-score of x̄ (z), chance the sample mean is at most x̄ (P)
Step 1 · Find standard error of x̄ (SE)
Use SE = σ ÷ √n
Means
of
n
values
spread
less
than
single
values:
divide
σ
by
the
root
of
n
.
SE
=
10
÷
√
25
SE
=
10
5
SE = 2 cm
Step 2 · Find z-score of x̄ (z)
Use z = (x̄ − μ) ÷ SE
How
many
standard
errors
x̄
is
from
μ
.
z
=
173 − 170
2
z
=
3
2
z
=
1.5
Step 3 · Find chance the sample mean is at most x̄ (P)
Use P = Φ(z)
Φ(
z
)
is
the
area
under
the
standard
normal
curve
left
of
z
.
P
=
Φ(1.5)
P
=
0.9332
→
more
than
x̄
:
1
−
0.9332
=
0.0668
Check
Put the numbers back in. Both sides should match.
2
=
10
÷
√
25
✓
1.5
=
173 − 170
2
✓
0.9332 = Φ(1.5) ✓
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