SME
Grade 12 Math · Problem type
Sample means and sums with any σ
Use this for “Bags of rice weigh 2 kg on average with σ = 0.12 kg. Find the chance that 36 bags average over 2.03 kg, or weigh over 73.08 kg in all.”
Chart
The population has μ = 2 kg and σ = 0.12 kg.
Means
of
samples
of
n
=
36:
σ/√n
=
0.12/
√
36
=
0.02.
P(X̄ ≥ 2.03) = 0.0668.
Mean of the total: 72 kg
Standard deviation of the total: 0.72 kg
A sample total: 73.08 kg
This is an example. Type your own values to start a new problem.
SE (kg)
=
σ (kg)
n
z
=
x̄ (kg)
−
μ (kg)
SE (kg)
μΣX
Mean of the total
worked out (not typed)
σΣX
Standard deviation of the total
worked out (not typed)
Σx
A sample total
calculated
P
Chance the sample mean is more than x̄
calculated
Formulas
SE
=
σ
÷
√
n
0.02
=
0.12
÷
√
36
μΣX
=
n
×
μ
72 = 36 × 2
σΣX
=
√
n
×
σ
0.72
=
√
36
×
0.12
Σx
=
n
×
x̄
73.08 = 36 × 2.03
z
=
x̄
−
μ
SE
1.5
=
2.03 − 2
0.02
z
=
Σx − μΣX
σΣX
1.5
=
73.08 − 72
0.72
P
=
1
−
Φ(
z
)
0.06681 = 1 − Φ(1.5)
Assumptions
•
The sample is random; x̄ and the total ΣX are close to normal when n ≥ 30 or the population is normal.
•
The mean x̄ centers on μ and spreads by σ ÷ √n; the total centers on nμ and spreads by √n × σ.
•
The mean is over x̄ exactly when the total is over n × x̄, so both chances are the same.
Step-by-step
WE KNOW
x̄ = 2.03 kg, μ = 2 kg, σ = 0.12 kg, n = 36
FIND
standard error of x̄ (SE), mean of the total (μΣX), standard deviation of the total (σΣX), a sample total (Σx), z-score of x̄ and of Σx (z), chance the sample mean is more than 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
=
0.12
÷
√
36
SE
=
0.12
6
SE = 0.02 kg
Step 2 · Find mean of the total (μΣX)
Use μΣX = n × μ
Each
of
the
n
values
adds
μ
to
the
total
on
average.
μΣX = 36 × 2
μΣX = 72 kg
Step 3 · Find standard deviation of the total (σΣX)
Use σΣX = √n × σ
Variances
add:
n
values
give
nσ²,
so
the
total
spreads
by
√
n
×
σ
.
σΣX
=
√
36
×
0.12
σΣX = 6 × 0.12
σΣX = 0.72 kg
Step 4 · Find a sample total (Σx)
Use Σx = n × x̄
A
mean
of
x̄
over
n
values
is
a
total
of
n
×
x̄
.
Σx = 36 × 2.03
Σx = 73.08 kg
Step 5 · Find z-score of x̄ and of Σx (z)
Use z = (x̄ − μ) ÷ SE
How
many
standard
errors
x̄
is
from
μ
.
z
=
2.03 − 2
0.02
z
=
0.03
0.02
z
=
1.5
Step 6 · Find chance the sample mean is more than x̄ (P)
Use P = 1 − Φ(z)
The
whole
area
is
1;
take
away
the
area
left
of
z
.
P
=
1
−
Φ(1.5)
P
=
0.06681
→
at
most
x̄
,
or
a
total
of
at
most
Σx:
1
−
0.06681
=
0.93319
Check
Put the numbers back in. Both sides should match.
0.02
=
0.12
÷
√
36
✓
72 = 36 × 2 ✓
0.72
=
√
36
×
0.12
✓
73.08 = 36 × 2.03 ✓
1.5
=
2.03 − 2
0.02
✓
1.5
=
73.08 − 72
0.72
✓
0.06681 = 1 − Φ(1.5) ✓
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