SME
Grade 11 Math · Problem type
The share between two values
Use this for “Scores have μ = 50 and σ = 5. What share is between 45 and 60, and how many of 400?”
Chart
X is normal with μ = 50 and σ = 5.
P(45 ≤ X ≤ 60) = 0.8186.
Values in all: 400
Expected count between a and b: 327.4
This is an example. Type your own values to start a new problem.
μ
Mean
example
σ
Standard deviation
example
a
Lower value
example
b
Upper value
example
z₁
z-score of a
calculated
z₂
z-score of b
calculated
P
Share between a and b
calculated
N
Values in all
example
E
Expected count between a and b
calculated
Formulas
z₁
=
a
−
μ
σ
−1
=
45 − 50
5
z₂
=
b
−
μ
σ
2
=
60 − 50
5
P
=
Φ(
z₂
)
−
Φ(
z₁
)
0.8186 = Φ(2) − Φ(−1)
E
=
N
×
P
327.4 = 400 × 0.8186
Assumptions
•
The data are roughly normal, with mean μ and standard deviation σ.
•
The share between a and b is the area left of b minus the area left of a.
•
Of N values, about N × P fall between a and b.
Step-by-step
WE KNOW
μ = 50, σ = 5, a = 45, b = 60, N = 400
FIND
z-score of a (z₁), z-score of b (z₂), share between a and b (P), expected count between a and b (E)
Step 1 · Find z-score of a (z₁)
Use z₁ = (a − μ) ÷ σ
How far the value is from the mean, in standard deviations.
z₁
=
45 − 50
5
z₁
=
−
5
5
z₁
=
−1
Step 2 · Find z-score of b (z₂)
Use z₂ = (b − μ) ÷ σ
How far the value is from the mean, in standard deviations.
z₂
=
60 − 50
5
z₂
=
10
5
z₂
=
2
Step 3 · Find share between a and b (P)
Use P = Φ(z₂) − Φ(z₁)
The
area
left
of
b
,
less
the
area
left
of
a
.
P
=
Φ(2)
−
Φ(−1)
P
=
0.8186
Step 4 · Find expected count between a and b (E)
Use E = N × P
The
share
P
of
the
N
values.
E
=
400
×
0.8186
E
=
327.4
(about
327
of
the
400)
Check
Put the numbers back in. Both sides should match.
−1
=
45 − 50
5
✓
2
=
60 − 50
5
✓
0.8186 = Φ(2) − Φ(−1) ✓
327.4 ≈ 400 × 0.8186 ✓
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Normal distributions, z-scores and margin of error
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The share more than a distance from the mean
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