SME
Grade 11 Math · Problem type
The share more than a distance from the mean
Use this for “Weights have μ = 500 g and σ = 2 g. How many of 500 packages are more than 3 g off?”
Chart
X is normal with μ = 500 g and σ = 2 g.
P(X ≤ 497 or X ≥ 503) = 0.1336.
Distance from the mean: 3 g
z-score of d: 1.5
Values in all: 500
Expected count outside: 66.81
This is an example. Type your own values to start a new problem.
μ
Mean
example
σ
Standard deviation
example
d
Distance from the mean
example
L
Lower cutoff
calculated
U
Upper cutoff
calculated
z
z-score of d
calculated
P
Share more than d from the mean
calculated
N
Values in all
example
E
Expected count outside
calculated
Formulas
L
=
μ
−
d
497 = 500 − 3
U
=
μ
+
d
503 = 500 + 3
z
=
d
σ
1.5
=
3
2
P
=
2
×
(1
−
Φ(
z
))
0.1336 = 2 × (1 − Φ(1.5))
E
=
N
×
P
66.81 = 500 × 0.1336
Assumptions
•
The data are roughly normal, with mean μ and standard deviation σ.
•
The curve is symmetric, so the two tails beyond d from the mean hold the same area.
•
Of N values, about N × P are more than d from the mean.
Step-by-step
WE KNOW
μ = 500 g, σ = 2 g, d = 3 g, N = 500
FIND
lower cutoff (L), upper cutoff (U), z-score of d (z), share more than d from the mean (P), expected count outside (E)
Step 1 · Find lower cutoff (L)
Use L = μ − d
Go
down
d
from
the
mean.
L
=
500
−
3
L
=
497
g
Step 2 · Find upper cutoff (U)
Use U = μ + d
Go
up
d
from
the
mean.
U
=
500
+
3
U
=
503
g
Step 3 · Find z-score of d (z)
Use z = d ÷ σ
The distance in standard deviations.
z
=
3
2
z
=
1.5
Step 4 · Find share more than d from the mean (P)
Use P = 2 × (1 − Φ(z))
One
tail
is
1
−
Φ(
z
);
the
other
side
has
the
same
area.
P
=
2
×
(1
−
Φ(1.5))
P
=
0.13361
Step 5 · Find expected count outside (E)
Use E = N × P
The
share
P
of
the
N
values.
E
=
500
×
0.13361
E
=
66.81
(about
67
of
the
500)
Check
Put the numbers back in. Both sides should match.
497 = 500 − 3 ✓
503 = 500 + 3 ✓
1.5
=
3
2
✓
0.13361 = 2 × (1 − Φ(1.5)) ✓
66.81 ≈ 500 × 0.13361 ✓
Topic check
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