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Z-score
Standard Normal Distribution
0.4
-3σ
-2σ σ
σ
2σ
3σ
Mean = 0
A Standard Normal Distribution is a normal distribution which has mean = 0 and
σ = 1.
(See statisticsTheory.doc)
Z - Scores
The Z-Score enable transformation of any normal distribution to standard normal
distribution by using a formula:
Z=
x-μ
σ
Formula (1)
The above formula is for obtaining a z-score for an entire population. Usability testing obviously
samples a very small subset of the population and thus the following formula is used:
Z=
x-x
σ
Formula (2)
and s are used as estimators for the population's true mean and standard deviation. Both
formulas essentially calculate the same thing:
Z=
data _ po int - mean
s tan dard _ deviation
Formula (3)
© Copyrigh (2007)William Mak – All rights are reserved
1
Z-score
How to Use the Z-score for Finding Probabilities
The best way is to use examples to show.
Consider a production process makes shafts which are normally distributed with an
average diameter of 2.0cm and a variance of 4 cm sq.. Find the probabilities that a single
bolt selected at random from the process will have a diameter of:
(a)
(b)
(c)
(d)
more than 3 cm;
exactly 2 cm;
less than 1 cm;
Between 1.5 and 3cm.
Solution:
Let x = diameter of a shaft ~ N (2, 4), where N (2, 4) means that the random
variable is normal distribution with mean value = 2, and variance = 4.
Remember that sample mean, x =
And
(a)
s
2
∑x
n
∑x
=
n
2
-
x
2
For P (x > 3):
Area = 0.1915
By Formula (1):
Z=
x-μ
= 0.5
σ
Area = 0.3085
Therefore, P(x > 3) = P (Z > 0.5)
Z = 0.5
Now, using Z = 0.5, we can find from Standard Normal Distribution Table that the area
between 0 and 0.5 (as the diagram above) is 0.1915. So, the area enclosed beyond Z = 0.5
is equal to 0.3085.
Therefore the probability that the diameter of a shaft is bigger than 3 cm is
0.3085 or 31%.
Remember that the total area enclosed by the curve is 1, or 100%.
(b) diameter is exactly = 2 cm:
© Copyrigh (2007)William Mak – All rights are reserved
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Z-score
P (x = 2)
Area = 0.000
By Formula (1):
Z=
2-2
=0
2
Z=0
Therefore, P (x = 2) = P (Z = 0)
For Z = 0.5, using Standard Normal Distribution Table the area between Z
= 0 and Mean (which is 0), as the diagram above, is 0.000.
Hence, P (x = 2) = P (Z = 0) = 0
(c) P (1.5 < x < 3) =?
When x = 1.5, using the transformation formula, Z =
Similarly, Z =
1.5 - 2
= -0.25
2
3-2
= 0.5
2
Therefore, P (1.5 < x < 3) = P (-0.25 < Z < 0.5)
= P (-0.25< Z < 0) plus P (0 < Z < 0.5)
the area = 0.987 + 0.1915
= 0.2902
So, the probability = 29%
-0.25
0.5
© Copyrigh (2007)William Mak – All rights are reserved
3
Z-score
Example again!
(2) For the process in Example 1, if 16 shafts are selected what is the probability that the
average diameter will exceed 2.1 cm?
Solution:
Let x bet the diameter of a shaft from a population of shafts ~ N (μ, σ )
2
Let
σ
x = average diameter, of 16 bolts ~ N (μ,
2
n
)
2
~ N (μ, 2
16
1
~ N (μ, )
4
Note:
σ
)
2
is the variance of the 16 pieces of shafts having the mean = x . Hence
n
the standard deviation of that 16 pieces of shaft =
σ
n
σ
n
2
=
σ
n
.
is termed as standard error
For P ( x > 2.1)
Z=
data_point - μ
s tan dard _ deviation
Z=
x-μ
σ
x
2.0
2.1
n
2.1 - 2
2.1 - 2
1
4
= 0.2
=
0
Z
?
So, P ( x > 2.1) = P (Z > 0.2)
From table, the area being = 0.5 – 0.0793 = 42%
© Copyrigh (2007)William Mak – All rights are reserved
4
Z-score
Theorem (Central Limit)
If
x , x ,x
1
2
3
…
x
4
are randomly selected and each of which has a
mean, μ, and variance, σ , then:
2
(a) x = (
(b)
x + x +x
1
2
∑x = ( x1 +
3
…
x +x
2
3
x )/n=
n
…
∑x
n
σ
---- N (μ,
n
2
)
x ) --- N (n μ, n σ )
2
n
Example (3)
The average pay of employees in a company is $100 per week, having standard deviation
of $4 per week. Find the probabilities that:
(a) the average salary of 9 employees is less than $98.5 per week
(b) the total salary bill of 25 employees is less than $2600 per week.
Solution:
σ
(a) Let x = average salary of 9 employees ~ N (100,
n
2
)
2
~ N (100, 4
9
)
For P ( x < 98.5)
data_point - μ
Z=
s tan dard _ deviation
=
=
x-μ
S .D
98.5 - 100
2
4
9
= - 1.125
0
-1.125
© Copyrigh (2007)William Mak – All rights are reserved
Z
5
Z-score
∴ P ( x < 98.5) = P (Z < -1.125)
= P (Z > 1.125) = 0.5 – 0.3697 (as from table)
= 13%
(b)
Let
∑x
be the total salary bill of 25 employees ~ (n μ, n σ )
2
2
~ (25x100, 16x 4 )
For P ( ∑x < 2600)
Z=
data_point - μ
s tan dard _ deviation
Area =?
=
2600 - 25 x100
2
=1
25 x 4
∴ P ( ∑x < 2600) = P (Z < 1)
0
1
Z
Area = 0.3413 + 0.5 = 0.8413 = 84.13%
© Copyrigh (2007)William Mak – All rights are reserved
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