Download Lecture 10 1 Definition 2 Standard Normal Distribution

Lecture 10
Normal Distribution
1
Definition
A continuous r.v. X is said to have a normal distribution with parameters µ (=mean) and σ (standard deviation) if its probability density
function is
2 1
1
exp − 2 x − µ
f (x) = √
2σ
2πσ 2
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Standard Normal Distribution
• Definition:
The normal distribution with parameter values µ = 0 and σ = 1
is called “standard normal” and is written as Z. Thus, its’ pdf is
1 1
f (z) = √ exp − z 2
2
2π
The cumulative distribution function of Z is
Z a
1 1
√ exp − z 2 dz = Φ a .
P Z≤a =
2
2π
−∞
• How to read Table A.3 (p. 668):
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• Example 1:
P Z ≤ 1.25 = Φ(1.25) =
P − 0.38 < Z < 1.25 = Φ(1.25) − Φ(−0.38) =
• Percentile and Zα
P Z > Zα = α
• Some important Zα values
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α
0.1
0.05
Zα
1.28 1.645
0.025 0.01 0.005 0.001
1.96t
2.33
2.58
3.08
Non-Standard Normal Distribution
• X=normal distribution with a mean of µ and a variance of σ is
denoted by X ∼ N µ, σ .
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• Standardization:
If X ∼ N µ, σ , then
•
X −µ
σ
has standard normal distribution; Z ∼ N 0, 1 .
Z=
X −µ
E
σ
• Thus,
P a≤X ≤b
=0
V
x−µ
σ
= 1.
X −µ
b−µ
a−µ
= P
≤
≤
σ
σ
σ
b − µ
a − µ
= Φ
−Φ
σ
σ
• Example 2
Reaction time ∼ N 1.25, 0.46 . Then what is the probability that
the reaction time is between 1.00 sec and 1.75 sec.
1.00 − 1.25 X − 1.25 1.75 − 1.25
P 1.00 ≤ X ≤ 1.75 = P
≤
≤
0.46
0.46
0.46
= P − 0.54 ≤ Z ≤ 1.09
= Φ 1.09 − Φ − 0.54
= 0.8621 − 0.2946 = 0.5675.
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Percentile of an Arbitrary Normal Distribution
• Formula:
100 × p−th percentile of N µ, σ is
µ + 100 × p−th percentile of the standard normal · σ
Let 100 × p−th percentile of the standard normal=ηp. Then,
X −µ
P X ≤ µ + ηp · σ = P
≤ etap = P Z ≤ ηp = p.
σ
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• Example 3:
Amount of distilled water is normal distribution with µ = 64 and
σ = 0.78Then what container size C is necessary to ensure that
overflow occurs only 0.5% of the distilled water.
P X > c = 0.005
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Normal Approximation to Binomial
• Formula:
X ∼ B n, p . Note that E X = np and V (X) = np(1 − p).
Then,
X − np
k + 0.5 − np
P X ≤k = P p
≤ p
np(1 − p)
np(1 − p)
k + 0.5 − np
≈ P Z≤ p
np(1 − p)
k − 0.5 − np
X − np
≤ p
P X <k = P p
np(1 − p)
np(1 − p)
k − 0.5 − np
≈ P Z≤ p
np(1 − p)
• Example 4:
25% of licensed drivers in a particular state do not buy insurance.
Suppose 50 drivers are randomly selected from the state. What is
probability that the number of drivers not having insurance are at
most 10? What is the probability that the number of drivers not
having insurance are between 5 and 15 (inclusive)?
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