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Normal Distribution

The document describes the normal distribution, its probability density function, and key properties. The normal distribution is symmetric and defined by its mean (μ) and variance (σ^2). It also describes how to transform a normal distribution to the standard normal distribution and calculate probabilities for events within, between, or outside given values using the normal distribution function F(z). An example calculates the probability of concrete block weights falling within, below, or above given values based on the normal distribution of block weights.

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Md. Nazmul Huda 1610805643
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60 views5 pages

Normal Distribution

The document describes the normal distribution, its probability density function, and key properties. The normal distribution is symmetric and defined by its mean (μ) and variance (σ^2). It also describes how to transform a normal distribution to the standard normal distribution and calculate probabilities for events within, between, or outside given values using the normal distribution function F(z). An example calculates the probability of concrete block weights falling within, below, or above given values based on the normal distribution of block weights.

Uploaded by

Md. Nazmul Huda 1610805643
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Normal Distribution

Normal Distribution:
The probability density function of normal distribution is:

−∞ < 𝑥 < ∞,
1 (𝑥−µ)2
−
𝑓(𝑥 ) = 𝑒 2𝜎 2 − ∞ < µ < ∞,
√2𝜋𝜎 2
0 ≤ 𝜎2 < ∞

Expectation: 𝐸 (𝑥 ) = µ
Variance: 𝑉(𝑥 ) = 𝜎 2

Standard normal distribution: When µ = 0 and 𝜎 2 = 1 then the


normal distribution is called standard normal distribution.
The probability density function of standard normal distribution is:
1 1
− 𝑧2
𝑓(𝑧) = 𝑒 2 −∞<𝑧<∞
√2𝜋

Property of a normal distribution:


1) It is symmetric.
2) Mean=Mode=Median.
3) It is unimodal.
4) The total area under the curve is equal to one.
5) The normal curve approaches, but never touches, the x-axis
Transformation:
(𝑥−µ)2
∞ ∞ 1 −
∫−∞ 𝑓(𝑥 ) 𝑑𝑥 = ∫−∞ √2𝜋𝜎2 𝑒 2𝜎2 𝑑𝑥
1 2
∞ 1 𝑥−µ
= ∫−∞ 𝑒 −2𝑧 𝜎 𝑑𝑧 Let, 𝑧 =
√2𝜋 𝜎 𝜎
1
∞ 1 − 𝑧2 𝑥 µ
= ∫−∞ √2𝜋 𝑒 2 𝑑𝑧 => 𝑧 = −
𝜎 𝜎
𝑑𝑧 1
=> =
𝑑𝑥 𝜎

=> 𝑑𝑥 = 𝜎 𝑑𝑧
𝑥 −∞ ∞
𝑧 −∞ ∞

That is if 𝑋 ~𝑁(µ, 𝜎 2 ) and if you want to transform the normal


distribution to standard distribution then the transform random variable
𝑥−µ
is 𝑍 = (Z score)
𝜎

Probability Calculations for Normal Distributions:

𝑃(𝑎 < 𝑥 < 𝑏)


𝑎−µ 𝑥−µ 𝑏−µ
= 𝑃( < < )
𝜎 𝜎 𝜎
𝑎−µ 𝑏−µ
= 𝑃( <𝑍< )
𝜎 𝜎
𝑏−µ 𝑎−µ
= 𝐹( ) − 𝐹( )
𝜎 𝜎
𝑃(𝑥 < 𝑏)
= 𝑃(−∞ < 𝑥 < 𝑏)
−∞ − µ 𝑥 − µ 𝑏 − µ
= 𝑃( < < )
𝜎 𝜎 𝜎
𝑏−µ
= 𝑃(−∞ < 𝑍 < )
𝜎
𝑏−µ
= 𝐹( )
𝜎

𝑃(𝑥 > 𝑎)
= 𝑃(𝑎 < 𝑥 < ∞)
𝑎−µ 𝑥−µ ∞−µ
= 𝑃( < < )
𝜎 𝜎 𝜎
𝑎−µ
= 𝑃( < 𝑍 < ∞)
𝜎
𝑎−µ
= 𝐹 (∞) − 𝐹( )
𝜎
𝑎−µ
= 1 − 𝐹( )
𝜎

Example: A company manufactures concrete blocks that are used for


construction purposes. Suppose that the weights of the individual concrete
blocks are normally distributed with a mean value of μ = 11.0 kg and a
standard deviation of σ = 0.3 kg.
i) Calculate the probability that a concrete block weight is less than
10.5kg.
ii) Calculate the probability that a concrete block weight is within 10kg to
12kg.
iii) Calculate the probability that a concrete block weight is greater than
10.5kg.
Solution:
i)
𝑃(𝑥 < 10.5)
= 𝑃(−∞ < 𝑥 < 10.5)
−∞ − 11 𝑥 − 11 10.5 − 11
= 𝑃( < < )
0.3 0.3 0.3
= 𝑃(−∞ < 𝑍 < −1.67)
= 𝐹(−1.67)
= 0.0475
ii)
𝑃(10 < 𝑥 < 12)
10 − 11 𝑥 − 11 12 − 11
= 𝑃( < < )
0.3 0.3 0.3
= 𝑃(−3.33 < 𝑍 < 3.33)
= 𝐹 (3.33) − 𝐹(−3.33)
= 0.99957 − 0.00043
= 0.99914

iii)
𝑃(𝑥 > 10.5)
= 𝑃(10.5 < 𝑥)
= 𝑃(10.5 < 𝑥 < ∞)
10.5 − 11 𝑥 − 11 ∞ − 11
= 𝑃( < < )
0.3 0.3 0.3
= 𝑃(−1.67 < 𝑍 < ∞)
= 𝐹 (∞) − 𝐹(−1.67)
= 1 − .04746
= 0.95254

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