Summary

Summary: 2DD80

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Course
Statistics For Industrial Engineering (2DD80)

Institution
Technische Universiteit Eindhoven (TUE)

Book
Applied Statistics and Probability for Engineers

This summary contains all the theory needed for the 2DD80 exam. In addition, the document contains an example assignment for each piece of theory to test your knowledge and see how the theory is applied in an assignment.

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Summarized whole book? No
Which chapters are summarized? Hoofdstuk 4 tot en met 11
Uploaded on October 18, 2022
Number of pages 47
Written in 2022/2023
Type Summary

probability
density function
mean
variance
hypothesis

Book Title:Applied Statistics and Probability for Engineers

Author(s):Douglas C. Montgomery, George C. Runger

Edition:Unknown
ISBN:9781119585596
Edition:7

Institution
Technische Universiteit Eindhoven (TUE)
Education
Industrial Engineering
Course
Statistics For Industrial Engineering (2DD80)

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Chapter 4

4.1

Probability density function:
For a continuous random variable X, a probability density function is a function such that
1. F ( x)≥0
∞
2. ∫ f ( x ) dx=1
−∞
b
3. P ( a ≤ X ≤ b )=∫ f ( x ) dx = area under f(x) from a to b for any a and b
a

If X is a continuous random variable, for any x1 and x2,
P ( x 1 ≤ X ≤ x 2 ) =P ( x 1< X ≤ x 2 )=P ( x 1 ≤ X < x 2 )=P(x 1< X < x 2)

Example
−x
f ( x ) ¿ e for 0< x
∞
a. P ( 1< x )=∫ e−x dx=[−e−x ]=e−1 =0.2858
1
2.5
P ( 1< x<2.5 )=∫ e dx=[−e ]=e −e
−x −x −1 −2.5
b. =0.2037
1
4
P ( x< 4 )=∫ e dx=[−e ]=1−e =0.9817
−x −x −4
c.
0
x
P ( X < x )=0.10 →∫ e dx=[−e ]=1−e =0.10→ x=−ln ( 0.9 )=0.1054
−x −x −x
d.
0

4.2

The cumulative distribution function of a continuous random variable is
x
F ( x )=P ( X ≤ x ) =∫ f ( u ) du for−∞ < x <∞
−∞

Probability density function from the cumulative distribution function
dF (x )
Given F(x), f ( x )= as long as the derivative exists
dx

Example

{
0 x <0
F ( x )= 0.25 x 0 ≤ x< 5
1 5≤ x

a. P ( x<2.8 )=P( x ≤2.8) because X is a continuous random variable.
Then, P( X< 2.8)=F(2.8)=0.2(2.8)=0.56
b. P ( x>5 )=1−P ( x ≤ 1.5 )=1−0.2 ( 1.5 )=0.7
c. P ( X ←2 )=Fx (−2 )=0

, d. P ( x> 6 )=1−Fx ( 6 )=0

4.3

The mean or expected value of X, denoted as u or E(X) is
∞
μ= E ( X )= ∫ f ( u ) du for−∞ < x< ∞
−∞

The variance of X, denoted as V(X) or σ2 is
∞
σ =V ( X )= ∫ ¿ ¿
2

−∞

The standard deviation of X is σ =√ σ 2

Expected value of a function of a continuous random variable
E¿

Example
2
f ( x )=1.5 x for−1< x< 1

[ ]
1 4
x
a. E ( X ) =∫ 1.5 x dx= 1.5
3
=0
−1 4
1
b. V ( X )=∫ 1.5 x ¿ ¿
3

−1

4.4

Continuous uniform distribution
A continuous random variable X with a probability density function
1
f ( x )= a≤x ≤b
( b−a )
Is a continuous uniform random variable

Mean and variance
If X is a continuous uniform random variable over a ≤ x ≤ b
a+b 2
μ= E ( X )= ∧σ =V ( X )=¿ ¿
2

Example
Uniform distribution over the interval [-1,1]
−1+1
a. E ( X ) =
2
V ( X )=¿ ¿
σ =0,577
x
1
b. P (−x< X < x )=0.90 → ∫ dt= [ 0,5 t ] =0.5 ( 2 x ) =x
−x 2

, {
0 x←1
c. Cumulative distribution function: F ( x ) = 0.5 x+ 5−1 ≤ x <1
1 1≤ x

4.5

Normal distribution
A random variable X with probability density function
1
f ( x )= e−¿¿¿
√2 π σ
Is a normal random variable with parameters μ where−∞< μ< ∞∧σ >0
Also, E ( X ) =μ∧V ( X ) =σ 2
And the notation N( μ , σ 2) is used to denote the distribution

Standard normal random variable
A normal random variable with μ=0 and σ2=1 is called a standard normal random variable and is
denoted as Z.
The cumulative distribution function of a standard random variable is denoted as
Ф ( z )=P(Z ≤ z)

Standardizing a normal random variable
If X is a normal variable with E(X)=μ and V(X)=σ 2, the random variable
X−μ
Z= is a normal random variable with E(Z)=0 and V(Z)=1.
σ
That is, Z is a standard normal random variable

Standardizing to calculate a probability
Suppose that X is a normal random variable with mean μ and variance σ 2. Then,
P ( X ≤ x )=P ( X−μ
σ
≤
σ )
x−μ
=P (Z ≤ z ) where Z is a standard normal random variable, and
x−μ
z= is the z-value obtained by standardizing X.
σ
x−μ
The probability is obtained by using Appendix table III with z=
σ

Example
X is normally distributed with a mean of 10 and a standard deviation of 2
a) P(X < 13) = P(Z < (13-10)/2) = P(Z < 1.5) = 0.93319
b) P(X > 9) = 1 - P(X < 9) = 1 - P(Z < (9-10)/2) = 1 - P(Z < -0.5) = 0.69146

c) P(6 < X < 14) = = P(-2 < Z < 2) = P(Z < 2) -P(Z < - 2)]= 0.9545

d) P(2 < X < 4) = = P(-4 < Z < -3) = P(Z < -3) - P(Z < -4) = 0.00132

e) P(-2 < X < 8) = P(X < 8) - P(X < -2) = = P(Z < -1) - P(Z < -6) = 0.15866

4.6

Normal approximation to the binomial distribution

, If X is a binomial random variable with parameters n and p,
X−np
Z= is approximately a standard normal random variable.
√np (1− p)
To approximate a binomial probability with a normal distribution, a continuity correction is applied
as follows:

(
P ( X ≤ x )=P ( X ≤ x+ 0.5 ) ≈ P Z ≤
x+ 0.5−np
√ np ( 1− p ) )
And

P ( x ≤ X )=P( x−0.5 ≤ X) ≈ P
(
x−0.5−np
√ np ( 1− p )
≤Z
)
The approximation is good for np> 5∧n (1− p)> 5

Normal approximation to the Poisson distribution
If X is a Poisson random variable with E(X)=λ and V(X)=λ,
X−λ
Z= is approximately a standard normal random variable.
√λ
The same continuity correction used for the binomial distribution can also be applied.
The approximation is good for λ>5

Example
X is a binomial random variable with n=200 and p=0.4

a) E(X) = 200(0.4) = 80, V(X) = 200(0.4)(0.6) = 48 and σ X =√ 48

Then,
P( X≤70)≃P Z≤
√ (
70. 5−80
48 )
=P( Z≤−1. 37 )=0 . 0853

P(70<X <90)≃P
√ 48 (
70 .5−80
<Z≤
89 . 5−80
√ 48
=P(−1 . 37<Z≤1. 37 )
)
b) =0 .9 1466-0 . 08534=0 . 8293

c)

4.7

Exponential distribution
The random variable X that equals the distance between successive events from Poisson process with
mean number of events λ>0 per unit interval is an exponential random variable with parameter λ.
The probability density function of X is,

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