Business Statistics Samenvatting Formules en Methoden
Distributions
Uniform Allemaal zelfde kans
Bernoulli distribution Event is binary ( yes/no)
Binomial distribution Verzameling binary events
Parameters: pie: probability
n = number of trials
Hypergeometric Sampling without replacement. (I.E. Quality control)
Distribution Parameters: N=population size
(discrete valued n=sample size
distribution) s=number of successes in N
Example in R:
Spot 1 or fewer defects in a sample of 7 products out of a batch
of 1,000 with defect probability 0.05.
P(x<_1|N=1000,n=7,s=50)
=phyper (1,50,950,7) = 0.9562
1e getal= P op hoeveel witte ballen pakken
2e getal = hoeveel witte ballen/kapotte dingen in population
3e getal=zwarte ballen/hele dingen
4e getal= hoeveel ballen pakke
Poisson Distribution Number of events in a particular time slot ( No upper limit)
(discrete valued Parameter: λ = intensity
distribution) Sum of two independent poissons is poisson:
λ 1+ λ2 (gewoon labdas optellen)
Expectation μ: λ
Variance σ 2: λ
Example in R:
P of no orders?
P ( X=0|λ=5 )=¿ dpois(0, lambda=5) = 0.0067
P of 4 to 10 orders?
P ( 4 ≤ X ≤ 10|λ=5 )=P ( X ≤10| λ=5 )−P( X ≤ 3∨ λ=5)
=ppois (10, lambda=5) – ppois(3, lambda=5)
=0.9863-0.2650=0.7213
P of more than 12 orders?
P ( X >12|λ=5 )=1−P ( X ≤ 12|λ=5 )
= 1-ppois (12, lambda=5) = 0.0020
, Exponential distribution Waiting times, time until an event happens
(continuous random Parameter: λ
variable) Expected waiting time is 1/ λ
Expectation μ= 1/ λ
Variance σ 2=1/ λ2
Example in R:
Waiting times are exponentially distributed with intensity λ =0.1
What is the probability of having to wait more than 5 minutes?
P ( X >5 ) =1−P( X ≤ 5)
=1-pexp(5, rate=0.1)= 0.6065307
What is the probability of having to wait 6 to 11 minutes?
P ( 6< X <11 )=P ( X <11 )−P( X <6)
=pexp(11,rate=0.1) – pexp(6,rate=0.1)=0.2159406
What is the maximum waiting time with a 99% probability?
P ( X < x )=0.99 = qexp(0.99, rate= 0.1)= 46.0517
Chi-squared ( x 2 ¿ ¿ Waiting times as well
distribution More in testing and statistical sense
Parameter: DF(degrees of freedom) orn
Expectation μ=n
Variance σ 2=2 n
Example in R:
2
P( x ¿ ¿ 6 ≤2.31) ¿
Pchisq(2.31, df=6)=0.1109037
P ( x 6 ≤ x ) =0.90
2
Qchisq(0.90, df=6)=10.64464
F-distribution Used in testing and statistical sense
Parameter: DF1 (degrees of freedom)
DF2
Example in R:
P( F 5,2 ≤ 2.31)
Pf(2.31, df1=5, df2=2)= 0.6708211
P¿
Qf(0.90, df1=5, df2=2)= 9.292
Tekens
Population size N
Sample size n
Population Mean μ
Sample mean x́
Expected value E(X)
2
Population Variance σ
Distributions
Uniform Allemaal zelfde kans
Bernoulli distribution Event is binary ( yes/no)
Binomial distribution Verzameling binary events
Parameters: pie: probability
n = number of trials
Hypergeometric Sampling without replacement. (I.E. Quality control)
Distribution Parameters: N=population size
(discrete valued n=sample size
distribution) s=number of successes in N
Example in R:
Spot 1 or fewer defects in a sample of 7 products out of a batch
of 1,000 with defect probability 0.05.
P(x<_1|N=1000,n=7,s=50)
=phyper (1,50,950,7) = 0.9562
1e getal= P op hoeveel witte ballen pakken
2e getal = hoeveel witte ballen/kapotte dingen in population
3e getal=zwarte ballen/hele dingen
4e getal= hoeveel ballen pakke
Poisson Distribution Number of events in a particular time slot ( No upper limit)
(discrete valued Parameter: λ = intensity
distribution) Sum of two independent poissons is poisson:
λ 1+ λ2 (gewoon labdas optellen)
Expectation μ: λ
Variance σ 2: λ
Example in R:
P of no orders?
P ( X=0|λ=5 )=¿ dpois(0, lambda=5) = 0.0067
P of 4 to 10 orders?
P ( 4 ≤ X ≤ 10|λ=5 )=P ( X ≤10| λ=5 )−P( X ≤ 3∨ λ=5)
=ppois (10, lambda=5) – ppois(3, lambda=5)
=0.9863-0.2650=0.7213
P of more than 12 orders?
P ( X >12|λ=5 )=1−P ( X ≤ 12|λ=5 )
= 1-ppois (12, lambda=5) = 0.0020
, Exponential distribution Waiting times, time until an event happens
(continuous random Parameter: λ
variable) Expected waiting time is 1/ λ
Expectation μ= 1/ λ
Variance σ 2=1/ λ2
Example in R:
Waiting times are exponentially distributed with intensity λ =0.1
What is the probability of having to wait more than 5 minutes?
P ( X >5 ) =1−P( X ≤ 5)
=1-pexp(5, rate=0.1)= 0.6065307
What is the probability of having to wait 6 to 11 minutes?
P ( 6< X <11 )=P ( X <11 )−P( X <6)
=pexp(11,rate=0.1) – pexp(6,rate=0.1)=0.2159406
What is the maximum waiting time with a 99% probability?
P ( X < x )=0.99 = qexp(0.99, rate= 0.1)= 46.0517
Chi-squared ( x 2 ¿ ¿ Waiting times as well
distribution More in testing and statistical sense
Parameter: DF(degrees of freedom) orn
Expectation μ=n
Variance σ 2=2 n
Example in R:
2
P( x ¿ ¿ 6 ≤2.31) ¿
Pchisq(2.31, df=6)=0.1109037
P ( x 6 ≤ x ) =0.90
2
Qchisq(0.90, df=6)=10.64464
F-distribution Used in testing and statistical sense
Parameter: DF1 (degrees of freedom)
DF2
Example in R:
P( F 5,2 ≤ 2.31)
Pf(2.31, df1=5, df2=2)= 0.6708211
P¿
Qf(0.90, df1=5, df2=2)= 9.292
Tekens
Population size N
Sample size n
Population Mean μ
Sample mean x́
Expected value E(X)
2
Population Variance σ
