Introduction to gamma distribution r:
A gamma distribution r is one of the common categories of statistical distribution that is connected to the beta distribution and arise physically in process for which is coming up times, connecting Poisson distributed actions are relevant. The gamma distribution is a two-parameter of constant probability distributions. It has a range parameter θ and a form parameter k.
Gamma Distribution (or) Erlang Distribution
Definition:
A continuous X is said to follow an Erlang distribution (or) gamma distribution with parameter υ > 0 and λ > 0 if its probability density function is given by
f(x) = `{(lambda^2/Gammanu 0 < x
Example Problem for Gamma Distribution:
Example 1:- using gamma distribution
The everyday expenditure of milk in a city in excess of 20,000 gallons is about distributed as a gamma variate with parameters υ = 2 and λ = 1/ 10,000. The metropolis has a each day stock of 30,000 gallons. What is the probability that the stock is lacking on a exacting day?
Solution:
Let X be the random variables denoting the daily consumption of milk in a city.
Then Y = X – 20,000 has gamma distribution with probability distribution function
G(y) = 1/(10,000)2 Γ(2) y2-1ey/10,000,y ≥ 0
G(y) = ye-y/10,000/(10,000)2, y ≥ 0
Since the daily stock of the city is 30,000 gallons, the required probability that the stock is insufficient on a particular day is given by Is this topic Mathematical Induction hard for you? Watch out for my coming posts.
P[X > 30,000] = P[Y > 10,000]
= `int_10000^oog(x)dy`
= `int_10000^oo` ye-y/10,000/(10,000)2 dy
Put z = y/ 10,000. Then dz = dy/ 10,000
P[ X>30,000] = `int_10000^oo` ze-z dz
= [-ze-z –e-z]
= 2/e
Example 2:- using gamma distribution
In a some city, the everyday expenditure of electric power in millions of kilowatt hours can be treat as a random variables having gamma distribution with parameters λ = 1 /2 and k= 3. If the power plant of this metropolis has a everyday power of 12 million kilowatt-hours, what is the probability that this power supply will be insufficient on some specified day?
Solution :
Let X denotes the daily consumption of electric power
In that case the probability distribution function of X is specified by
F(x) = (1 /2)3/(Γ(3)) x2e-x/2, x ≥ 0
P[ the power supply is inadequate]
=P[X>12] = `int_12^oo` f(x)dx
= `int_12^oo` 1/ Γ(3) (1/8) x2e-x/2 dx
= 1/16 [ x2(-2e-x/2)- (2x) (4e-x/2)+2(-8e-x/2)]
= e/16 [ 288 +96 +16] = 25e-6=0.0625
A gamma distribution r is one of the common categories of statistical distribution that is connected to the beta distribution and arise physically in process for which is coming up times, connecting Poisson distributed actions are relevant. The gamma distribution is a two-parameter of constant probability distributions. It has a range parameter θ and a form parameter k.
Gamma Distribution (or) Erlang Distribution
Definition:
A continuous X is said to follow an Erlang distribution (or) gamma distribution with parameter υ > 0 and λ > 0 if its probability density function is given by
f(x) = `{(lambda^2/Gammanu 0 < x
Example Problem for Gamma Distribution:
Example 1:- using gamma distribution
The everyday expenditure of milk in a city in excess of 20,000 gallons is about distributed as a gamma variate with parameters υ = 2 and λ = 1/ 10,000. The metropolis has a each day stock of 30,000 gallons. What is the probability that the stock is lacking on a exacting day?
Solution:
Let X be the random variables denoting the daily consumption of milk in a city.
Then Y = X – 20,000 has gamma distribution with probability distribution function
G(y) = 1/(10,000)2 Γ(2) y2-1ey/10,000,y ≥ 0
G(y) = ye-y/10,000/(10,000)2, y ≥ 0
Since the daily stock of the city is 30,000 gallons, the required probability that the stock is insufficient on a particular day is given by Is this topic Mathematical Induction hard for you? Watch out for my coming posts.
P[X > 30,000] = P[Y > 10,000]
= `int_10000^oog(x)dy`
= `int_10000^oo` ye-y/10,000/(10,000)2 dy
Put z = y/ 10,000. Then dz = dy/ 10,000
P[ X>30,000] = `int_10000^oo` ze-z dz
= [-ze-z –e-z]
= 2/e
Example 2:- using gamma distribution
In a some city, the everyday expenditure of electric power in millions of kilowatt hours can be treat as a random variables having gamma distribution with parameters λ = 1 /2 and k= 3. If the power plant of this metropolis has a everyday power of 12 million kilowatt-hours, what is the probability that this power supply will be insufficient on some specified day?
Solution :
Let X denotes the daily consumption of electric power
In that case the probability distribution function of X is specified by
F(x) = (1 /2)3/(Γ(3)) x2e-x/2, x ≥ 0
P[ the power supply is inadequate]
=P[X>12] = `int_12^oo` f(x)dx
= `int_12^oo` 1/ Γ(3) (1/8) x2e-x/2 dx
= 1/16 [ x2(-2e-x/2)- (2x) (4e-x/2)+2(-8e-x/2)]
= e/16 [ 288 +96 +16] = 25e-6=0.0625
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