The Gamma and Normal DistributionsThe Gamma and Normal Distributions. 3.2, 3.3. The Gamma...
Transcript of The Gamma and Normal DistributionsThe Gamma and Normal Distributions. 3.2, 3.3. The Gamma...
The Gamma and NormalDistributions
3.2, 3.3
The Gamma DistributionConsider a Poisson process with rate ππ:Let a random variable, ππ, denote the waiting time until the πΌπΌth occurrence.
ππ follows a Gamma Distribution.
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The Gamma Function, Ξ
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This is the definition of the gamma function
Gamma Distribution X~Gamma(πΌπΌ,ππ)
ππ π₯π₯ = 1Ξ(πΌπΌ)πππΌπΌ
π₯π₯πΌπΌβ1ππβπ₯π₯/ππ, 0 β€ x < β
πΈπΈ[ππ] = πΌπΌππ
ππππππ[ππ] = πΌπΌππ2
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Gamma ExampleCustomers arrive in a shop according to a Poisson process with a mean rate of 20 per hour. What is the probability that the shopkeeper will have to wait more than 10 minutes for the arrival of the 4th customer?
οΏ½10
β1
Ξ 4 34π₯π₯4β1ππβπ₯π₯/3 πππ₯π₯ = 0.57
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Gamma ExampleCustomers arrive in a shop according to a Poisson process with a mean rate of 20 per hour. What is the probability that the shopkeeper will have to wait more than 10 minutes for the arrival of the 4th customer?
οΏ½10
β1
Ξ 4 34π₯π₯4β1ππβπ₯π₯/3 πππ₯π₯ = 0.57
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Normal Distribution
Normal Distributionβ« Most important distribution in statisticsβ« Fits many natural phenomena such as IQ,
measurement error, height, etc.β« A symmetric distribution with a central peak, and tails
that taper off.
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Normal Distribution β Empirical Rule
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In a normal distribution, approximately 68/95/99.7% of the data falls within 1/2/3 standard deviations of the mean.
Normal Distribution X~N(ππ,ππ2)
ππ(π₯π₯) = 12ππππ2
ππβ(π₯π₯βππ)2
2ππ2 , -β < π₯π₯ < β
πΈπΈ[ππ] = ππ
ππππππ[ππ] = ππ2
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Normal DistributionLet X ~ Normal(ππ,ππ2)β« To find the P[a < X < b], one would need to evaluate
the integral:
οΏ½ππ
ππ1
2ππππ2ππβ
(π₯π₯βππ)22ππ2 πππ₯π₯.
β« A closed-form expression for this integral does not exist, so we need to use numerical integration techniques.
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Notes about the Normal DistributionThe Normal Distribution is symmetric with a central peak:
β« P[X > c] = P[X < -c]β« Mean = Median = Modeβ« Half of the area is to the left/right of 0.
Examples: if ππ ~ ππ(0,1)β« ππ[ππ β€ 0.2] = 0.5 + ππ[0 β€ ππ β€ 0.2]β« ππ[ππ β€ 0.3] = ππ[ππ β₯ 0.7]
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ExamplesLet ππ ~ ππ(0,1)
a) Find P[Z >2] (0.0228)b) Find P[ -2 < Z < 2] (0.9544)c) Find P[0 < Z < 1.73] (0.4582)
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Linear Transformation TheoremLet X βΌ N(Β΅, Ο2). Then Y = Ξ±X + Ξ² follows also a normal distribution.
ππ βΌ ππ(Ξ±Β΅+Ξ², Ξ±2Ο2) Can convert any normal distribution to standard normal by subtracting mean and dividing sd:
β« Z = ππβππππ
Using this theorem, we can see that ππ ~ ππ(0,1)
15(Recall) Let ππ have mean, πΈπΈ[ππ], and variance, ππ2.Let Y = ππππ + ππ. Then, ππ has mean πππΈπΈ[ππ] + ππ, and variance ππ2ππ2.
Example
Suppose the mass of Thorβs hammers in kg (he has an infinite number) are distributed X βΌ N(10, 32). Find the proportion of Thorβs hammers that have mass larger than 13.4 kg. (if we randomly select a hammer, find the probability that its mass > 13.4 kg).
ans. P[X > 13.4] = P[Z > 1.13] = 0.129216
What is z?β« The value of z gives the number of standard
deviations the particular value of X lies above or below the mean Β΅.
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Examples
Normal Distribution
1 Cream and Fluttr knows that the daily demand for cupcakes is a random variable which follows the normal distribution with mean 43.3 cupcakes and standard deviation 4.6. They would like to make enough so that there is only a 5% chance of demand exceeding the number of cupcakes made. (How many should they make?)
z=1.645 x = 5119
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2 Suppose again that Thorβs hammers are normally distributed with: E[X] = 10, Var[X] = 9.
Find the 25th percentile of X. (How much mass should a hammer have, in order to have more than 25% of all hammers)
Ans. z = -0.675 ππ0.25 = 7.975 22
3 Stapletonβs Auto Park of Urbana believes that total sales for next month will follow the normal distribution, with mean, ππ, and a standard deviation, ππ= $300,000. What is the probability that Stapletonβs sales will fall within $150000 of the mean next month?
Ans. 0.6915 β 0.3085 = .38323
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