the gamma and normal distributionsthe gamma and normal distributions. 3.2, 3.3. the gamma...
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![Page 1: The Gamma and Normal DistributionsThe Gamma and Normal Distributions. 3.2, 3.3. The Gamma Distribution. Consider a Poisson process with rate ππ: Let a random variable, ππ,](https://reader035.vdocuments.site/reader035/viewer/2022071415/610fbc4b8ee67f284325a5e1/html5/thumbnails/1.jpg)
The Gamma and NormalDistributions
3.2, 3.3
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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
3When n is an integer,
This is the definition of the gamma function
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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
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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.
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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.
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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
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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
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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
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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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