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Probability and Statistics Exercise: Normal Distribution and Rayleigh Distribution - Prof., Exámenes de Estadística

Normal DistributionRayleigh DistributionStatistical InferenceProbability Theory

Solutions to two statistics exercises from a textbook. The first exercise involves finding the probability that the sample mean of a normally distributed population is greater than a certain value and determining the sample quasivariance such that 5% of the sample variances are less than this value. The second exercise deals with the rayleigh distribution, specifically calculating the bias, mean square error, and consistency of the maximum likelihood estimator for a random variable with a rayleigh distribution. The document also includes some background information on the rayleigh distribution.

Qué aprenderás

  • What is the probability that the sample mean is greater than 50 in a normally distributed population?
  • What is the bias and mean square error for the estimator ^θM in the Rayleigh distribution?

Tipo: Exámenes

2014/2015

Subido el 31/10/2015

bbrg16
bbrg16 🇪🇸

4 documentos

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¡Descarga Probability and Statistics Exercise: Normal Distribution and Rayleigh Distribution - Prof. y más Exámenes en PDF de Estadística solo en Docsity! Statistics II 30 October 2015 Exercise 1 (5 points) A random sample of size n = 18 is obtained from a normally distributed population with a population mean of μ = 46 and a variance of σ2 = 50. (a) What is the probability that the sample mean is greater than 50? (b) What is the value of the sample quasivariance such that 5% of the sample variances would be less than this value? (From: Newbold, P., W. Carlson and B. Thorne. Statistics for Business and Economics. Pearson-Prentice Hall.) (a) P( X̄ > 50)=P( X̄−μ √σ 2 n > 50−μ √σ 2 n ) =P(T > 50−46 √ 5018 ) =P(T > 2.4)=1−P (T ≤ 2.4)=1−0.9918=0.0082 (b) For the sample quasivariance 0.05 = P ( S2 < a )=P((n−1)S 2 σ 2 < (n−1)a σ 2 )=P(T < (18−1)a50 ) → 0.95 = P(T ≥ (18−1)a 50 ) → (18−1)a 50 = 8.67 → a = 8.67⋅50 17 =25.5 For the sample variance 0.05 = P ( s2 < a )=P( ns 2 σ 2 < na σ 2 )=P(T < 18 a50 ) → 0.95 = P(T ≥ 18 a 50 ) → 18 a 50 = 8.67 → a = 8.67⋅50 18 =24.08 Exercise 2 (5 points) Let X be a random variable with probability function f (x ;θ)= x θ 2 e − x2 2θ2 , x ≥0, (θ>0) such that E ( X )=θ √ π 2 and Var ( X )= 4−π 2 θ 2 . Let X = (X1,...,Xn) be a simple random sample. The application of the method of the moments provides the estimator θ̂M=√ 2π X̄ . For this estimator, calculate the bias and the mean square error, and study the consistency. Cultural note: In probability theory and statistics, the Rayleigh distribution is a continuous probability distribution for positive-valued random variables. A Rayleigh distribution is often observed when the overall magnitude of a vector is related to its directional components. One example where the Rayleigh distribution naturally arises is when wind velocity is analyzed into its orthogonal 2-dimensional vector components. Assuming that the magnitudes of each component are uncorrelated, normally distributed with equal variance, and zero mean, then the overall wind speed (vector magnitude) will be characterized by a Rayleigh distribution. A second example of the distribution arises in the case of random complex numbers whose real and imaginary components are i.i.d. (independently and identically distributed) Gaussian with equal variance and zero mean. In that case, the absolute value of the complex number is Rayleigh-distributed. The distribution is named after Lord Rayleigh. (From: Wikipedia.) Bachelor's Degree in Economics 1 Complutense University of Madrid
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