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4. Suppose that the probability density function of X is f(x) = {3x² 0

4. Suppose that the probability density function of X is f(x) = {3x² 0

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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4. Suppose that the probability density function of X is 0 < x < 1 elsewhere Determine P (X < ¹), P ( ≤ x < ²), and P ( x ≥ 3). Determine the cumulative function X of X. Determine the mean, variance, and standard deviation of X. f(x) = {³ = {³x² 0
Transcribed Image Text:4. Suppose that the probability density function of X is 0 < x < 1 elsewhere Determine P (X < ¹), P ( ≤ x < ²), and P ( x ≥ 3). Determine the cumulative function X of X. Determine the mean, variance, and standard deviation of X. f(x) = {³ = {³x² 0
•Types of functions to express probability distribution (a) Probability Mass Function (PMF) f(x) = P(x = x) (b) Cumulative Distribution Functions (CDF) f(x) = P(X ≤ x) Mean M = { xf(x) Variance 0² = [x²f (x) - μ² Standard Deviation. 0x Tot Discrete Uniform Distribution f(x) = //12 M = 0² = b+ a 2 (b-a+1)² 12 •Binomial Distribution (f(x) = (~) p² (1-P)^-x с пр np(1-P) 1• Hypergeometric Distribution (*) (*) (~) f(x) = и = пр 8² = np(1-P/(N=1) where K/N P= ·Poisson Distribution -^+ (^T)* X! f(x) = e M = AT 0² = NT Where N = population size sample size /no. of trials P probability - p = probability of success on a single trial K = no. of successes in the population in the sample of X - no. successes
Transcribed Image Text:•Types of functions to express probability distribution (a) Probability Mass Function (PMF) f(x) = P(x = x) (b) Cumulative Distribution Functions (CDF) f(x) = P(X ≤ x) Mean M = { xf(x) Variance 0² = [x²f (x) - μ² Standard Deviation. 0x Tot Discrete Uniform Distribution f(x) = //12 M = 0² = b+ a 2 (b-a+1)² 12 •Binomial Distribution (f(x) = (~) p² (1-P)^-x с пр np(1-P) 1• Hypergeometric Distribution (*) (*) (~) f(x) = и = пр 8² = np(1-P/(N=1) where K/N P= ·Poisson Distribution -^+ (^T)* X! f(x) = e M = AT 0² = NT Where N = population size sample size /no. of trials P probability - p = probability of success on a single trial K = no. of successes in the population in the sample of X - no. successes
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