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## Week 8 Stat Notes

by: Emmy Thornsberry

12

0

2

# Week 8 Stat Notes MATH 2600

Emmy Thornsberry
GC&SU

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Covers discrete random variable, Bernoulli principle, and probabilities and expectations of discrete random variables.
COURSE
Introduction To Statistics
PROF.
James Baugh
TYPE
Class Notes
PAGES
2
WORDS
CONCEPTS
stat
KARMA
25 ?

## Popular in Math

This 2 page Class Notes was uploaded by Emmy Thornsberry on Tuesday October 11, 2016. The Class Notes belongs to MATH 2600 at Georgia College & State University taught by James Baugh in Fall 2016. Since its upload, it has received 12 views. For similar materials see Introduction To Statistics in Math at Georgia College & State University.

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Date Created: 10/11/16
Week 8 Stat Notes Given X, a discrete random variable - Population mean μ=E[x]= ∑ x·P(X=x) all x - Population variance σ​ =E[(x-μ)​ ]=E[x​ ]-μ​ 2​ 2 For sums of independent random variables - The population mean of the sum is the sum of the means - Variance of the sum is the sum of the variances x=sum of two fair dice - μ=7 2​ - σ​ = 35/6 x=3 - μ= 3.5 2​ - σ = 35/12 - σ= 35/12 Bernoulli Distribution c​ - Let A be an event with the probability for A being p (P(A)=p) and q=1-p=P(A​ ) - x= {1 if A occurs, 0 if A​ occurs} - X 0 1 PDF q=1-p p - σ= pq √ - σ = E[x​ ]-μ​ =p-p​ = p(1-p)=pq - μ= ∑ xP(X=x)=0·q+1·p=p all x 2​ 2 2 - E[x​ ]=0​ ·q+1​ ·p=p Given x is a sum of n equivalent but not independent, the population mean of the sum is the sum of the means and the variance of the sum is the sum of the variances. Bernoulli random variables - μ= p+p+p+p+.. ..p=n·p 2​ - σ =n-pq - Binomial random variable x is the number of A outcomes in n independent x​n-x - P(X=x)=​ Cn​px​q​ - Sample space= {0,1,2….n} Example of binomial random variable - By hand - There’s a p=20% chance that a random board will have a knot in it. You buy 8 boards. x= number of knots in your lumber - p(x=2) (probability that two boards have knots in them) - C​ p​ q​-x n​ x​ 2​ 8-2​ - 8​​2​.20)​ (.8)​ =0.2936 X is the number of widgets in a bin of 50 which have a manufacturing defect. There is a 1% chance of a given widget being defective. - μ=50·0.01=1/2 - σ= 0.605=0.703 - σ = npq=(50)(.01)(.99)=0.495 - n=50 - p=.01 - q=1-p=.99 - P(x=0)=0.605 - Finding on calculator - 2nd VARS - binompdf( - Trial: n; P:p; x value: 0 - CPQ method: ​ C​ p​n​​ x​x​ n-x 0​ 50-0 - 50​​0​​ q​ - 1​ 1​ .99​50-0 50​ 0​ - P(1< ​ ​x<10)= 0.395 - CDF=P(X​<x ​ ) - 2nd vars - binomcdf(50,.01,9)-binomcdf(50,.01,0) - Gives the probabilities of anything from 1-9 (bc 10 is not included) - Anytime you’re working with inequalities, use a CDF

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