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海量文库 文档专家

发布时间：2013-11-25 13:03:05

Random Variables and Probability

Distributions

?Random Variables-Random responses corresponding to subjects randomly selected from a population.?Probability Distributions -A listing of the possible

soutcomes and their probabilities (discrete r.v.) or their

densities (continuous r.v.s)

?Normal Distribution-Bell-shaped continuous distribution widely used in statistical inference

?Sampling Distributions -Distributions corresponding to sample statistics (such as mean and proportion) computed from random samples

Normal Distribution

?Bell-shaped, symmetric family of distributions?Classified by 2 parameters: Mean (m) and standard deviation (s). These represent locationand spread?Random variables that are approximately normal have the following properties wrt individual measurements:–

–

–

–Approximately half (50%) fall above (and below) meanApproximately 68% fall within 1 standard deviation of meanApproximately 95% fall within 2 standard deviations of meanVirtually all fall within 3 standard deviations of mean?Notation when Yis normally distributed with mean mand standard deviation s:

Y~N(m,s)

Normal Distribution

P(Y?m)?0.50P(m?s?Y?m?s)?0.68P(m?2s?Y?m?2s)?0.95

Example -Heights of U.S. Adults?Female and Male adult heights are well approximated by normal distributions: YF~N(63.7,2.5) YM~N(69.1,2.6)

Source: Statistical Abstract of the U.S. (1992)

Standard Normal (Z) Distribution?Problem: Unlimited number of possible normal distributions (-?< m< ?, s> 0)

?Solution: Standardize the random variable to have mean 0 and standard deviation 1

Y~N(m,s)?Z?Y?m

s~N(0,1)

?Probabilities of certain ranges of values and specific percentiles of interest can be obtained through the standard normal (Z) distribution

Standard Normal (Z) Distribution?Standard Normal Distribution Characteristics:–

–

–

–P(Z ?0) = P(Y ?m) = 0.5000P(-1 ?Z ?1) = P(m-s ?Y?m+s ) = 0.6826P(-2 ?Z ?2) = P(m-2s ?Y?m+2s ) = 0.9544P(Z ?za) = P(Z ?-za) = a (using Z

-table)

Finding Probabilities of Specific Ranges?Step 1 -Identify the normal distribution of interest (e.g. its mean (m) and standard deviation (s) )

?Step 2 -Identify the range of values that you wish to determine the probability of observing (YL, YU), where often the upper or lower bounds are ?or -?

?Step 3 -Transform YLand YUinto Z-values:

ZL?YL?m

sZU?YU?m

s

?Step 4 -Obtain P(ZL?Z ?ZU) from Z-table

Example -Adult Female Heights?What is the probability a randomly selected female is 5’10” or taller (70 inches)?

?Step 1 -Y~ N(63.7 , 2.5)

?Step 2 -YL = 70.0 YU = ?

?Step 3-70.0?63.7ZL??2.52ZU??2.5

?Step 4 -P(Y ?70) = P(Z?2.52) = .0059 ( ?1/170)

Finding Percentiles of a Distribution?Step 1 -Identify the normal distribution of interest (e.g. its mean (m) and standard deviation (s) )

?Step 2-Determine the percentile of interest 100p% (e.g. the 90th percentile is the cut-off where only 90% of scores are below and 10% are above)

?Step 3 -Turn the percentile of interest into a tail probability a and corresponding z-value (zp):–If 100p?50 then a= 1-p and zp= za

–If 100p< 50 then a= p and zp= -za

?Step 4 -Transform zpback to original units:

Yp?m?zsp

Example -Adult Male Heights?

?

?

?Above what height do the tallest 5% of males lie above?Step 1 -Y~ N(69.1 , 2.6)Step 2 -Want to determine 95thpercentile (p= .95)Step 3 -Since 100p> 50, a = 1-p= 0.05

zp= za= z.05= 1.645

?Step 4 -Y.95= 69.1 + (1.645)(2.6) = 73.4

Statistical Models

?When making statistical inference it is useful to write random variables in terms of model

parameters and random errors

Y?m?(Y?m)?m????Y?m?Here mis a fixed constant and ?is a random variable

?In practice mwill be unknown, and we will use sample data to estimate or make statements regarding its value

Sampling Distributions and the Central

Limit Theorem

?Sample statistics based on random samples are also

random variables and have sampling distributionsthat are probability distributions for the statistic (outcomes that would vary across samples)

?When samples are large and measurements independent then many estimators have normal sampling

distributions (CLT):?s?

–Sample Mean:

–Sample Proportion: ^Y~N?m,?n????(1??)???~N??,??n??

Example -Adult Female Heights?Random samples of n= 100 females to be selected?For each sample, the sample mean is computed?Sampling distribution:

2.5??Y~N?63.5,??N(63.5,0.25)??

?Note that approximately 95% of all possible random samples of 100 females will have sample means between 63.0 and 64.0 inches