Random Variables and Probability DistributionsNormal DistributionSlide 3Example - Heights of U.S. AdultsStandard Normal (Z) DistributionSlide 6Finding Probabilities of Specific RangesExample - Adult Female HeightsFinding Percentiles of a DistributionExample - Adult Male HeightsStatistical ModelsSampling Distributions and the Central Limit TheoremSlide 13Random Variables and Probability Distributions•Random Variables - Random responses corresponding to subjects randomly selected from a population.•Probability Distributions - A listing of the possible outcomes and their probabilities (discrete r.v.s) 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 samplesNormal Distribution•Bell-shaped, symmetric family of distributions•Classified by 2 parameters: Mean () and standard deviation (). These represent location and spread•Random variables that are approximately normal have the following properties wrt individual measurements:–Approximately half (50%) fall above (and below) mean–Approximately 68% fall within 1 standard deviation of mean–Approximately 95% fall within 2 standard deviations of mean–Virtually all fall within 3 standard deviations of mean•Notation when Y is normally distributed with mean  and standard deviation  :),(~NYNormal Distribution95.0)22(68.0)(50.0)( YPYPYPExample - 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)INCHESM76.575.574.573.572.571.570.569.568.567.566.565.564.563.562.561.560.559.5Cases weighted by PCTM20100Std. Dev = 2.61 Mean = 69.1N = 99.23INCHESF70.569.568.567.566.565.564.563.562.561.560.559.558.557.556.555.5Cases weighted by PCTF20181614121086420Std. Dev = 2.48 Mean = 63.7N = 99.68Source: Statistical Abstract of the U.S. (1992)Standard Normal (Z) Distribution•Problem: Unlimited number of possible normal distributions (- <  <  ,  > 0)•Solution: Standardize the random variable to have mean 0 and standard deviation 1)1,0(~),(~ NYZNY• Probabilities of certain ranges of values and specific percentiles of interest can be obtained through the standard normal (Z) distributionStandard Normal (Z) Distribution•Standard Normal Distribution Characteristics:–P(Z  0) = P(Y   ) = 0.5000–P(-1  Z  1) = P(-Y  +) = 0.6826–P(-2  Z  2) = P(-2Y  +2) = 0.9544–P(Z  za) = P(Z  -za) = a (using Z-table)a 0.500 0.100 0.050 0.025 0.010 0.005za0.000 1.282 1.645 1.960 2.326 2.576Finding Probabilities of Specific Ranges•Step 1 - Identify the normal distribution of interest (e.g. its mean () and standard deviation () )•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 YL and YU into Z-values:UULLYZYZ• Step 4 - Obtain P(ZL Z  ZU) from Z-tableExample - 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 - ULZZ 52.25.27.630.70• Step 4 - P(Y  70) = P(Z  2.52) = .0059 (  1/170) z .00 .01.02.032.4 .0082 .0080 .0078 .00752.5.0062 .0060.0059.00572.6 .0047 .0045 .0044 .0043Finding Percentiles of a Distribution•Step 1 - Identify the normal distribution of interest (e.g. its mean () and standard deviation () )•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 zp back to original units:ppzY 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 95th percentile (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) = 73.4z .03.04 .05.061.5 .0630 .0618 .0606 .05941.6.0516.0505 .0495.04851.7 .0418 .0409 .0401 .0392Statistical Models•When making statistical inference it is useful to write random variables in terms of model parameters and random errors YYY )(• Here  is a fixed constant and  is a random variable• In practice  will be unknown, and we will use sample data to estimate or make statements regarding its valueSampling Distributions and the Central Limit Theorem•Sample statistics based on random samples are also random variables and have sampling distributions that 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):–Sample Mean:–Sample Proportion: nNY,~nN)1(,~^Example - Adult Female Heights•Random samples of n = 100 females to be selected•For each sample, the sample mean is computed•Sampling distribution:)25.0,5.63(1005.2,5.63~ NNY • Note that approximately 95% of all possible random samples of 100 females will have sample means between 63.0 and 64.0