4.7 The Standard Normal DistributionDefinition:A continuous random variable with pdf:2/21( )2zf zewith supportz , is saidto be a Standard Normal Random Variable and the pdf is the Standard Normal Distribution.We always denote this special random variableZ. Showing that this is indeed a pdf, we would need toshow that this integrates to 1. Good luck. This will be shown in the appendix.Theorem:The mean of the standard normal is 0. That is,[ ]0E Z.Proof:2220/2/2/20111[ ]222zzzE Zzedzzedzzedz. It should be clear that thetwo integrals are equal in magnitude and opposite in sign. So the expected value is 0 as long as eachconverges. Focusing on the second integral, we have222/2/2/200011111( 1)(01)22222zzzzedzzedze.The other integral will equal12. So, the expected value equals 0.Theorem:The variance of the standard normal distribution is 1. That is,Var[ ]1Z.2222/2/211[]22zzE Zzedzzzedz. This allows us to integrate by parts withuzdudzand 22/2/21122zzdvzedzve. This givesWe define the CDF of Z in the usual manner:[ ]( )( )zE Zzf t dt . The antiderivative of( )f zdoesnot exist in closed form. So for any axis valuez,( )zmust be computed numerically. Since we cannotdo this by hand, we will use a computer generated CDF chart.Suppose that we wish to determine(.651.07)PZ. This asks todetermine the probability that a randomly selected item from thestandard normal distribution will be between .65 and 1.07.The answer tothis question is the amount of area under the pdf between .65 and 1.07.The answer to this is clearly equal to(1.07)(.65) . We will go withthe z-chart to determine these two CDF probabilities.