The final exam solutions

Part I, #1, Central limit theorem Let X1,X2, …, Xn be a sequence of i.i.d. random variables each having mean μ and variance σ2 The sum of a large number of independent random variables has a distribution that is approximately normal

Part II #2 If the successful probability of trial is very small, then the accumulatively many successful trials distribute as a Poisson. The Poisson parameter λ is the mean frequency of interest within a large bundle of experiment The exponential distribution is often used to describe the distribution of the amount of time until the specific event first occurs. The exponential distribution seems to partition the Poisson distribution with a parameter λ by every two occurrences into several small time intervals and focuses on a specific time interval. The same distributional parameter λ, The exponential distribution emphasizes the cycle time, 1/λ, while the Poisson on the frequency λ.

Part I, #3 If Z and Xn2 are independent random variables, with Z having a std. normal dist. And Xn2 having a chi-square dist. with n degrees of freedom, For large n, the t distribution approximates to the standard normal distribution

Part I, #4 When these two population variances are unknown but equal, we calculate the pooled variance estimator, Sp2, by means of weighted average of individual sample variance, S12 and S22

Part I, #5 It is not correct. He had better say that the statistic or random variables used to obtain this confidence interval is such that 95% of the time that it is employed it will result in an interval covered the μ. He can assert with probability 0.95 the interval will cover the μ only before the sample data are observed. Whereas after the sample data are observed and computed the interval, he can only assert that the resultant interval indeed contains μ with confidence 0.95.

Part I, #6

Part I, #7 (a) if d(X) is the estimator of θ, and E[d(X)]=θ, then the d(x) is unbiased. (b) the MLE estimator is not unbiased. (c) the sample variance is an unbiased estimator of σ2

Part II #1 Mean=np=90, Var=np(1-p)=150(.6)(.4)=36, ∴ standard deviation=6 P{X ≦ 80}=p{X<80.5}=P{(X- 90)/6<( )/6}=P{Z<-1.583} =1-P{Z ≦ 1.583} (= =0.057 )

Part II #2,3

Part II #4

Part II #5,6