Question

The number of programming errors in HW assignments by computer science students is assumed to follow a Poisson distribution with mean l (so that the density function is f(x) = lambda ^ x * (e ^ - lambda) / x! ).

a) Recall that the mean and variance of a Poisson distribution are both equal to l. A professor examines the distribution of the number of programming errors for one of her assignments. There were 50 assignments turned in. Suppose that lambda = 5 in this case. What are the mean and variance of Xbar?

b) What does the central limit theorem tell us about the distribution of Xbar?

c) Assuming that n=50 is large enough for the CLT to be valid, what is the probability that the average number of errors is 5.0 or higher? What is the probability that the average number of errors is 6.0 or higher?

d) In past years the average number of errors for this assignment has been 5.0. This year the average is 6.0. The teaching assistant argues that the class is just as good, or better, as past years and the higher average is just due to chance. What do you think? Justify your answer. (Hint: find the p-value for H0: lambda = 5 vs H1: lambda > 5)

Answer 1

Answer: number of Programming Let a denotes the errors in HW asslagments is assumed to follow a Poisson distribution with mean a. 0,!,2... XN poisson (2) density function is f(1) = a! mean and variance of X: A = 5 given u= d = 5 and 22=1=5 There are so assignments and Professor examine on of the assignment fro Poisson distribution mean Ex) = x=5 Variance, Vax (x) ==5 Hence EG) = M =5 var () >> Foto Tixi in 50 h limit theorem, if x Accosding to central with mean Mand follows any distribution variance ? for large ni X N N Cumh)

c) By using CLT, INN(5,5/6) ; where xn Poisson (5) mean u=5 n=5 XNN (5,410) P (735) = P(ante = P(230) 30.5 5-5 P (X>6) = P(12-14 h vikso) P(233.16) - 1-P (723.16) c: using standardi normal tabie) -1-0.99921 > 0.00079 d) Ho :1=5 and X= 6 given H135 བྱ་ 1 10 na 50 Ho. CECR)) >5 and var () - under 7-M uN Co.) ta olun

Test statistics : to 6-5 1515150 T50 Vड =5103.16 P-value = P(Z >3-16) = 0.00079. critical value 70.05 - 1.645 and at a = 0.05 P-value <0.05) we reject to and conclude ass to cr. the class has more arg which means that than s. af errors number 0