Let X1,X2, . . . ,X18 be a random sample of size 18 from a

QUESTION:

Let \(X_{1}, X_{2}, \ldots, X_{18}\) be a random sample of size 18 from a chi-square distribution with \(r=1\). Recall that \(\mu=1\) and \(\sigma^{2}=2\).

(a) How is \(Y=\sum_{i=1}^{18} X_{i}\) distributed?

(b) Using the result of part (a), we see from Table IV in Appendix B that

\(P(Y \leq 9.390)=0.05 \text { and } P(Y \leq 34.80)=0.99\) .

Compare these two probabilities with the approximations found with the use of the central limit theorem.

Equation Transcription:

Text Transcription:

X_1,X_2,…,X_18  

r=1  

mu=1  

sigma^2=2  

Y=sum_i=1^18  X_i  

P(Y < or = 9.390)=0.05  

P(Y < or = 34.80)=0.99