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Let X1, X2, ... , X18 be a random sample of size 18 from a chi-square distribution with
Chapter 5, Problem 5.6-5(choose chapter or problem)
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
\(P(Y \leq 9.390)=0.05\) and \(P(Y \leq 34.80)=0.99\).
Compare these two probabilities with the approximations found with the use of the central limit theorem.
Questions & Answers
(1 Reviews)
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
\(P(Y \leq 9.390)=0.05\) and \(P(Y \leq 34.80)=0.99\).
Compare these two probabilities with the approximations found with the use of the central limit theorem.
ANSWER:Step 1
Given that,
\(n = 18\)
\({X_i} \sim {\chi ^2}\left( 1 \right)\)
\(\mu = 1\)
\({\sigma ^2} = 2\)
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