Please answer the following question(s). Please show all your working/solution(s).

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Suppose that Y is a normally distributed with mean 0 and variance 02 (a) Argue that YZ/a2 ~ Xi- (b) The quantity Y2/02 is said to be a pivot for 02 because its distribution, which is , does not depend on 02 (even if the random variable itself is a function of 02! Interesting huh?). Use the pivotal quantity Y2/02 to nd an random interval [L3, U B] such that P([LB, U B] contains 02) is 0.95. This interval is called a 95% - condence interval for 02. Hint: F1721 (1)) can be calculated in R using qchisq(v, 1). 1