= Solution
The standardized observations $X_i/\sigma$ are independent standard <normal random variables>. By the definition of the <chi-squared distribution>,
$$
Q=\frac{\sum_iX_i^2}{\sigma^2}\sim\chi_n^2.
$$
Its distribution is continuous, so $F_n(c_2)-F_n(c_1)=1-\alpha$ gives the required <probability>. Put $S=\sum_iX_i^2$. Solving $c_1<S/\sigma^2<c_2$ for the positive <variance> reverses the reciprocal order. The exact <confidence interval> is
$$
\boxed{\left(\frac S{F_n^{-1}(1-\alpha/2)},\frac S{F_n^{-1}(\alpha/2)}\right).}
$$
The mean is known to be zero, so the degrees of freedom are $n$, not $n-1$. Coverage is exactly $1-\alpha$ under the stated normal model.
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