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Paciicfifthist
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Update 12.tex
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src/chapters/6/sections/means_medians_modes_moments/problems/12.tex

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&= \frac{1}{n^2} \cdot n\sigma^2 + \frac{1}{n^2} \cdot (nc)^2 \\
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&= \frac{\sigma^2}{n} + c^2 \\
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\end{flalign*}
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Hence the bias of \(T_1\) is \[\frac{\sigma^2}{n} + c^2 - c^2 = \frac{\sigma^2}{n}\] \\
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Hence the bias of \(T_1\) is \[\frac{\sigma^2}{n} + c^2 - c^2 = \frac{\sigma^2}{n}.\]
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Finding the bias of \(T_2\), we need to find \(\text{E}(T_2) - c^2\). \[ \text{E}(T_2) = \text{E}(\frac{1}{n}\sum_{j=1}^{n}{X}_j^2) \]
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By linearity of expectation,

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