(iii) \( \left\{\begin{aligned} x_{1}-5 x_{2}-x_{3}+12 x_{4}-x_{5} & =-4 \\ -2 x_{1}+10 x_{2}+3 x_{3}-33 x_{4}+2 x_{5}+x_{6} & =4 \\ x_{3}-9 x_{4}+x_{5} & =8 \\ x_{1}-5 x_{2}+2 x_{3}-15 x_{4}+2 x_{5}+x_{6} & =14\end{aligned}\right. \)

\( \left. \begin{array} { r } { f ( x ) = \sin ^ { 2 } ( x ) , a = 0 } \\ { L ( x ) = \square } \end{array} \right. \)

3. \( 1 5 \longdiv { 3 4 4 } \)

A \( 90 \% \) confidence interval for \( \mu \) is \( (11.5,12.3) \). What is a correct interpretation of this interval? The probability is 0.95 that the value of \( \mu \) is between 11.5 and 12.3 . We are \( 90 \% \) confident that the value of \( \mu \) is between 11.5 and 12.3 . \( 90 \% \) of the time the value of \( \mu \) is between 11.5 and 12.3 .

(2.) \( \begin{array}{r}3,933 \\ \times \quad 4 \\ \hline\end{array} \)

The plot of the data clearly indicates outliers. Why is the resulting confidence interval still valid? Because there are several outliers, not just one. Because the standard deviation of 10 is greater than the sample mean \( \bar{x} \) of 9.34 . Because the sample size of 121 is large enough to apply the Central Limit Theorem.