12. Find the \( z \)-score boundaries that separate a normal distribution as described in each of the following. a. The middle \( 20 \% \) from the \( 80 \% \) in the tails b. The middle \( 25 \% \) from the \( 75 \% \) in the tails c. The middle \( 70 \% \) from the \( 30 \% \) in the tails d. The middle \( 90 \% \) from the \( 10 \% \) in the tails

To find the \( z \)-score boundaries that separate the specified portions of a normal distribution, we need to look up values in the standard normal distribution table or use a calculator that provides the cumulative distribution function (CDF) for the normal distribution. Here are the boundaries for each case: a. For the middle \( 20\% \), which leaves \( 80\% \) in the tails, we want to find the \( z \)-scores that correspond to the \( 40\% \) (the lower tail area is \( \frac{80\%}{2} = 40\% \)). This gives \( z \)-scores of approximately \( -0.8416 \) and \( 0.8416 \). b. For the middle \( 25\% \), leaving \( 75\% \) in the tails, we seek the \( z \)-scores corresponding to \( 37.5\% \) (the lower tail area becomes \( \frac{75\%}{2} = 37.5\% \)). This yields \( z \)-scores of approximately \( -0.524 \) and \( 0.524 \). c. For the middle \( 70\% \), leaving \( 30\% \) in the tails, we find \( 15\% \) for the lower tail area (\( \frac{30\%}{2} \)). The corresponding \( z \)-scores are approximately \( -1.036 \) and \( 1.036 \). d. For the middle \( 90\% \), removing \( 10\% \) in the tails translates to \( 5\% \) in the lower tail area. The \( z \)-scores here are roughly \( -1.645 \) and \( 1.645 \). Always remember to check the specific \( z \)-values using either a standard normal distribution table or a statistical calculator to ensure accuracy!