Question

find the value of $z_{\alpha}$. $z_{0.28}$ $z_{0.28}=\square$ (round to two decimal places as needed.)

Explanation:

Step1: Understand the meaning of \( z_{\alpha} \)

\( z_{\alpha} \) is the \( z \)-score such that the area to the right of \( z_{\alpha} \) under the standard normal curve is \( \alpha \). So for \( z_{0.28} \), the area to the right of \( z_{0.28} \) is \( 0.28 \), and the area to the left is \( 1 - 0.28=0.72 \).

Step2: Use the standard - normal table (or z - table)

We look for the value in the body of the standard - normal table (which gives the cumulative probability \( P(Z\leq z) \)) closest to \( 0.72 \).
Looking up in the standard - normal table:
If we use a standard - normal table (the \( z \)-table), we find that \( P(Z\leq0.58)=0.7190 \) and \( P(Z\leq0.59) = 0.7224 \). Using linear interpolation (or more accurately, relying on the table values), the \( z \)-score corresponding to a cumulative probability of \( 0.72 \) is approximately \( 0.58 \).

Answer:

\( 0.58 \)