Using the paper version of the statistical table, find the value of \( b \). New Statistics Tables,pdf State answers to two decimal places. \[ \begin{aligned} P(z \leq b) & =0.0853 \\ b & =\text { type your answer... } \end{aligned} \] \[ P(zb)=0.8888 \\ b=\text { type your answer... } \end{array} \]

We are given four probability statements involving the standard normal variable z. Using the standard normal (z) table, we find the z‐value corresponding to the given probabilities (to two decimal places): 1. For P(z ≤ b) = 0.0853:   We need b such that the cumulative probability is 0.0853. Looking in the z table, we find that when z = –1.37, P(z ≤ –1.37) ≈ 0.0853.   Answer: b = –1.37 2. For P(z < b) = 0.9082:   Now, we look for the z value giving a cumulative probability of 0.9082. The z table shows that when z = 1.33, P(z < 1.33) ≈ 0.9082.   Answer: b = 1.33 3. For P(z ≥ b) = 0.5910:   Recall that P(z ≥ b) = 1 – P(z ≤ b). Thus,     P(z ≤ b) = 1 – 0.5910 = 0.4090.   Find the z value corresponding to 0.4090. The table indicates that when z = –0.23, P(z ≤ –0.23) ≈ 0.4090.   Answer: b = –0.23 4. For P(z > b) = 0.8888:   Again, notice P(z > b) = 1 – P(z ≤ b), so     P(z ≤ b) = 1 – 0.8888 = 0.1112.   The z table shows that when z = –1.22, P(z ≤ –1.22) ≈ 0.1112.   Answer: b = –1.22 Thus, the answers (rounded to two decimals) are: • P(z ≤ b) = 0.0853 ⟹ b = –1.37 • P(z < b) = 0.9082 ⟹ b = 1.33 • P(z ≥ b) = 0.5910 ⟹ b = –0.23 • P(z > b) = 0.8888 ⟹ b = –1.22