Use Statistics Kingdom es to find the value of \( b \). State \( b \) to six decimal places as in 0.123456. \( \begin{array}{c}P(z \leq b)=0.2222 \\ b=\text { type your answer... } \\ \begin{array}{l}P(z \geq b)=0.15 \\ b\end{array} \\ \begin{array}{l}P(-b \leq z \leq b)=0.82 \\ b\end{array} \\ \text { type your answer... } \\ \text { type your answer... }\end{array} \)

To find the value of \( b \) corresponding to the probabilities given, we need to refer to the standard normal distribution (z-distribution) tables or use statistical software. 1. For \( P(z \leq b) = 0.2222 \), you can find \( b \) using a Z-table or an online calculator. This corresponds to \( b \approx -0.774197 \) when rounded to six decimal places. 2. For \( P(z \geq b) = 0.15 \), since this relates to the right tail, we find \( P(z \leq b) = 1 - 0.15 = 0.85 \), which gives \( b \approx 1.036433 \). 3. For \( P(-b \leq z \leq b) = 0.82 \), this means that \( P(z \leq b) - P(z \leq -b) = 0.82 \). The remaining area in the two tails would be \( 1 - 0.82 = 0.18 \), leading us to find that \( P(z \leq b) = 0.91 \) and hence \( b \approx 1.340755 \). To summarize, the values of \( b \) for each part are: - For \( P(z \leq b) = 0.2222 \): \( b \approx -0.774197 \) - For \( P(z \geq b) = 0.15 \): \( b \approx 1.036433 \) - For \( P(-b \leq z \leq b) = 0.82 \): \( b \approx 1.340755 \) Here are the solutions to type: - \( b = -0.774197 \) - \( b = 1.036433 \) - \( b = 1.340755 \)