1. **Problem Statement:** Find the probabilities for the standard normal random variable $z$ for the given intervals. 2. **Formula and Rules:** For a standard normal variable $z$, probabilities are found using the cumulative distribution function (CDF) $\Phi(z)$, which gives $P(Z \leq z)$. Important rules: - $P(a \leq Z \leq b) = \Phi(b) - \Phi(a)$ - $P(Z > a) = 1 - \Phi(a)$ 3. **Calculations:** (a) $P(-0.17 \leq z \leq 1.99) = \Phi(1.99) - \Phi(-0.17)$ Using standard normal tables or a calculator: $\Phi(1.99) \approx 0.9767$ $\Phi(-0.17) = 1 - \Phi(0.17) \approx 1 - 0.5675 = 0.4325$ So, $$P(-0.17 \leq z \leq 1.99) = 0.9767 - 0.4325 = 0.5442$$ (b) $P(-0.19 \leq z \leq 0.32) = \Phi(0.32) - \Phi(-0.19)$ $\Phi(0.32) \approx 0.6255$ $\Phi(-0.19) = 1 - \Phi(0.19) \approx 1 - 0.5753 = 0.4247$ So, $$P(-0.19 \leq z \leq 0.32) = 0.6255 - 0.4247 = 0.2008$$ (c) $P(z \leq 1.21) = \Phi(1.21) \approx 0.8869$ (d) $P(z > -1.47) = 1 - P(z \leq -1.47) = 1 - \Phi(-1.47)$ $\Phi(-1.47) = 1 - \Phi(1.47) \approx 1 - 0.9292 = 0.0708$ So, $$P(z > -1.47) = 1 - 0.0708 = 0.9292$$ **Final answers:** - (a) 0.5442 - (b) 0.2008 - (c) 0.8869 - (d) 0.9292