1. **Problem Statement:** Find the probabilities for a standard normal variable $Z$ for the following cases: (i) $P(Z > 1.09)$ (ii) $P(Z < -1.65)$ (iii) $P(-1.00 < Z < 1.96)$ (iv) $P(1.25 < Z < 2.75)$ 2. **Formula and Rules:** The standard normal distribution $Z$ has mean 0 and standard deviation 1. The cumulative distribution function (CDF) $\Phi(z)$ gives $P(Z \leq z)$. To find probabilities: - $P(Z > a) = 1 - \Phi(a)$ - $P(Z < a) = \Phi(a)$ - $P(a < Z < b) = \Phi(b) - \Phi(a)$ 3. **Calculations:** (i) $P(Z > 1.09) = 1 - \Phi(1.09)$ From standard normal tables or calculator, $\Phi(1.09) \approx 0.8621$ So, $P(Z > 1.09) = 1 - 0.8621 = 0.1379$ (ii) $P(Z < -1.65) = \Phi(-1.65)$ Using symmetry, $\Phi(-1.65) = 1 - \Phi(1.65)$ $\Phi(1.65) \approx 0.9505$ So, $P(Z < -1.65) = 1 - 0.9505 = 0.0495$ (iii) $P(-1.00 < Z < 1.96) = \Phi(1.96) - \Phi(-1.00)$ $\Phi(1.96) \approx 0.9750$ $\Phi(-1.00) = 1 - \Phi(1.00) \approx 1 - 0.8413 = 0.1587$ So, $P(-1.00 < Z < 1.96) = 0.9750 - 0.1587 = 0.8163$ (iv) $P(1.25 < Z < 2.75) = \Phi(2.75) - \Phi(1.25)$ $\Phi(2.75) \approx 0.9970$ $\Phi(1.25) \approx 0.8944$ So, $P(1.25 < Z < 2.75) = 0.9970 - 0.8944 = 0.1026$ 4. **Final Answers:** (i) 0.1379 (ii) 0.0495 (iii) 0.8163 (iv) 0.1026