1. The problem asks to find the area under the standard normal curve for various ranges of the variable $z$. 2. The standard normal distribution is a normal distribution with mean $0$ and standard deviation $1$. The area under the curve between two $z$-values corresponds to the probability that a standard normal variable falls within that range. 3. We use the cumulative distribution function (CDF) of the standard normal distribution, denoted as $\Phi(z)$, which gives the area to the left of $z$. 4. For each part, we calculate the area using $\Phi(z)$ values from standard normal tables or a calculator. 5. Part 1: Area from $z=0$ to $z=1.35$. $$\text{Area} = \Phi(1.35) - \Phi(0)$$ Using standard normal tables, $\Phi(1.35) \approx 0.9115$ and $\Phi(0) = 0.5$. $$\text{Area} = 0.9115 - 0.5 = 0.4115$$ 6. Part 2: $P(-2.05 \leq z \leq 1.96)$. $$\text{Area} = \Phi(1.96) - \Phi(-2.05)$$ $\Phi(1.96) \approx 0.9750$, $\Phi(-2.05) = 1 - \Phi(2.05) \approx 1 - 0.9798 = 0.0202$. $$\text{Area} = 0.9750 - 0.0202 = 0.9548$$ 7. Part 3: $P(z \leq -1.47)$. $$\text{Area} = \Phi(-1.47) = 1 - \Phi(1.47)$$ $\Phi(1.47) \approx 0.9292$. $$\text{Area} = 1 - 0.9292 = 0.0708$$ 8. Part 4: Area to the left of $z=2.01$. $$\text{Area} = \Phi(2.01) \approx 0.9778$$ 9. Part 5: $P(-3.1 \leq z \leq -1.27)$. $$\text{Area} = \Phi(-1.27) - \Phi(-3.1)$$ $\Phi(-1.27) = 1 - \Phi(1.27) \approx 1 - 0.8980 = 0.1020$. $\Phi(-3.1) = 1 - \Phi(3.1) \approx 1 - 0.9990 = 0.0010$. $$\text{Area} = 0.1020 - 0.0010 = 0.1010$$ Final answers: 1. $0.4115$ 2. $0.9548$ 3. $0.0708$ 4. $0.9778$ 5. $0.1010$