1. **Problem Statement:** Find the areas corresponding to given Z-values using the Z-Table. 2. **Formula and Rules:** The Z-Table gives the area to the left of a Z-value in a standard normal distribution. To find the area to the right, use $1 - \text{area to the left}$. 3. **Calculations for part 1:** - To the left of $Z=0.70$: From Z-Table, area $=0.7580$. - To the right of $Z=-2.32$: Area to left of $-2.32$ is $0.0102$, so area to right $=1-0.0102=0.9898$. - To the right of $Z=1.06$: Area to left of $1.06$ is $0.8554$, so area to right $=1-0.8554=0.1446$. - To the left of $Z=-0.96$: From Z-Table, area $=0.1685$. - To the right of $Z=2.17$: Area to left of $2.17$ is $0.9850$, so area to right $=1-0.9850=0.0150$. - To the right of $Z=0$: Area to left of $0$ is $0.5$, so area to right $=1-0.5=0.5$. 4. **Challenge: Find area between two Z-values (area between $Z=a$ and $Z=b$) = $|\text{area to left of } b - \text{area to left of } a|$** - Between $Z=0$ and $Z=0.53$: Area left of $0.53=0.7019$, left of $0=0.5$, so area $=0.7019-0.5=0.2019$. - Between $Z=0$ and $Z=-0.96$: Area left of $0=0.5$, left of $-0.96=0.1685$, so area $=0.5-0.1685=0.3315$. - Between $Z=0.53$ and $Z=1.32$: Area left of $1.32=0.9066$, left of $0.53=0.7019$, so area $=0.9066-0.7019=0.2047$. - Between $Z=-0.53$ and $Z=-0.70$: Area left of $-0.53=0.2981$, left of $-0.70=0.24196$, so area $=0.2981-0.24196=0.05614$. - Between $Z=1.2$ and $Z=-0.96$: Area left of $1.2=0.8849$, left of $-0.96=0.1685$, so area $=0.8849-0.1685=0.7164$. - Between $Z=-1.2$ and $Z=-2.58$: Area left of $-1.2=0.1151$, left of $-2.58=0.0049$, so area $=0.1151-0.0049=0.1102$. **Final answers:** 1. Left of 0.70: $0.7580$ 2. Right of -2.32: $0.9898$ 3. Right of 1.06: $0.1446$ 4. Left of -0.96: $0.1685$ 5. Right of 2.17: $0.0150$ 6. Right of 0: $0.5$ Challenge areas: 1. $0.2019$ 2. $0.3315$ 3. $0.2047$ 4. $0.05614$ 5. $0.7164$ 6. $0.1102$