 ## D.12 Appendix A2 Solutions

(LCA2.1) What proportion of the area under the normal curve is less than 3? Greater than 12? Between 0 and 12?

• Less than 3: 3 is one standard deviation less than the mean of 6, since $$\frac{3 - \mu}{\sigma} = \frac{3-6}{3} = -1$$. Thus we compute the area to the left of z = -1 in Figure A.2: 0.15% + 2.35% + 13.5% = 16%.
• Greater than 12: 12 is two standard deviations greater than the mean of 6, since $$\frac{12 - \mu}{\sigma} = \frac{12-6}{3} = +2$$, we compute the area to the right of z = +2 in Figure A.2: 2.35% + 0.15% = 2.5%.
• Between 0 and 12: 0 is two standard deviations less than the mean of 6, since $$\frac{0 - \mu}{\sigma} = \frac{0-6}{3} = -2$$. Thus we compute the area between z = -2 and z = +2 in Figure A.2: 13.5% + 34% + 34% + 13.5% = 95%.

(LCA2.2) What is the 2.5th percentile of the area under the normal curve? The 97.5th percentile? The 100th percentile?

• 2.5th percentile: Starting from the left of Figure A.2, since 0.15% + 2.35% = 2.5%, the 2.5th percentile is z = - 2. However, this is in standard units. Thus we need the value in the normal distribution that is two standard deviations lower than the mean: $$\mu - 2 \cdot\sigma = 6 - 2 \cdot 3 = 0$$.
• 97.5th percentile: Starting from the left of Figure A.2, since 0.15% + 2.35% + 13.5% + 34% + 34% + 13.5% = 97.5%, the 97.5th percentile is z = +2. However, this is in standard units. Thus we need the value in the normal distribution that is two standard deviations higher than the mean: $$\mu + 2 \cdot\sigma = 6 + 2 \cdot 3 = 12$$.
• 100th percentile: $$+\infty$$. In other words, 100% of values will be less than positive infinity.