Let's solve these probabilities and percentile points under the normal distribution curve, following the same process as the provided example.

The process involves:

1. Expressing the percentile as a probability.
2. Finding the corresponding z-score from the z-table or inverse normal distribution function.
3. Computing the raw score using the formula $x = z \cdot \sigma + \mu$.

Assumptions for all computations:

• Mean ($\mu$) = 80
• Standard deviation ($\sigma$) = 15

Solution:

1. $P_{99}$

• Percentile: 99th → $P(z) = 0.9900$
• $z$-value: $z = 2.33$ (from z-table)
• Raw score:
  $x = z \cdot \sigma + \mu = 2.33 \cdot 15 + 80 = 34.95 + 80 = 114.95$

2. $P_{90}$

• Percentile: 90th → $P(z) = 0.9000$
• $z$-value: $z = 1.28$ (from z-table)
• Raw score:
  $x = z \cdot \sigma + \mu = 1.28 \cdot 15 + 80 = 19.2 + 80 = 99.2$

3. $P_{68}$

• Percentile: 68th → $P(z) = 0.6800$
• $z$-value: $z = 0.47$ (from z-table)
• Raw score:
  $x = z \cdot \sigma + \mu = 0.47 \cdot 15 + 80 = 7.05 + 80 = 87.05$

4. $P_{40}$

• Percentile: 40th → $P(z) = 0.4000$
• $z$-value: $z = -0.25$ (from z-table)
• Raw score:
  $x = z \cdot \sigma + \mu = -0.25 \cdot 15 + 80 = -3.75 + 80 = 76.25$

5. $P_{32}$

• Percentile: 32nd → $P(z) = 0.3200$
• $z$-value: $z = -0.47$ (from z-table)
• Raw score:
  $x = z \cdot \sigma + \mu = -0.47 \cdot 15 + 80 = -7.05 + 80 = 72.95$

6. $P_{21}$

• Percentile: 21st → $P(z) = 0.2100$
• $z$-value: $z = -0.81$ (from z-table)
• Raw score:
  $x = z \cdot \sigma + \mu = -0.81 \cdot 15 + 80 = -12.15 + 80 = 67.85$

7. $P_{50}$

• Percentile: 50th → $P(z) = 0.5000$
• $z$-value: $z = 0.00$ (from z-table)
• Raw score:
  $x = z \cdot \sigma + \mu = 0 \cdot 15 + 80 = 80$

8. $P_{14}$

• Percentile: 14th → $P(z) = 0.1400$
• $z$-value: $z = -1.08$ (from z-table)
• Raw score:
  $x = z \cdot \sigma + \mu = -1.08 \cdot 15 + 80 = -16.2 + 80 = 63.8$

9. $P_{27}$

• Percentile: 27th → $P(z) = 0.2700$
• $z$-value: $z = -0.61$ (from z-table)
• Raw score:
  $x = z \cdot \sigma + \mu = -0.61 \cdot 15 + 80 = -9.15 + 80 = 70.85$

10. $P_{88}$

• Percentile: 88th → $P(z) = 0.8800$
• $z$-value: $z = 1.18$ (from z-table)
• Raw score:
  $x = z \cdot \sigma + \mu = 1.18 \cdot 15 + 80 = 17.7 + 80 = 97.7$

Final Results:

1. $P_{99}$: $x = 114.95$
2. $P_{90}$: $x = 99.2$
3. $P_{68}$: $x = 87.05$
4. $P_{40}$: $x = 76.25$
5. $P_{32}$: $x = 72.95$
6. $P_{21}$: $x = 67.85$
7. $P_{50}$: $x = 80.00$
8. $P_{14}$: $x = 63.8$
9. $P_{27}$: $x = 70.85$
10. $P_{88}$: $x = 97.7$

Let me know if you need details on any specific calculation!

Here are some questions for you to expand understanding:

1. What is the significance of z-scores in normal distributions?
2. How does changing the mean ($\mu$) affect raw scores for the same percentile?
3. Why do percentiles below 50 have negative z-scores?
4. Can you explain why $P_{50}$ corresponds exactly to the mean?
5. How can this process be applied in real-world data interpretation?

Tip: Always ensure to double-check z-values from a reliable z-table or statistical tool for accuracy!