To solve these questions, the given IQ model follows a normal distribution $N(100, 16)$, where the mean ($\mu$) is 100 and the standard deviation ($\sigma$) is 16. Let’s calculate step-by-step.

Step-by-step solutions:

(a) Percent of IQs above 85

1. Compute the z-score for 85: $z = \frac{x - \mu}{\sigma} = \frac{85 - 100}{16} = -0.9375$
2. Use a z-table or calculator to find $P(Z > -0.9375)$. From symmetry, $P(Z > -0.9375) = 1 - P(Z < -0.9375)$. Using a z-table: $P(Z < -0.9375) \approx 0.1747$, so $P(Z > -0.9375) \approx 1 - 0.1747 = 0.8253$ or 82.5%.

(b) Percent of IQs below 95

1. Compute the z-score for 95: $z = \frac{x - \mu}{\sigma} = \frac{95 - 100}{16} = -0.3125$
2. Find $P(Z < -0.3125)$ using a z-table or calculator. Using a z-table: $P(Z < -0.3125) \approx 0.3783$ or 37.8%.

(c) Percent of IQs between 116 and 136

1. Compute z-scores for 116 and 136:
   • For 116: $z = \frac{116 - 100}{16} = 1.0$
   • For 136: $z = \frac{136 - 100}{16} = 2.25$
2. Find probabilities:
   • $P(Z < 1.0) \approx 0.8413$ (from z-table)
   • $P(Z < 2.25) \approx 0.9878$
   • Difference: $P(116 < X < 136) = P(Z < 2.25) - P(Z < 1.0) = 0.9878 - 0.8413 = 0.1465$ or 14.7%.

Would you like me to delve deeper into how to use z-tables, or do you want further clarification?

Questions for Practice:

1. How do z-scores relate to probability in a normal distribution?
2. What is the process to calculate $P(Z > z)$ using a z-table?
3. How does symmetry in a normal curve simplify calculations?
4. Can you explain why $P(Z < z) + P(Z > z) = 1$?
5. How would changing $\sigma$ to 20 affect the results?

Tip: Always sketch a normal curve to visualize probabilities—it enhances understanding!