Answer:a. 2.14% should have IQ scores between 40 and 60b. 15.87% should have IQ scores below 80Step-by-step explanation:* Lets explain how to solve the problem- For the probability that a < X < b (X is between two numbers, a and b), convert a and b into z-scores and use the table to find the area between the two z-values.- Lets revise how to find the z-score- The rule the z-score is z = (x - μ)/σ , where# x is the score# μ is the mean# σ is the standard deviation* Lets solve the problem- IQS are normally distributed with a mean of 100 and standard deviation of 20∴ μ = 100 and σ = 20a.- The IQS is between 40 and 60∴ 40 < X < 60∵ z = (x - μ)/σ∴ z = (40 - 100)/20 = -60/20 = -3∴ z = (60 - 100)/20 = -40/20 = -2- Use the z table to find the corresponding area∵ P(z > -3) = 0.00135∵ P(z < -2) = 0.02275∴ P(-3 < z < -2) = 0.02275 - 0.00135 = 0.0214∵ P(40 < X < 60) = P(-3 < z < -2)∴ P(40 < X < 60) = 0.0214 = 2.14%* 2.14% should have IQ scores between 40 and 60b.- The IQS is below 80∴ X < 80∵ z = (x - μ)/σ∴ z = (80 - 100)/20 = -20/20 = -1- Use the z table to find the corresponding area∵ P(z < -1) = 0.15866∵ P(X < 80) = P(z < -1)∴ P(X < 80) = 0.15866 = 15.87%* 15.87% should have IQ scores below 80