Final answer:
To find the proportion of people who score between 9 and 12 on the test, calculate the z-scores for x1 = 9 and x2 = 12, look up the cumulative probabilities for these z-scores in a standard normal distribution table, and subtract the probabilities to find the proportion.
Step-by-step explanation:
To find the proportion of people who score between 9 and 12 on the test, we need to find the area under the normal distribution curve between those two scores. First, we calculate the z-scores for x1 = 9 and x2 = 12 using the formula: z = (x - μ) / σ. With a mean (μ) of 10 and a standard deviation (σ) of 2, we get z1 = (9 - 10) / 2 = -0.5 and z2 = (12 - 10) / 2 = 1. Next, we look up the cumulative probability associated with these z-scores in a standard normal distribution table.
The cumulative probability for z1 is P(Z ≤ -0.5) ≈ 0.3085. The cumulative probability for z2 is P(Z ≤ 1) ≈ 0.8413. To find the proportion between 9 and 12, we subtract the cumulative probability for z1 from the cumulative probability for z2: P(9 ≤ X ≤ 12) = P(Z ≤ 1) - P(Z ≤ -0.5) ≈ 0.8413 - 0.3085 ≈ 0.5328.
Therefore, about 53.28% of people score between 9 and 12 on the test.