A normal population has mean μ =33 and standard deviation σ =7.(a) What proportion of the population is between 18 and 29?

To solve this problem, we need to convert the raw scores to z-scores and then look up the proportions in the standard normal distribution table.

Step 1: Convert the raw scores to z-scores. The formula for calculating a z-score is:

Z = (X - μ) / σ

where X is the raw score, μ is the population mean, and σ is the standard deviation.

For X = 18, the z-score is:

Z = (18 - 33) / 7 = -15 / 7 = -2.14

For X = 29, the z-score is:

Z = (29 - 33) / 7 = -4 / 7 = -0.57

Step 2: Look up the proportions in the standard normal distribution table.

The proportion for Z = -2.14 is 0.0162, which means 1.62% of the population is below a raw score of 18.

The proportion for Z = -0.57 is 0.2843, which means 28.43% of the population is below a raw score of 29.

Step 3: Subtract the two proportions to find the proportion of the population between the two raw scores.

0.2843 - 0.0162 = 0.2681

So, approximately 26.81% of the population is between 18 and 29.