Assume that the random variable X is normally distributed with mean μ = 120 and standard deviation σ = 13. Find P(110 < X < 145).

To solve this problem, we need to convert the raw scores to z-scores and then use the standard normal distribution to find the probabilities.

Step 1: Convert the raw scores to z-scores. The formula for converting a raw score x to a z-score is z = (x - μ) / σ.

For x = 110, the z-score is z = (110 - 120) / 13 = -0.769. For x = 145, the z-score is z = (145 - 120) / 13 = 1.923.

Step 2: Use the standard normal distribution to find the probabilities. We want to find P(-0.769 < Z < 1.923).

You can look up these values in a standard normal distribution table or use a calculator that can calculate normal distribution probabilities.

The value for P(Z < -0.769) is approximately 0.221. The value for P(Z < 1.923) is approximately 0.973.

Step 3: Subtract the probabilities. P(110 < X < 145) = P(Z < 1.923) - P(Z < -0.769) = 0.973 - 0.221 = 0.752.

So, the probability that X is between 110 and 145 is approximately 0.752 or 75.2%.