Quantitative Literacy Practice Exam

To find the percentage of buyers who paid less than $145,600 in a normal distribution, we first need to determine how many standard deviations this price is away from the mean. The mean price of a home is $150,000, and the standard deviation is $2,200. To calculate the z-score for $145,600, we use the formula: \[ z = \frac{(X - \mu)}{\sigma} \] Where: - \(X\) is the value we are assessing ($145,600), - \(\mu\) is the mean ($150,000), - \(\sigma\) is the standard deviation ($2,200). Substituting the values into the formula: \[ z = \frac{(145600 - 150000)}{2200} = \frac{-4400}{2200} = -2 \] Next, we refer to a standard normal distribution table or use a calculator to find the probability associated with a z-score of -2. A z-score of -2 indicates that $145,600 is 2 standard deviations below the mean. In the context of a normal distribution, approximately 97.5% of values lie