The Correct Answer and Explanation is:
Based on the statistical tools shown, the correct answer to a typical problem on this page would be a probability found using a z-score. For an illustrative example where a test has a mean of 80 and a standard deviation of 5, the probability of scoring less than 87 is 0.9192.
Explanation
Unfortunately, the specific question on the provided document is unreadable due to the image quality. However, the page contains all the necessary components for solving a statistics problem involving a normal distribution. These components are the z-score formula and a standard normal distribution table, also known as a z-table. I will explain the process using a common type of problem this page is designed to solve.
The purpose of a z-score is to standardize a data point. The formula z = (x – μ) / σ transforms a value (x) from any normal distribution into a standard score. This score indicates how many standard deviations the value is from its population mean (μ). Once a z-score is calculated, it can be used with the standard normal distribution, which has a mean of 0 and a standard deviation of 1.
The large table connects a z-score to its cumulative probability. This probability is the area under the standard normal curve to the left of that z-score, representing the proportion of all data that falls below that value.
Let’s use a hypothetical example. Suppose test scores are normally distributed with a mean (μ) of 80 and a standard deviation (σ) of 5. We want to find the probability a student scores less than 87.
First, we calculate the z-score for x = 87:
z = (87 – 80) / 5 = 7 / 5 = 1.40
Next, we look up this z-score in the table. We find the row for 1.4 and the column for .00. The number where they intersect is 0.9192.
This means there is a 0.9192 probability, or a 91.92% chance, that a randomly selected student will score less than 87.
