Answer; approximately 0.9
Step-by-step explanation: The shaded area in the following graph indicates the area to the left of x. This area could represent the percentage of students scoring less than a particular grade on a final exam. This area is represented by the probability P(X x). Normal tables, computers, and calculators are used to provide or calculate the probability P(X x).
The area to the right is then P(X > x) = 1 – P(X x). Remember, P(X x) = Area to the left of the vertical line through x. P(X x) = 1 – P(X x) = Area to the right of the vertical line through x. P(X x) is the same as P(X ≤ x) and P(X > x) is the same as P(X ≥ x) for continuous distributions.
Suppose the graph above was to represent the percentage of students scoring less than 75 on a final exam, with this probability equal to 0.39. This would also indicate that the percentage of students scoring higher than 75 was equal to 1 minus 0.39 or 0.61.
Find the 90th percentile. For each problem or part of a problem, draw a new graph. Draw the x-axis. Shade the area that corresponds to the 90th percentile. This time, we are looking for a score that corresponds to a given area under the curve.
Let k = the 90th percentile. The variable k is located on the x-axis. P(x k) is the area to the left of k. The 90th percentile k separates the exam scores into those that are the same or lower than k and those that are the same or higher. Ninety percent of the test scores are the same or lower than k, and 10 percent are the same or higher. The variable k is often called a critical value.
We know the mean, standard deviation, and area under the normal curve. We need to find the z-score that corresponds to the area of 0.9 and then substitute it with the mean and standard deviation into our z-score formula. The z-table shows a z-score of approximately 1.28, for an area under the normal curve to the left of z (larger portion) of approximately 0.9. Thus, we can write the following:
Find the 90th percentile. For each problem or part of a problem, draw a new graph. Draw the x-axis. Shade the area that corresponds to the 90th percentile. This time, we are looking for a score that corresponds to a given area under the curve.
Let k = the 90th percentile. The variable k is located on the x-axis. P(x k) is the area to the left of k. The 90th percentile k separates the exam scores into those that are the same or lower than k and those that are the same or higher. Ninety percent of the test scores are the same or lower than k, and 10 percent are the same or higher. The variable k is often called a critical value.
We know the mean, standard deviation, and area under the normal curve. We need to find the z-score that corresponds to the area of 0.9 and then substitute it with the mean and standard deviation into our z-score formula. The z-table shows a z-score of approximately 1.28, for an area under the normal curve to the left of z (larger portion) of approximately 0.9.