Using the Normal Distribution Quiz

PDF page 367: 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).

PDF page 367: 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 were 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.

PDF page 367: 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 were 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.

PDF page 367: 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 were 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.

PDF page 368: If the area to the left is 0.0228, then the area to the right is 1 – 0.0228 = 0.9772.

PDF page 368: To calculate the probability, use the probability tables provided in Appendix H Tables without the use of technology. The tables include instructions for how to use them. The probability is represented by the area under the normal curve. To find the probability, calculate the z-score and look up the z-score in the z-table under the z-column. Most z-tables show the area under the normal curve to the left of z. Others show the mean to z area. The method used will be indicated on the table. We will discuss the z-table that represents the area under the normal curve to the left of z. Once you have located the z-score, locate the corresponding area. This will be the area under the normal curve, to the left of the z-score. This area can be used to find the area to the right of the z-score, or by subtracting from 1 or the total area under the normal curve. These areas can also be used to determine the area between two z-scores.

PDF page 368: To calculate the probability, use the probability tables provided in Appendix H Tables without the use of technology. The tables include instructions for how to use them. The probability is represented by the area under the normal curve. To find the probability, calculate the z-score and look up the z-score in the z-table under the z-column. Most z-tables show the area under the normal curve to the left of z. Others show the mean to z area. The method used will be indicated on the table. We will discuss the z-table that represents the area under the normal curve to the left of z. Once you have located the z-score, locate the corresponding area. This will be the area under the normal curve, to the left of the z-score. This area can be used to find the area to the right of the z-score, or by subtracting from 1 or the total area under the normal curve. These areas can also be used to determine the area between two z-scores.

PDF page 368: a. Let X = a score on the final exam. X ~ N(63, 5), where μ = 63 and σ = 5.

PDF page 368: The z-table shows that the area to the left of z is 0.6554. Subtracting this area from 1 gives 0.3446. Then, find P(x > 65).

PDF page 369: The probability that any student selected at random scores more than 65 is 0.3446.

PDF page 369: The syntax for the instructions is as follows: normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. 99 You get 1E99 (= 10 ) by pressing 1, the EE key—a 2nd key—and then 99. Or, you can enter 10^99 instead. The 99 99 number 10 is way out in the right tail of the normal curve. We are calculating the area between 65 and 10 . In 99 99 some instances, the lower number of the area might be –1E99 (= –10 ). The number –10 is way out in the left tail of the normal curve. We chose the exponent of 99 because this produces such a large number that we can reasonably expect all of the values under the curve to fall below it. This is an arbitrary value and one that works well, for our purpose.

PDF page 369: Then find P(x < 85), and shade the graph. Using a computer or calculator, find P(x < 85) = 1. normalcdf(0,85,63,5) = 1 (rounds to one) The probability that one student scores less than 85 is approximately one, or 100 percent.

PDF page 370: Thus, our score, k, is 69.4.

PDF page 370: th The 90 percentile is 69.4. This means that 90 percent of the test scores fall at or below 69.4 and 10 percent fall at or above. To get this answer on the calculator, follow this next step: