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Section 6.3

Section 6.3. Finding Probability Using a Normal Distribution. Objectives . Calculate probabilities for a normal distribution. . Example 6.12: Finding the Probability that a Normally Distributed Random Variable Will Be Less Than a Given Value .

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Section 6.3

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  1. Section 6.3 Finding Probability Using a Normal Distribution

  2. Objectives Calculate probabilities for a normal distribution.

  3. Example 6.12: Finding the Probability that a Normally Distributed Random Variable Will Be Less Than a Given Value Body temperatures of adults are normally distributed with a mean of 98.60 °F and a standard deviation of 0.73 °F. What is the probability of a healthy adult having a body temperature less than 96.9 °F? Solution We are interested in finding the probability that a person’s body temperature X is less than 96.9 °F, written mathematically as P(X < 9.69). The following graphs illustrate this. First, we need to convert 96.9 to a standard score.

  4. Example 6.12: Finding the Probability that a Normally Distributed Random Variable Will Be Less Than a Given Value (cont.)

  5. Example 6.12: Finding the Probability that a Normally Distributed Random Variable Will Be Less Than a Given Value (cont.) Second, we need to find the area to the left of z −2.33, written mathematically as P(z < −2.33). The picture on the right shows this portion under the standard normal curve. Either looking up -2.33 in the cumulative normal table or using a calculator, we see that the shaded region has an area of approximately 0.0099. Thus, the probability of a healthy adult having a body temperature less than 96.9 °F is approximately 0.0099.

  6. Example 6.12: Finding the Probability that a Normally Distributed Random Variable Will Be Less Than a Given Value (cont.) Alternate Calculator Method Note that when using a TI-83/84 Plus calculator to solve this problem, it is not necessary to first convert 96.9 to a standard score. The probability can be found in one step by entering normalcdf(-1û99, 96.9,98.60,0.73), as shown in the screenshot.

  7. Example 6.13: Finding the Probability that a Normally Distributed Random Variable Will Be Greater Than a Given Value Body temperatures of adults are normally distributed with a mean of 98.60 °F and a standard deviation of 0.73 °F. What is the probability of a healthy adult having a body temperature greater than 97.6 °F? Solution This time, we are interested in finding the probability that a person’s body temperature X is greater than 97.6 °F, written mathematically as P(X > 9.76). The following graph illustrates this. First, convert 97.6 to a standard score.

  8. Example 6.13: Finding the Probability that a Normally Distributed Random Variable Will Be Greater Than a Given Value (cont.)

  9. Example 6.13: Finding the Probability that a Normally Distributed Random Variable Will Be Greater Than a Given Value (cont.) Second, we need to find the area under the standard normal curve to the right of z −1.37, which we denote mathematically as P(z > −1.37). The right graph in the previous figure shows this region under the curve. Using a calculator or the cumulative normal tables, we find that this area is approximately 0.9147. Hence, the probability of a healthy adult having a body temperature greater than 97.6°F is approximately 0.9147.

  10. Example 6.13: Finding the Probability that a Normally Distributed Random Variable Will Be Greater Than a Given Value (cont.) Alternate Calculator Method When using a TI-83/84 Plus calculator to solve this problem, it is not necessary to first convert 97.6 to a standard score. The probability can be found in one step by entering normalcdf (97.6,1û99, 98.60,0.73), as shown in the screenshot in next slide. Note that by solving for the probability in one step on the calculator, we eliminate the rounding error that is introduced when we first round the standard score to z −1.37.

  11. Example 6.13: Finding the Probability that a Normally Distributed Random Variable Will Be Greater Than a Given Value (cont.) Thus, the calculator gives the more accurate value of approximately 0.9146 for the probability when this method is used.

  12. Example 6.14: Finding the Probability that a Normally Distributed Random Variable Will Be between Two Given Values Body temperatures of adults are normally distributed with a mean of 98.60 °F and a standard deviation of 0.73 °F. What is the probability of a healthy adult having a body temperature between 98 °F and 99 °F? Solution We are now interested in finding the probability that a person’s body temperature X is between 98 °F and 99 °F, written mathematically as P(98 < X < 99). The left graph in the following figure illustrates this region under the normal curve. First, convert both 98 and 99 to standard scores.

  13. Example 6.14: Finding the Probability that a Normally Distributed Random Variable Will Be between Two Given Values (cont.)

  14. Example 6.14: Finding the Probability that a Normally Distributed Random Variable Will Be between Two Given Values (cont.) We now need to calculate the area between z1 -0.82 and z2 0.55, written mathematically as P( −0.82 < z < 0.55). The picture on the right illustrates this area under the standard normal curve. Finally, remember that to find area between two z-scores you simply enter the two z-scores in a TI-83/84 Plus calculator as normalcdf(-0.82,0.55), which yields a value of approximately 0.5027, as shown in the screenshot on next slide.

  15. Example 6.14: Finding the Probability that a Normally Distributed Random Variable Will Be between Two Given Values (cont.) Thus, the probability of a healthy adult having a body temperature between 98 °F and 99 °F is approximately 0.5027.

  16. Example 6.14: Finding the Probability that a Normally Distributed Random Variable Will Be between Two Given Values (cont.) Alternate Calculator Method When using a TI-83/84 Plus calculator to solve this problem, it is not necessary to first convert 98 and 99 to standard scores. The probability can be found in one step by entering normalcdf(98,99,98.60, 0.73), as shown in the screenshot on next slide. Note that by solving for the probability in one step on the calculator, we eliminate the rounding error that is introduced when we first round the standard scores to z1 −0.82 and z2 0.55.

  17. Example 6.14: Finding the Probability that a Normally Distributed Random Variable Will Be between Two Given Values (cont.) Thus, the calculator gives the more accurate value of approximately 0.5026 for the probability when this method is used.

  18. Example 6.15: Finding the Probability that a Normally Distributed Random Variable Will Be in the Tails Defined by Two Given Values Body temperatures of adults are normally distributed with a mean of 98.60 °F and a standard deviation of 0.73 °F. What is the probability of a healthy adult having a body temperature less than 99.6 °F or greater than 100.6 °F?

  19. Example 6.15: Finding the Probability that a Normally Distributed Random Variable Will Be in the Tails Defined by Two Given Values (cont.) Solution We are interested in finding the probability that a person’s body temperature X is either less than 99.6 °F or greater than 100.6 °F. We can denote this mathematically as P(X < 99.6 or X > 100.6). The picture on the left below on next slide shows the area under the normal curve in which we are interested. First, convert each temperature to a standard score.

  20. Example 6.15: Finding the Probability that a Normally Distributed Random Variable Will Be in the Tails Defined by Two Given Values (cont.)

  21. Example 6.15: Finding the Probability that a Normally Distributed Random Variable Will Be in the Tails Defined by Two Given Values (cont.) We now need to calculate the sum of the areas to the left of z1 1.37 and to the right of z2 2.74. This may be denoted mathematically as P(z < 1.37 or z > 2.74). The right graph in the previous figure illustrates these areas under the standard normal curve.

  22. Example 6.15: Finding the Probability that a Normally Distributed Random Variable Will Be in the Tails Defined by Two Given Values (cont.) Using a TI-83/84 Plus calculator, we find each area required as shown below and in the screenshot. normalcdf(-1û99,1.37)≈ 0.914656 normalcdf(2.74,1û99)≈ 0.003072

  23. Example 6.15: Finding the Probability that a Normally Distributed Random Variable Will Be in the Tails Defined by Two Given Values (cont.) Adding these together, we have 0.914656 + 0.003072 ≈ 0.9177. Thus, the probability of a healthy adult having a body temperature either less than 99.6 °F or greater than 100.6 °F is approximately 0.9177.

  24. Example 6.15: Finding the Probability that a Normally Distributed Random Variable Will Be in the Tails Defined by Two Given Values (cont.) Alternate Calculator Method Note that when using a TI-83/84 Plus calculator to solve this problem, it is not necessary to first convert 99.6 and 100.6 to standard scores. The probability can be found in one step by entering 1Þnormalcdf(99.6,100.6,98.60,0.73), as shown in the screenshot.

  25. Example 6.16: Finding the Probability that a Normally Distributed Random Variable Will Differ from the Mean by More Than a Given Value Body temperatures of adults are normally distributed with a mean of 98.60 °F and a standard deviation of 0.73 °F. What is the probability of a healthy adult having a body temperature that differs from the population mean by more than 1 °F? Solution Because we want the temperatures to differ from the population mean by more than 1 °F, we need to add 1 °F to and subtract 1 °F from the mean of 98.6 °F to get the temperatures in which we are interested.

  26. Example 6.16: Finding the Probability that a Normally Distributed Random Variable Will Differ from the Mean by More Than a Given Value (cont.) Doing this tells us we are interested in finding the probability that a person’s body temperature X is less than 97.6 °F or greater than 99.6 °F. We can denote this mathematically as P(X < 97.6 or X> 99.6). The left graph in the next slide shows the area under the normal curve in which we are interested. Just as we did in the previous example, we first convert each temperature to a standard score.

  27. Example 6.16: Finding the Probability that a Normally Distributed Random Variable Will Differ from the Mean by More Than a Given Value (cont.)

  28. Example 6.16: Finding the Probability that a Normally Distributed Random Variable Will Differ from the Mean by More Than a Given Value (cont.) We now need to calculate the sum of the areas to the left of z1 −1.37 and to the right of z2 1.37. This may be denoted mathematically as P(z < −1.37 or z > 1.37). The graph on the right in the previous figure illustrates these areas under the standard normal curve. Notice that these z-scores are equal distances from the mean. Because of the symmetry of the curve, the shaded areas must then be equal.

  29. Example 6.16: Finding the Probability that a Normally Distributed Random Variable Will Differ from the Mean by More Than a Given Value (cont.) The easiest way to calculate the total area is to find the area to the left of z1 −1.37 and simply double it. We then have (0.0853)(2) = 0.1706. Thus, the probability of a healthy adult having a body temperature either less than 97.6 °F or greater than 99.6 °F is approximately 0.1706.

  30. Example 6.16: Finding the Probability that a Normally Distributed Random Variable Will Differ from the Mean by More Than a Given Value (cont.) Alternate Calculator Method When using a TI-83/84 Plus calculator to solve this problem, it is not necessary to first convert 97.6 and 99.6 to standard scores. The probability can be found in one step by entering 1Þnormalcdf(97.6,99.6, 98.60,0.73), as shown in the screenshot on next slide. Note that by solving for the probability in one step on the calculator, we eliminate the rounding error that is introduced when we first round the standard scores to z1 −1.37 and z2 1.37.

  31. Example 6.16: Finding the Probability that a Normally Distributed Random Variable Will Differ from the Mean by More Than a Given Value (cont.) Thus, when this one-step method is used, the calculator gives the more accurate value of approximately 0.1707 for the probability.

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