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Section 6.4

Section 6.4. Finding Values of a Normally Distributed Random Variable. Objectives. Identify z -values for given areas under the standard normal curve. Example 6.17: Finding the z -Value with a Given Area to Its Left. What z -value has an area of 0.7357 to its left? Solution

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Section 6.4

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  1. Section 6.4 Finding Values of a Normally Distributed Random Variable

  2. Objectives Identify z-values for given areas under the standard normal curve.

  3. Example 6.17: Finding the z-Value with a Given Area to Its Left What z-value has an area of 0.7357 to its left? Solution In order to sketch a graph of where the area under the curve lies, we need to shade under more than half of the curve because 0.7357 > 0.5. Therefore, we know that the z-score will be a positive number.

  4. Example 6.17: Finding the z-Value with a Given Area to Its Left (cont.)

  5. Example 6.17: Finding the z-Value with a Given Area to Its Left (cont.) Scan through the interior of Table B from Appendix A for an area of 0.7357. Look to the left and up to find the corresponding z-value. Looking at the row and column titles, you will see that z = 0.63.

  6. Example 6.17: Finding the z-Value with a Given Area to Its Left (cont.)

  7. Example 6.17: Finding the z-Value with a Given Area to Its Left (cont.) Let’s also demonstrate how to find the solution by using a TI-83/84 Plus calculator. Begin by pressing and then . Choose option 3:invNorm(. This option requires you to enter the area to the left of the z-value that you are interested in finding. For problems where the area to the left is given, such as this one, knowing the value to enter in the calculator is easy. This will become more challenging with the other types of problems. The area to the left of the z-score we are interested in is 0.7357.

  8. Example 6.17: Finding the z-Value with a Given Area to Its Left (cont.) Enter invNorm(0.7357), as shown in the screenshot, and press . The answer is z ≈ 0.63.

  9. Example 6.18: Finding the z-Value with a Given Area to Its Left Find the value of z such that the area to the left of z is 0.2000. Solution First, sketching a graph of the curve shows us that since 0.2000 < 0.5, our z-score will be a negative value.

  10. Example 6.18: Finding the z-Value with a Given Area to Its Left (cont.)

  11. Example 6.18: Finding the z-Value with a Given Area to Its Left (cont.) We can then begin to find 0.2000 in the body of the cumulative normal distribution table for negative z-values, Table A from Appendix A. Notice that none of the areas is exactly 0.2000. Our area falls between the areas of 0.2005 and 0.1977 in the table. One method for finding the z-value for an area not in the table is interpolation; however, for our purposes, we will use the area closest to the one we want. The value of 0.2005 is the closest one to 0.2000. Its z-value is -0.84. So, the approximate z-score for an area of 0.2000 is z = -0.84.

  12. Example 6.18: Finding the z-Value with a Given Area to Its Left (cont.)

  13. Example 6.18: Finding the z-Value with a Given Area to Its Left (cont.) Since the area we are given is the area to the left of the unknown z-value, using a TI-83/84 Plus calculator is straightforward. Begin by pressing and then . Choose option 3:invNorm(. The area to the left of the z-score we are interested in is 0.2000.

  14. Example 6.18: Finding the z-Value with a Given Area to Its Left (cont.) Enter invNorm(0.2000), as shown in the screenshot, and press . The answer is z ≈ -0.84.

  15. Example 6.19: Finding the z-Value That Represents a Given Percentile What z-value represents the 90th percentile? Solution The 90th percentile is the z-value for which 90% of the area under the standard normal curve is to the left of z. So, we need to find the value of z that has an area of 0.9000 to its left.

  16. Example 6.19: Finding the z-Value That Represents a Given Percentile (cont.)

  17. Example 6.19: Finding the z-Value That Represents a Given Percentile (cont.) Looking for 0.9000 (or an area extremely close to it) in the interior of the cumulative normal tables, we find 0.8997, which corresponds to a z-value of 1.28. Thus z ≈ 1.28 represents the 90th percentile.

  18. Example 6.19: Finding the z-Value That Represents a Given Percentile (cont.) Using a TI-83/84 Plus calculator, we can find a value of z with a given area to its left. As noted previously, the 90th percentile is the z-value that has an area of 0.9000 to its left. Enter invNorm(0.9000), as shown in the screenshot, and press . The answer is z ≈ 1.28.

  19. Example 6.20: Finding the z-Value with a Given Area to Its Right What z-value has an area of 0.0096 to its right? Solution First, sketch a graph of the curve. Because 0.0096 is such a small area on the right, we know we should sketch the z-score on the positive side of the graph. Since we need the area to the left, not the area to the right, in order to use the tables, subtract the area to the right from 1. This gives us an area to the left of z equal to 0.9904.

  20. Example 6.20: Finding the z-Value with a Given Area to Its Right (cont.)

  21. Example 6.20: Finding the z-Value with a Given Area to Its Right (cont.) Scan through the interior of the table for an area of 0.9904. Look to the left and up to find the corresponding z-value. What we find is that z = 2.34.

  22. Example 6.20: Finding the z-Value with a Given Area to Its Right (cont.) Since the area we are given is area to the right of the unknown z-value, using a TI‑83/84 Plus calculator requires that we first find the area to the left of z. We have already calculated that the area to the left is 0.9904, but we can have the calculator find this area at the same time that it finds the z-score.

  23. Example 6.20: Finding the z-Value with a Given Area to Its Right (cont.) Enter invNorm(1Þ0.0096), as shown in the screenshot, and press . The answer is z ≈ 2.34.

  24. Example 6.21: Finding the z-Value with a Given Area between -z and z Find the value of z such that the area between -z and z is 0.90. Solution If the area between -z and z is 0.90, then the total area in the two tails must be 1 - 0.90 = 0.10. In our sketch, the shaded area will take up most of the area under the curve. Because the curve is symmetric, the area left after subtracting from 1 is divided equally between the two tails. So, half of that area, which is isin the left tail.

  25. Example 6.21: Finding the z-Value with a Given Area between -z and z (cont.)

  26. Example 6.21: Finding the z-Value with a Given Area between -z and z (cont.) Because the cumulative normal tables give area to the left, we can look up this tail area and find the corresponding z-value. Since the probabilities in the normal tables are given to four decimal places, we look for the probability that is closest to the left tail area of 0.0500.

  27. Example 6.21: Finding the z-Value with a Given Area between -z and z (cont.)

  28. Example 6.21: Finding the z-Value with a Given Area between -z and z (cont.) In the interior of the table, we find two areas that are equally close to 0.0500, so use the z-value exactly halfway between the z-scores corresponding to those two areas. The z-value halfway between z = -1.64 and z = -1.65 is -1.645. Thus the positive z-value such that the area between -z and z is 0.90 is z = 1.645. Since the area we are given is the area between the unknown z-values, using a TI-83/84 calculator requires that we first find the area to the left of -z. We have already calculated that the area to the left of -z is 0.05.

  29. Example 6.21: Finding the z-Value with a Given Area between -z and z (cont.) Enter invNorm(0.05), as shown in the screenshot, and press . The answer is -z ≈ -1.644854. Rounded to three decimal places, the positive z-value is 1.645.

  30. Example 6.22: Finding the z-Value with a Given Area in the Tails to the Left of -z and to the Right of z Find the value of z such that the area to the left of -z plus the area to the right of z is 0.1616. Solution If the total area in both tails is 0.1616, then the area in one of the tails must be half of that or

  31. Example 6.22: Finding the z-Value with a Given Area in the Tails to the Left of -z and to the Right of z (cont.)

  32. Example 6.22: Finding the z-Value with a Given Area in the Tails to the Left of -z and to the Right of z (cont.) Scan through the interior of the cumulative normal table for an area of 0.0808. The corresponding z-value is -z = -1.40. Thus, the z-value such that the area to the left of -z plus the area to the right of z is 0.1616 is z = 1.40.

  33. Example 6.22: Finding the z-Value with a Given Area in the Tails to the Left of -z and to the Right of z (cont.) Since the area we are given is the total area in the two tails, we must first find the area to the left of -z. We have already calculated that the area to the left of -z is 0.0808, but we can have the calculator find this area at the same time that it finds the z-score.

  34. Example 6.22: Finding the z-Value with a Given Area in the Tails to the Left of -z and to the Right of z (cont.) Enter invNorm(0.1616/2), as shown in the screenshot, and press . The answer is -z ≈ -1.399711. Rounded to two decimal places, the positive value of z is 1.40.

  35. Example 6.23: Finding the Value of a Normally Distributed Random Variable with a Given Area to Its Right If a normal distribution has a mean of 28.0 and a standard deviation of 2.5, what is the value of the random variable X that has an area to its right equal to 0.6700? Solution Remember that we need to use the steps in reverse. Begin by using the given area to find the value of z. We will use a TI-83/84 Plus calculator to find the solution to this question.

  36. Example 6.23: Finding the Value of a Normally Distributed Random Variable with a Given Area to Its Right (cont.)

  37. Example 6.23: Finding the Value of a Normally Distributed Random Variable with a Given Area to Its Right (cont.) Since we know that the area to the right of x is 0.6700, enter invNorm(1Þ0.6700), as shown in the screenshot, and press . The answer is z ≈ -0.439913.

  38. Example 6.23: Finding the Value of a Normally Distributed Random Variable with a Given Area to Its Right (cont.) The final step is to use this z-score, along with the given values of the mean and standard deviation, to solve for x. Substituting these values into the formula for x we have the following.

  39. Example 6.23: Finding the Value of a Normally Distributed Random Variable with a Given Area to Its Right (cont.) Alternate Calculator Method It is also possible to have the calculator solve this problem directly without doing any intermediate calculations. Instead of finding the z-score first and then solving for x, we can have the calculator find the value of the normally distributed value in one step. The only difference when working with a normal distribution that is not the standard normal distribution is that we must enter the mean and standard deviation in addition to the area to the left of the value in which we are interested.

  40. Example 6.23: Finding the Value of a Normally Distributed Random Variable with a Given Area to Its Right (cont.) The function syntax is invNorm(area,m,s). Enter invNorm(1Þ0.6700,28.0,2.5), as shown in the screenshot, and press . The answer is x ≈ 26.9.

  41. Example 6.24: Finding the Value of a Normally Distributed Random Variable That Represents a Given Percentile The body temperatures of adults are normally distributed with a mean of 98.60 °F and a standard deviation of 0.73 °F. What temperature represents the 90th percentile? Solution In order to determine the temperature that represents the 90th percentile, we first need to find the z-value that represents the 90th percentile. In Example 6.19, we found this value to be z ≈ 1.281552. Once we have the value of z, we can substitute z, s, and m into the formula for x that we have from earlier in this section.

  42. Example 6.24: Finding the Value of a Normally Distributed Random Variable That Represents a Given Percentile (cont.) Thus, a temperature of approximately 99.54 °F represents the 90th percentile.

  43. Example 6.24: Finding the Value of a Normally Distributed Random Variable That Represents a Given Percentile (cont.) Alternate Calculator Method When using a TI-83/84 Plus calculator to solve this problem, it is not necessary to first find the z-value that represents the 90th percentile. The temperature that represents the 90th percentile can be found in one step by entering invNorm(0.90,98.60,0.73), as shown in the screenshot.

  44. Example 6.25: Finding the Value of a Normally Distributed Random Variable That Represents a Given Quartile Let’s assume that the lengths of newborn full-term babies in the United States are normally distributed with a mean length of 20.0 inches and a standard deviation of 1.2 inches. What is the minimum length that a baby could be and still have a length that is amongst the top 25% of baby lengths? Solution If we want to find the minimum length for the top 25% of baby lengths, we can simply find the value of the 75th percentile, or Q3.

  45. Example 6.25: Finding the Value of a Normally Distributed Random Variable That Represents a Given Quartile (cont.)

  46. Example 6.25: Finding the Value of a Normally Distributed Random Variable That Represents a Given Quartile (cont.) A percentage of 75% is the same as an area of 0.75 under the normal distribution. We will begin by finding the z-score that has an area of 0.75 to its left. Enter invNorm(0.75), as shown in the screenshot, and press . The answer is z ≈ 0.674490.

  47. Example 6.25: Finding the Value of a Normally Distributed Random Variable That Represents a Given Quartile (cont.) Now that we know the value of z, we must use it along with the mean and standard deviation to calculate the corresponding value of the random variable X that represents the length of the baby. We can use the formula for x that we saw previously.

  48. Example 6.25: Finding the Value of a Normally Distributed Random Variable That Represents a Given Quartile (cont.) Rounding this to one decimal place (the same as in the standard deviation), we see that the minimum length a baby can be and still be in the top 25% of lengths of full-term newborn babies is approximately 20.8 inches. Alternate Calculator Method When using a TI-83/84 Plus calculator to solve this problem, it is not necessary to first find the z-score that has an area of 0.75 to its left.

  49. Example 6.25: Finding the Value of a Normally Distributed Random Variable That Represents a Given Quartile (cont.) The length that represents the 75th percentile can be found in one step by entering invNorm(0.75, 20.0,1.2), as shown in the screenshot.

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