Lesson 5 Using Technology to find areas_values - Copy

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May 1, 2024

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Chapter 2 Modeling Distributions of Data Lesson 5 Using Technology to find Areas in a Normal Distribution Z-Score Table / Table A / The Standard Normal Probability Table: We have a Normal Distribution with mean 6.84 and SD 1.55. We want to find the area to the left of 6. Technology / TI-84/83: 1. Find Areas from Values: 1. Using the Standard Normal Distribution: What proportion of observations are less than z = -0.54? Step 1:

Step 2: Four inputs: the left bound, the right bound, the mean and the SD Step 3: We chose -1000 as the lower bound because it’s many, many SDs less than the mean. 2. Using the unstandardized Normal Distribution: What proportion of observations are less than x = 6? Step 1: Step 2: Step 3: This answer is slightly different from the one we got using the standard Normal Distribution because we rounded the standardized score to two decimal places. 3. Finding Areas between two values: Using the table:

Using Technology: What proportion of observations are less than x = 6? Step 1: Step 2: Lower: -0.54, upper: 1.39, Step 3: 2. Finding values from areas: Calculate the value corresponding to a given percentile Using the Table:

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We can use table to find the z-score with an area of 0.9 to its left. We need to find the value that is closest to 0.90 in the middle of the table. Using TI-833/84: Let’s use this command to confirm the value for the 90th percentile. https://savi-cdn.macmillantech.com/brightcove/index.html?videoId=5731635175001 1. Using the Standard Normal Distribution: What is the 90th percentile value? Step 1: Step 2: Step 3: Step 4:

2. Using the unstandardized Normal Distribution: What is the 90th percentile value? Step 1: Step 2: Step 3: Step 4: Practice: Use technology to find areas and desired values. d. Find the 20 th percentile of Mrs. Starnes’s speeds. (a)

The proportion of z-scores below −2.56 is 0.0052. About 53 out of every 10,000 times Mrs. Starnes completes an easy Sudoku puzzle, she does so in less than 3 minutes. This is a proportion of 0.0053. (b) The proportion of z-scores above 0.78 is 1 − 0.7823 = 0.2177. About 21.83% of the time, it takes Mrs. Starnes more than 6 minutes to complete an easy puzzle. (c) The proportion of z-scores between 0.78 and 3.00 is 0.9987 − 0.7823 = 0.2164. About 21.7% of easy puzzles take Mrs. Starnes between 6 and 8 minutes to complete. (d) 0.20 area to the left of z → z=−0.84. Solving −0.84=x−5.30.9gives x = 4.544. The 20th percentile of Mrs. Starnes’s Sudoku times for easy problems is about 4.54 minutes.

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