{"id":503,"date":"2022-07-11T19:37:31","date_gmt":"2022-07-11T19:37:31","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/alphamodule\/?post_type=chapter&p=503"},"modified":"2022-07-11T19:37:31","modified_gmt":"2022-07-11T19:37:31","slug":"z-score-and-the-empirical-rule-background-youll-need-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/alphamodule\/chapter\/z-score-and-the-empirical-rule-background-youll-need-2\/","title":{"raw":"Z-Score and the Empirical Rule: Background You'll Need 2","rendered":"Z-Score and the Empirical Rule: Background You’ll Need 2"},"content":{"raw":"Let's try another couple of sets of questions like that to observe what's happening.\r\n
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question 3<\/h3>\r\n[ohm_question hide_question_numbers=1]241192[\/ohm_question]\r\n\r\n[reveal-answer q=\"425088\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"425088\"]The mean is\u00a0[latex]173.40[\/latex] cm and one standard deviation is\u00a0[latex]12.21[\/latex] cm, so you can add or subtract [latex]2\\times12.21[\/latex] to the mean as needed.[\/hidden-answer]\r\n\r\n<\/div>\r\n
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question 4<\/h3>\r\n[ohm_question hide_question_numbers=1]241193[\/ohm_question]\r\n\r\n[reveal-answer q=\"316771\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"316771\"]Follow the example formula, replacing the value you obtained from the previous question, first for the value above the mean, then separately for the value below the mean. That is, make two separate calculations.[\/hidden-answer]\r\n\r\n<\/div>\r\nHopefully you obtained [latex]2[\/latex] and [latex]-2[\/latex] for your answers to Question 4. Have you caught on to what's happening in these question pairs yet?\u00a0 Let's try one more pair. In Question 5, you'll identify values one and a half standard deviations above and below the mean. Can you predict what the answers to Question 6 should be?\r\n
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question 5<\/h3>\r\n[ohm_question hide_question_numbers=1]241194[\/ohm_question]\r\n\r\n[reveal-answer q=\"483705\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"483705\"]The mean is\u00a0[latex]173.40[\/latex] cm and one standard deviation is\u00a0[latex]12.21[\/latex] cm, so you can add or subtract [latex]1.5\\times12.21[\/latex] to the mean as needed.[\/hidden-answer]\r\n\r\n<\/div>\r\n
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question 6<\/h3>\r\n[ohm_question hide_question_numbers=1]241195[\/ohm_question]\r\n\r\n[reveal-answer q=\"125875\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"125875\"]Follow the example formula, replacing the value you obtained from the previous question, first for the value above the mean, then separately for the value below the mean. That is, make two separate calculations.[\/hidden-answer]\r\n\r\n<\/div>\r\nIn Questions 2, 4, and 6, you calculated values that were a given number of standard deviations above and below the mean. You discovered when you divided the difference between the value and the mean by the standard deviation, that the result was a positive number of standard deviations (for values above the mean) or a negative number of standard deviations (for values below the mean). That is, a resulting negative can be thought of as indicating a value that lies\u00a0to the left<\/em> of the mean, and the positive indicates a value that lies\u00a0to the right<\/em> of the mean.","rendered":"

Let’s try another couple of sets of questions like that to observe what’s happening.<\/p>\n

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question 3<\/h3>\n