{"id":54,"date":"2021-01-12T22:19:32","date_gmt":"2021-01-12T22:19:32","guid":{"rendered":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/standard-normal-distribution\/"},"modified":"2023-06-26T23:55:37","modified_gmt":"2023-06-26T23:55:37","slug":"standard-normal-distribution","status":"publish","type":"chapter","link":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/standard-normal-distribution\/","title":{"rendered":"Standard Normal Distribution"},"content":{"raw":"\r\n[latexpage]\r\n<\/span>\r\n
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The standard normal distribution<\/span> is a normal distribution of standardized values called<\/strong> z<\/em>-scores<\/span>. A z<\/em>-score is measured in units of the standard deviation.<\/strong> For example, if the mean of a normal distribution is five and the standard deviation is two, the value 11 is three standard deviations above (or to the right of) the mean. The calculation is as follows:<\/p>\r\n

x<\/em> = \u03bc<\/em> + (z<\/em>)(\u03c3<\/em>) = 5 + (3)(2) = 11<\/p>\r\n

The z<\/em>-score is three.<\/p>\r\n

The mean for the standard normal distribution is zero, and the standard deviation is one. The transformation $z=\\frac{x-\\mu}{\\sigma}$ produces the distribution Z<\/em> ~ N<\/em>(0, 1). The value x<\/em> in the given equation comes from a normal distribution with mean \u03bc<\/em> and standard deviation \u03c3<\/em>.<\/p>\r\n\r\n

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Z<\/em>-Scores<\/h3>\r\n

If X<\/em> is a normally distributed random variable and X<\/em> ~ N(\u03bc, \u03c3)<\/em>, then the z<\/em>-score is:<\/p>\r\n$$z=\\frac{x-\\mu}{\\sigma}$$\r\n\r\nThe z<\/em>-score tells you how many standard deviations the value x<\/em> is above (to the right of) or below (to the left of) the mean, \u03bc<\/em>.<\/strong> Values of x<\/em> that are larger than the mean have positive z<\/em>-scores, and values of x<\/em> that are smaller than the mean have negative z<\/em>-scores. If x<\/em> equals the mean, then x<\/em> has a z<\/em>-score of zero.\r\n

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Example <\/span>5.7<\/span><\/h3>\r\n<\/header>
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Suppose X<\/em> ~ N(5, 6)<\/em>. This says that X<\/em> is a normally distributed random variable with mean \u03bc<\/em> = 5 and standard deviation \u03c3<\/em> = 6. Suppose x<\/em> = 17. Then:<\/p>\r\n$$z=\\frac{x-\\mu}{\\sigma} = \\frac{17-5}{6} = 2$$\r\n\r\nThis means that x<\/em> = 17 is two standard deviations<\/strong> (2\u03c3<\/em>) above or to the right of the mean \u03bc<\/em> = 5.\r\n

Notice that: 5 + (2)(6) = 17 (The pattern is \u03bc<\/em> + z\u03c3<\/em> = x<\/em>)<\/p>\r\n

Now suppose x<\/em> = 1. Then: $z=\\frac{x-\\mu}{\\sigma} = \\frac{1-5}{6} = -0.67$ (rounded to two decimal places)<\/p>\r\n

This means that x<\/em> = 1 is 0.67 standard deviations (\u20130.67\u03c3<\/em>) below or to the left of the mean \u03bc<\/em> = 5. Notice that:<\/strong> 5 + (\u20130.67)(6) is approximately equal to one (This has the pattern \u03bc<\/em> + (\u20130.67)\u03c3 = 1)<\/p>\r\n

Summarizing, when z<\/em> is positive, x<\/em> is above or to the right of \u03bc<\/em> and when z<\/em> is negative, x<\/em> is to the left of or below \u03bc<\/em>. Or, when z<\/em> is positive, x<\/em> is greater than \u03bc<\/em>, and when z<\/em> is negative x<\/em> is less than \u03bc<\/em>.<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n

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Try It <\/span>5.7<\/span><\/h3>\r\n<\/header>
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What is the z<\/em>-score of x<\/em>, when x<\/em> = 1 and X<\/em> ~ N<\/em>(12,3)?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

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Example <\/span>5.8<\/span><\/h3>\r\n<\/header>
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Some doctors believe that a person can lose five pounds, on the average, in a month by reducing his or her fat intake and by exercising consistently. Suppose weight loss has a normal distribution. Let X<\/em> = the amount of weight lost (in pounds) by a person in a month. Use a standard deviation of two pounds. X<\/em> ~ N<\/em>(5, 2). Fill in the blanks.<\/p>\r\n \r\n

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  1. a. Suppose a person lost<\/strong> ten pounds in a month. The z<\/em>-score when x<\/em> = 10 pounds is z<\/em> = 2.5 (verify). This z<\/em>-score tells you that x<\/em> = 10 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).<\/li>\r\n \t
  2. Suppose a person gained<\/strong> three pounds (a negative weight loss). Then z<\/em> = __________. This z<\/em>-score tells you that x<\/em> = \u20133 is ________ standard deviations to the __________ (right or left) of the mean.<\/li>\r\n \t
  3. Suppose the random variables X<\/em> and Y<\/em> have the following normal distributions: X<\/em> ~ N<\/em>(5, 6) and Y<\/em> ~ N<\/em>(2, 1). If x<\/em> = 17, then z<\/em> = 2. (This was previously shown.) If y<\/em> = 4, what is z<\/em>?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n

    Solution <\/span>5.8<\/span><\/h4>\r\n
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    1. This z<\/em>-score tells you that x<\/em> = 10 is 2.5<\/strong> standard deviations to the right<\/strong> of the mean five<\/strong>.<\/li>\r\n \t
    2. z<\/em> = \u20134<\/strong>. This z<\/em>-score tells you that x<\/em> = \u20133 is four<\/strong> standard deviations to the left<\/strong> of the mean.<\/li>\r\n \t
    3. $z=\\frac{y-\\mu}{\\sigma} - \\frac{4-2}{1} = 2$ where $\\mu =2$ and $\\sigma = 1$.\r\n

      The z<\/em>-score for y<\/em> = 4 is z<\/em> = 2. This means that four is z<\/em> = 2 standard deviations to the right of the mean. Therefore, x<\/em> = 17 and y<\/em> = 4 are both two (of their own<\/strong>) standard deviations to the right of their<\/strong> respective means.<\/p>\r\n

      The z<\/em>-score allows us to compare data that are scaled differently.<\/strong> To understand the concept, suppose X<\/em> ~ N<\/em>(5, 6) represents weight gains for one group of people who are trying to gain weight in a six week period and Y<\/em> ~ N<\/em>(2, 1) measures the same weight gain for a second group of people. A negative weight gain would be a weight loss. Since x<\/em> = 17 and y<\/em> = 4 are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means<\/strong>.<\/p>\r\n<\/li>\r\n<\/ol>\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n

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      Try It <\/span>5.8<\/span><\/h3>\r\n<\/header>
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      Fill in the blanks.<\/p>\r\n

      Jerome averages 16 points a game with a standard deviation of four points. X<\/em> ~ N<\/em>(16,4). Suppose Jerome scores ten points in a game. The z<\/em>\u2013score when x<\/em> = 10 is \u20131.5. This score tells you that x<\/em> = 10 is _____ standard deviations to the ______(right or left) of the mean______(What is the mean?).<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

      The Empirical Rule<\/span><\/h3>\r\nIf X<\/em> is a random variable and has a normal distribution with mean \u00b5<\/em> and standard deviation \u03c3<\/em>, then the Empirical Rule<\/span> states the following:\r\n