{"id":256,"date":"2022-05-18T16:38:01","date_gmt":"2022-05-18T16:38:01","guid":{"rendered":"https:\/\/pressbooks.ccconline.org\/accintrostats\/chapter\/the-standard-normal-distribution\/"},"modified":"2022-11-09T16:14:25","modified_gmt":"2022-11-09T16:14:25","slug":"the-standard-normal-distribution","status":"publish","type":"chapter","link":"https:\/\/pressbooks.ccconline.org\/accintrostats\/chapter\/the-standard-normal-distribution\/","title":{"raw":"Chapter 7.2: The Standard Normal Distribution","rendered":"Chapter 7.2: The Standard Normal Distribution"},"content":{"raw":" \r\n

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<\/em> = \\(\\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\r\n

\\(z=\\frac{x\\text{\u00a0}\u2013\\text{\u00a0}\\mu }{\\sigma }\\)<\/div>\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
\r\n\r\nSuppose 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:\r\n
\\(z=\\frac{x\u2013\\mu }{\\sigma }=\\frac{17\u20135}{6}=2\\)<\/div>\r\n

This means that x<\/em> = 17 is two standard deviations<\/strong> (2\u03c3<\/em>) above or to the right of the mean \u03bc<\/em> = 5.<\/p>\r\nNotice that: 5 + (2)(6) = 17 (The pattern is \u03bc<\/em> + z\u03c3<\/em> = x<\/em>)\r\n

Now suppose x<\/em> = 1. Then: z<\/em> = \\(\\frac{x\u2013\\mu }{\\sigma }\\) = \\(\\frac{1\u20135}{6}\\) = \u20130.67 (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\nSummarizing, 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>.\r\n\r\n<\/div>\r\n

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Try It<\/div>\r\n
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\r\n\r\nWhat is the z<\/em>-score of x<\/em>, when x<\/em> = 1 and X<\/em> ~ N<\/em>(12,3)?\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n\r\nSome 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.\r\n\r\n \r\n
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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?).<\/p>\r\n\r\n<\/div>\r\n

\r\n\r\na. 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>.\r\n\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n
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\r\n\r\nb. 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.\r\n\r\n<\/div>\r\n
\r\n\r\nb. 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.\r\n\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n
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\r\n\r\nc. 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>?\r\n\r\n<\/div>\r\n
\r\n\r\nc. z<\/em> = \\(\\frac{y-\\mu }{\\sigma }\\) = \\(\\frac{4-2}{1}\\) = 2 where \u00b5<\/em> = 2 and \u03c3<\/em> = 1.\r\n\r\n<\/div>\r\n<\/div>\r\nThe 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.\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\r\n<\/div>\r\n

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Try It<\/div>\r\n
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Fill in the blanks.<\/p>\r\nJerome 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?).\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nThe Empirical Rule<\/span>If 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