{"id":519,"date":"2024-10-18T02:21:09","date_gmt":"2024-10-18T02:21:09","guid":{"rendered":"https:\/\/pressbooks.ccconline.org\/mat1260\/?post_type=chapter&p=519"},"modified":"2024-12-16T17:15:14","modified_gmt":"2024-12-16T17:15:14","slug":"6-2-z-scores","status":"publish","type":"chapter","link":"https:\/\/pressbooks.ccconline.org\/mat1260\/chapter\/6-2-z-scores\/","title":{"raw":"6.2: Z-Scores","rendered":"6.2: Z-Scores"},"content":{"raw":"
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Finding Probabilities for a Normal Random Variable<\/span><\/h2>\r\n

As we saw, the Standard Deviation Rule is very limited in helping us answer probability questions, and basically limited to questions involving values that fall exactly 1, 2, and 3 standard deviations away from the mean. How do we answer probability questions in general? The key is the position of the value relative to the mean, measured in standard deviations.<\/p>\r\n

We can approach the answering of probability questions two possible ways: a table and technology. In the next sections, you will learn how to use the \u201cstandard normal table,\u201d and then how the same calculations can be done with technology.<\/p>\r\n\r\n

Standardizing Values<\/h2>\r\n<\/div>\r\n<\/div>\r\n
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\r\n

The first step to assessing a probability associated with a normal value is to determine the\u00a0relative<\/em>\u00a0value with respect to all the other values taken by that normal variable. This is accomplished by determining how many standard deviations below or above the mean that value is.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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Example<\/h3>\r\n<\/header>\r\n
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Foot Length<\/h4>\r\n
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How many standard deviations below or above the mean male foot length is 13 inches? Since the mean is 11 inches, 13 inches is 2 inches above the mean. Since a standard deviation is 1.5 inches, this would be 2 \/ 1.5 = 1.33 standard deviations above the mean. Combining these two steps, we could write:<\/p>\r\n

(13 in. \u2013 11 in.) \/ (1.5 inches per standard deviation) = (13 \u2013 11) \/ 1.5 standard deviations = +1.33 standard deviations.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nIn the language of statistics, we have just found the\u00a0z-score\u00a0<\/em>for a male foot length of 13 inches to be z = +1.33. Or, to put it another way, we have\u00a0standardized<\/em>\u00a0the value of 13. In general, the standardized value z tells how many standard deviations below or above the mean the original value is, and is calculated as follows:\r\n\r\n<\/div>\r\n<\/div>\r\n

z-score = (value \u2013 mean)\/standard deviation<\/em><\/p>\r\n

The convention is to denote a value of our normal random variable X with the letter \u201cx.\u201d Since the mean is written\u00a0\u03bc<\/span><\/span><\/span><\/span><\/span>\u00a0and the standard deviation\u00a0\u03c3<\/span><\/span><\/span><\/span><\/span>, we may write the standardized value as<\/p>\r\n

[latex]\\mathcal{z}=\\frac{\\mathcal{x}-\\mu}{\\sigma}[\/latex]<\/p>\r\n

Notice that since\u00a0\u03c3<\/span><\/span><\/span><\/span><\/span>\u00a0is always positive, for values of x above the mean (\u03bc<\/span><\/span><\/span><\/span><\/span>), z will be positive; for values of x below\u00a0\u03bc<\/span><\/span><\/span><\/span><\/span>, z will be negative.<\/p>\r\n\"The<\/span><\/span>\"The<\/span><\/span>\r\n

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Example<\/h3>\r\n<\/header>\r\n
\r\n

Standardizing Foot Measurements<\/h4>\r\n
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Let\u2019s go back to our foot length example, and answer some more questions.<\/p>\r\n

(a)<\/em>\u00a0What is the standardized value for a male foot length of 8.5 inches? How does this foot length relate to the mean?<\/p>\r\n

z = (8.5 \u2013 11) \/ 1.5 = -1.67. This foot length is 1.67 standard deviations\u00a0below<\/em>\u00a0the mean.<\/p>\r\n

(b)<\/em>\u00a0A man\u2019s standardized foot length is +2.5. What is his actual foot length in inches? If z = +2.5, then his foot length is 2.5 standard deviations above the mean. Since the mean is 11, and each standard deviation is 1.5, we get that the man\u2019s foot length is: 11 + 2.5(1.5) = 14.75 inches.<\/p>\r\n

z-scores also allow us to compare values of different normal random variables. Here is an example:<\/p>\r\n

(c)<\/em>\u00a0In general, women\u2019s foot length is shorter than men\u2019s. Assume that women\u2019s foot length follows a normal distribution with a mean of 9.5 inches and standard deviation of 1.2. Ross\u2019 foot length is 13.25 inches, and Candace\u2019s foot length is only 11.6 inches. Which of the two has a longer foot relative to his or her gender group?<\/p>\r\n

To answer this question, let\u2019s find the z-score of each of these two normal values, bearing in mind that each of the values comes from a different normal distribution.<\/p>\r\n

Ross: z-score = (13.25 \u2013 11) \/ 1.5 = 1.5 (Ross\u2019 foot length is 1.5 standard deviations above the mean foot length for men).<\/p>\r\n

Candace: z-score = (11.6 \u2013 9.5) \/ 1.2 = 1.75 (Candace\u2019s foot length is 1.75 standard deviations above the mean foot length for women).<\/p>\r\n

Note that even though Ross\u2019 foot is longer than Candace\u2019s, Candace\u2019s foot is longer relative to their respective genders.<\/p>\r\nTo Sum Up \u2026 <\/span><\/strong>Part (c)<\/em>\u00a0illustrates how z-scores become crucial when you want to\u00a0compare distributions<\/em>.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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Learn by Doing<\/h3>\r\n<\/header>\r\n
\r\n\r\nThe hourly salary rate for accountants at the \u201cWe are the Best Accounting Firm\u201d follow a normal distribution, with a mean of $27 and a standard deviation of $4.\r\n\r\n[h5p id=\"132\"]\r\n\r\n<\/div>\r\n<\/div>\r\n
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Did I get this?<\/h3>\r\n<\/header>\r\n
\r\n\r\nScores on the final exam in Professor Meyer's statistics class follow a normal distribution, with a mean of 82 and a standard deviation of 5.\r\n\r\n[h5p id=\"133\"]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"
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\n

Finding Probabilities for a Normal Random Variable<\/span><\/h2>\n

As we saw, the Standard Deviation Rule is very limited in helping us answer probability questions, and basically limited to questions involving values that fall exactly 1, 2, and 3 standard deviations away from the mean. How do we answer probability questions in general? The key is the position of the value relative to the mean, measured in standard deviations.<\/p>\n

We can approach the answering of probability questions two possible ways: a table and technology. In the next sections, you will learn how to use the \u201cstandard normal table,\u201d and then how the same calculations can be done with technology.<\/p>\n

Standardizing Values<\/h2>\n<\/div>\n<\/div>\n
\n
\n

The first step to assessing a probability associated with a normal value is to determine the\u00a0relative<\/em>\u00a0value with respect to all the other values taken by that normal variable. This is accomplished by determining how many standard deviations below or above the mean that value is.<\/p>\n<\/div>\n<\/div>\n

\n
\n
\n
\n

Example<\/h3>\n<\/header>\n
\n

Foot Length<\/h4>\n
\n

How many standard deviations below or above the mean male foot length is 13 inches? Since the mean is 11 inches, 13 inches is 2 inches above the mean. Since a standard deviation is 1.5 inches, this would be 2 \/ 1.5 = 1.33 standard deviations above the mean. Combining these two steps, we could write:<\/p>\n

(13 in. \u2013 11 in.) \/ (1.5 inches per standard deviation) = (13 \u2013 11) \/ 1.5 standard deviations = +1.33 standard deviations.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

In the language of statistics, we have just found the\u00a0z-score\u00a0<\/em>for a male foot length of 13 inches to be z = +1.33. Or, to put it another way, we have\u00a0standardized<\/em>\u00a0the value of 13. In general, the standardized value z tells how many standard deviations below or above the mean the original value is, and is calculated as follows:<\/p>\n<\/div>\n<\/div>\n

z-score = (value \u2013 mean)\/standard deviation<\/em><\/p>\n

The convention is to denote a value of our normal random variable X with the letter \u201cx.\u201d Since the mean is written\u00a0\u03bc<\/span><\/span><\/span><\/span><\/span>\u00a0and the standard deviation\u00a0\u03c3<\/span><\/span><\/span><\/span><\/span>, we may write the standardized value as<\/p>\n

[latex]\\mathcal{z}=\\frac{\\mathcal{x}-\\mu}{\\sigma}[\/latex]<\/p>\n

Notice that since\u00a0\u03c3<\/span><\/span><\/span><\/span><\/span>\u00a0is always positive, for values of x above the mean (\u03bc<\/span><\/span><\/span><\/span><\/span>), z will be positive; for values of x below\u00a0\u03bc<\/span><\/span><\/span><\/span><\/span>, z will be negative.<\/p>\n

\"The<\/span><\/span>\"The<\/span><\/span><\/p>\n

\n
\n
\n
\n

Example<\/h3>\n<\/header>\n
\n

Standardizing Foot Measurements<\/h4>\n
\n

Let\u2019s go back to our foot length example, and answer some more questions.<\/p>\n

(a)<\/em>\u00a0What is the standardized value for a male foot length of 8.5 inches? How does this foot length relate to the mean?<\/p>\n

z = (8.5 \u2013 11) \/ 1.5 = -1.67. This foot length is 1.67 standard deviations\u00a0below<\/em>\u00a0the mean.<\/p>\n

(b)<\/em>\u00a0A man\u2019s standardized foot length is +2.5. What is his actual foot length in inches? If z = +2.5, then his foot length is 2.5 standard deviations above the mean. Since the mean is 11, and each standard deviation is 1.5, we get that the man\u2019s foot length is: 11 + 2.5(1.5) = 14.75 inches.<\/p>\n

z-scores also allow us to compare values of different normal random variables. Here is an example:<\/p>\n

(c)<\/em>\u00a0In general, women\u2019s foot length is shorter than men\u2019s. Assume that women\u2019s foot length follows a normal distribution with a mean of 9.5 inches and standard deviation of 1.2. Ross\u2019 foot length is 13.25 inches, and Candace\u2019s foot length is only 11.6 inches. Which of the two has a longer foot relative to his or her gender group?<\/p>\n

To answer this question, let\u2019s find the z-score of each of these two normal values, bearing in mind that each of the values comes from a different normal distribution.<\/p>\n

Ross: z-score = (13.25 \u2013 11) \/ 1.5 = 1.5 (Ross\u2019 foot length is 1.5 standard deviations above the mean foot length for men).<\/p>\n

Candace: z-score = (11.6 \u2013 9.5) \/ 1.2 = 1.75 (Candace\u2019s foot length is 1.75 standard deviations above the mean foot length for women).<\/p>\n

Note that even though Ross\u2019 foot is longer than Candace\u2019s, Candace\u2019s foot is longer relative to their respective genders.<\/p>\n

To Sum Up \u2026 <\/span><\/strong>Part (c)<\/em>\u00a0illustrates how z-scores become crucial when you want to\u00a0compare distributions<\/em>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

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Learn by Doing<\/h3>\n<\/header>\n
\n

The hourly salary rate for accountants at the \u201cWe are the Best Accounting Firm\u201d follow a normal distribution, with a mean of $27 and a standard deviation of $4.<\/p>\n

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