{"id":3797,"date":"2022-03-11T21:06:56","date_gmt":"2022-03-11T21:06:56","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/?post_type=chapter&p=3797"},"modified":"2022-03-22T17:43:03","modified_gmt":"2022-03-22T17:43:03","slug":"unused-pbj-items","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/chapter\/unused-pbj-items\/","title":{"raw":"Unused PBJ Items","rendered":"Unused PBJ Items"},"content":{"raw":"
\r\n

interactive example<\/h3>\r\nA particular dataset has a mean of\u00a0[latex]61[\/latex] and a standard deviation of\u00a0[latex]3.7[\/latex]. Calculate the following values.\r\n
    \r\n \t
  1. A value that is one standard deviation above<\/strong> the mean.<\/li>\r\n \t
  2. A value that is one standard deviation below<\/strong> the mean.<\/li>\r\n \t
  3. A value that is one-and-a-half standard deviations\u00a0below<\/strong> the mean.<\/li>\r\n \t
  4. A value that is two standard deviations\u00a0above\u00a0<\/strong>the mean.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"600825\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"600825\"]\r\n
      \r\n \t
    1. [latex]61+3.7=64.7[\/latex]<\/li>\r\n \t
    2. [latex]61-3.7=57.3[\/latex]<\/li>\r\n \t
    3. [latex]61-(1.5)(3.7)=55.45[\/latex]<\/li>\r\n \t
    4. [latex]61+(2)(3.7)=68.4[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
      \r\n

      interactive Example<\/h3>\r\nRecall, to calculate a z-score given an observation, use the formula [latex]z=\\dfrac{x-\\mu}{\\sigma}[\/latex], where [latex]x[\/latex] represents the value of the observation, [latex]\\mu[\/latex] represents the population mean, [latex]\\sigma[\/latex] represents the population standard deviation, and [latex]z[\/latex] represents the standardized value, or z-score.\r\n\r\nFor a dataset with a mean of\u00a0[latex]132[\/latex] and standard deviation of\u00a0[latex]9.8[\/latex], calculate the z-scores of the following observations, [latex]x[\/latex], and indicate if the given value is above or below the mean. Round answers to\u00a0[latex]2[\/latex] decimal places.\r\n
        \r\n \t
      1. [latex]x=112[\/latex]<\/li>\r\n \t
      2. [latex]x=141[\/latex]<\/li>\r\n \t
      3. [latex]x=158[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"172851\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"172851\"]\r\n
          \r\n \t
        1. [latex]z=\\dfrac{112-132}{9.8}\\approx -2.04[\/latex]. This value is [latex]2.04[\/latex] standard deviations\u00a0below<\/strong> the mean.<\/li>\r\n \t
        2. [latex]z=\\dfrac{141-132}{9.8}\\approx 0.92[\/latex]. This value is\u00a0[latex]0.92[\/latex] standard deviations\u00a0above\u00a0<\/strong>the mean.<\/li>\r\n \t
        3. [latex]z=\\dfrac{158-132}{9.8}\\approx 2.65[\/latex]. This value is [latex]2.65[\/latex] standard deviations\u00a0above<\/strong> the mean.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
          \r\n

          Example<\/h3>\r\nThis was written for 1C coreq -- don't use as an OHM problem for course objectives.<\/span>\r\n\r\nLet's say the following table lists the hair color of 12 customers at a salon.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
          Hair Color<\/td>\r\nNumber of People<\/td>\r\n<\/tr>\r\n
          Brown or Black<\/td>\r\n8<\/td>\r\n<\/tr>\r\n
          Blonde<\/td>\r\n3<\/td>\r\n<\/tr>\r\n
          Red<\/td>\r\n1<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe can discuss how much of each color is present out of the all the colors present by using a\u00a0proportion<\/strong>, which is a number that represents some part out of some whole. For example, one out of the 12 customers has red hair, so we could say that the proportion of customers with red hair is [latex]\\dfrac{1}{12}[\/latex].\r\n\r\nSince the proportion is a fraction, depending on our context, we may sometimes wish to reduce the fraction. For example, we could represent the number of customers with blonde hair as either 3 of the 12 total, or, by using a reduced fraction, as one fourth of the total. Let's see how. ...\r\n\r\n<\/div>","rendered":"
          \n

          interactive example<\/h3>\n

          A particular dataset has a mean of\u00a0[latex]61[\/latex] and a standard deviation of\u00a0[latex]3.7[\/latex]. Calculate the following values.<\/p>\n

            \n
          1. A value that is one standard deviation above<\/strong> the mean.<\/li>\n
          2. A value that is one standard deviation below<\/strong> the mean.<\/li>\n
          3. A value that is one-and-a-half standard deviations\u00a0below<\/strong> the mean.<\/li>\n
          4. A value that is two standard deviations\u00a0above\u00a0<\/strong>the mean.<\/li>\n<\/ol>\n
            Show Solution<\/span><\/p>\n
            \n
              \n
            1. [latex]61+3.7=64.7[\/latex]<\/li>\n
            2. [latex]61-3.7=57.3[\/latex]<\/li>\n
            3. [latex]61-(1.5)(3.7)=55.45[\/latex]<\/li>\n
            4. [latex]61+(2)(3.7)=68.4[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n
              \n

              interactive Example<\/h3>\n

              Recall, to calculate a z-score given an observation, use the formula [latex]z=\\dfrac{x-\\mu}{\\sigma}[\/latex], where [latex]x[\/latex] represents the value of the observation, [latex]\\mu[\/latex] represents the population mean, [latex]\\sigma[\/latex] represents the population standard deviation, and [latex]z[\/latex] represents the standardized value, or z-score.<\/p>\n

              For a dataset with a mean of\u00a0[latex]132[\/latex] and standard deviation of\u00a0[latex]9.8[\/latex], calculate the z-scores of the following observations, [latex]x[\/latex], and indicate if the given value is above or below the mean. Round answers to\u00a0[latex]2[\/latex] decimal places.<\/p>\n

                \n
              1. [latex]x=112[\/latex]<\/li>\n
              2. [latex]x=141[\/latex]<\/li>\n
              3. [latex]x=158[\/latex]<\/li>\n<\/ol>\n
                Show Solution<\/span><\/p>\n
                \n
                  \n
                1. [latex]z=\\dfrac{112-132}{9.8}\\approx -2.04[\/latex]. This value is [latex]2.04[\/latex] standard deviations\u00a0below<\/strong> the mean.<\/li>\n
                2. [latex]z=\\dfrac{141-132}{9.8}\\approx 0.92[\/latex]. This value is\u00a0[latex]0.92[\/latex] standard deviations\u00a0above\u00a0<\/strong>the mean.<\/li>\n
                3. [latex]z=\\dfrac{158-132}{9.8}\\approx 2.65[\/latex]. This value is [latex]2.65[\/latex] standard deviations\u00a0above<\/strong> the mean.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n
                  \n

                  Example<\/h3>\n

                  This was written for 1C coreq — don’t use as an OHM problem for course objectives.<\/span><\/p>\n

                  Let’s say the following table lists the hair color of 12 customers at a salon.<\/p>\n\n\n\n\n\n\n
                  Hair Color<\/td>\nNumber of People<\/td>\n<\/tr>\n
                  Brown or Black<\/td>\n8<\/td>\n<\/tr>\n
                  Blonde<\/td>\n3<\/td>\n<\/tr>\n
                  Red<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                  We can discuss how much of each color is present out of the all the colors present by using a\u00a0proportion<\/strong>, which is a number that represents some part out of some whole. For example, one out of the 12 customers has red hair, so we could say that the proportion of customers with red hair is [latex]\\dfrac{1}{12}[\/latex].<\/p>\n

                  Since the proportion is a fraction, depending on our context, we may sometimes wish to reduce the fraction. For example, we could represent the number of customers with blonde hair as either 3 of the 12 total, or, by using a reduced fraction, as one fourth of the total. Let’s see how. …<\/p>\n<\/div>\n","protected":false},"author":25777,"menu_order":6,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3797","chapter","type-chapter","status-publish","hentry"],"part":3388,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/3797","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/users\/25777"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/3797\/revisions"}],"predecessor-version":[{"id":4161,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/3797\/revisions\/4161"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/parts\/3388"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/3797\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/media?parent=3797"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapter-type?post=3797"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/contributor?post=3797"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/license?post=3797"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}