{"id":655,"date":"2020-07-20T22:50:53","date_gmt":"2020-07-21T02:50:53","guid":{"rendered":"https:\/\/openbooks.macewan.ca\/rcommander\/?post_type=chapter&p=655"},"modified":"2024-02-08T13:39:50","modified_gmt":"2024-02-08T18:39:50","slug":"5-5-working-with-any-normal-distribution","status":"publish","type":"chapter","link":"https:\/\/openbooks.macewan.ca\/introstats\/chapter\/5-5-working-with-any-normal-distribution\/","title":{"raw":"5.5 Working With Any Normal Distribution","rendered":"5.5 Working With Any Normal Distribution"},"content":{"raw":"Through standardization, we can solve problems related to any normal distribution [latex]N(\\mu, \\sigma)[\/latex]\u00a0using the standard normal table. The following diagram shows the procedure.<\/a>\r\n\r\n[caption id=\"attachment_1698\" align=\"aligncenter\" width=\"1410\"]\"Diagram Figure 5.9<\/strong>: Diagram for Working with Any Normal Distribution [Image description (See Appendix D Figure 5.9)<\/a>][\/caption]\r\n
\r\n

Example: Working With Any Normal Distribution Using the Standard Normal Table (Table II)<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSuppose the grades in a Statistics class are approximately normally distributed with a mean 70 and a standard deviation 10, i.e., [latex]X \\sim N(\\mu=70, \\sigma =10)[\/latex].\r\n
    \r\n \t
  1. Find the percentage of students whose grades are below 60.<\/li>\r\n<\/ol>\r\n[latex] P(X < 60) = P \\left( \\frac{X - \\mu}{\\sigma} < \\frac{60-\\mu}{\\sigma} \\right) = P\\left(Z < \\frac{60-70}{10}\\right) = P(Z < -1) = 0.1587. [\/latex]\r\n

    15.87% of the students have a grade below 60.<\/p>\r\n\r\n

      \r\n \t
    1. Find the percentage of students whose grades are above 95.<\/li>\r\n<\/ol>\r\n[latex]\\begin{align*} P(X \\: > \\: 95) &= P\\left(\\frac{X-\\mu}{\\sigma} \\: > \\: \\frac{95 - \\mu}{\\sigma}\\right) \\\\ \u00a0&= P\\left(Z \\: > \\: \u00a0\\frac{95-70}{10}\\right) \\\\ \u00a0&= P(Z \\: > \\: 2.5) \\\\ &= 1 - P(Z < 2.5) \\\\ &= 1 - 0.9938 = 0.0062.\u00a0 \\end{align*}[\/latex]\r\n

      or [latex]P(X \\: > \\: \u00a095) = P(Z \\: > \\: 2.5) = P(Z < -2.5) = 0.0062[\/latex].<\/p>\r\n

      Only 0.62% of students have a grade above 95.<\/p>\r\n\r\n

        \r\n \t
      1. Find the percentage of students whose grades are between 60 and 90.The [latex]z[\/latex]-score for 60 is [latex]z = \\frac{x - \\mu}{\\sigma} = \\frac{60-70}{10} = -1[\/latex]; the [latex]z[\/latex]-score for 90 is [latex]z = \\frac{x - \\mu}{\\sigma} = \\frac{90-70}{10} = 2[\/latex]. [latex]\\begin{align*} P(60 < X < 90) &= P(-1 < Z < 2) \\\\ &= P(Z < 2) - P(Z < -1) \\\\ \u00a0&= 0.9772 - 0.1597 \u00a0\\\\ &= 0.8185. \\end{align*}[\/latex]\r\n

        81.85% of the students have a grade between 60 and 90.<\/p>\r\n<\/li>\r\n<\/ol>\r\n

          \r\n \t
        1. Suppose the bottom 5% of students fail. Find the minimum grade needed in order to pass the course.<\/li>\r\n<\/ol>\r\n

          Find the grade x<\/em>\u00a0that bounds the bottom 5% of grades, i.e., [latex]P(X \\leq x) = 0.05[\/latex]. The [latex]z[\/latex]-score that captures the bottom 5% of the standard normal distribution is [latex]z=-1.645[\/latex].\u00a0Therefore, the corresponding x<\/em>\u00a0value is<\/p>\r\n[latex] x = \\mu +z \\times \\sigma = 70 + (-1.645) \\times 10 = 70 - 16.45 = 53.55.[\/latex]\r\n

          The passing grade is 53.55.<\/p>\r\n\r\n

            \r\n \t
          1. Suppose the top 2% of students will get an A. Find the minimum grade needed in order to obtain an A.<\/li>\r\n<\/ol>\r\n

            Find the grade x<\/em> that bounds the top 2% of grades, i.e., [latex]P(X \\ge x) = 0.02[\/latex]. The [latex]z[\/latex]-score that captures the top 2% of the standard normal distribution is [latex]z_{0.02}=2.05[\/latex]. Therefore, the corresponding x<\/em>\u00a0value is<\/p>\r\n[latex] x = \\mu +z \\times \\sigma = 70 + (2.05) \\times 10 = 70 + 20.5 = 90.5.[\/latex]\r\n

            The cutoff is 90.5.<\/p>\r\n\r\n

              \r\n \t
            1. Find the quartiles of student grades.<\/li>\r\n<\/ol>\r\n

              First, observe that the first, second and third quartiles of the standard normal distribution are [latex]z_1 = -0.67, z_2 = 0[\/latex] and [latex]z_3=0.67[\/latex],\u00a0 since [latex]P(Z < -0.67)\\approx 0.25, P(Z<0)=0.5[\/latex], and [latex]P(Z < 0.67)\\approx 0.75[\/latex]. Thus, the first, second and third quartiles of student grades are:<\/p>\r\n

              [latex]Q_1 = \\mu + z_1 \\times \\sigma = 70 + (-0.67) \\times 10 = 63.3,[\/latex]<\/p>\r\n

              [latex]Q_2 = \\mu + z_2 \\times \\sigma = 70 + (0) \\times 10 = 70,[\/latex]<\/p>\r\n

              [latex]Q_3 = \\mu + z_3 \\times \\sigma = 70 + (0.67) \\times 10 = 76.7.[\/latex]<\/p>\r\n

              Thus, the quartiles of student grades are:<\/p>\r\n

              [latex] Q_1= 63.3, \\quad Q_2 = 70, \\quad Q_3 =76.7.[\/latex]<\/p>\r\n

              Note: Recall that one of the properties of a symmetric distribution is that the mean and median are equal. So, we could have alternatively used the symmetry of a normal distribution to establish that [latex]Q_2 = \\mu =70[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

              \r\n

              Exercise: Work With Any Normal Distribution Using Standard Normal Table (Table II)<\/p>\r\n\r\n<\/header>\r\n

              \r\n\r\nSuppose the weight of boxes of a certain brand of cereal follows a normal distribution with a mean of 1,000 grams and a standard deviation of 40 grams.\r\n
                \r\n \t
              1. A box is rejected by the quality control department if its weight is below 950 grams. What percentage of boxes will be rejected?<\/li>\r\n \t
              2. Find the percentage of boxes with weight in between 980 grams and 1,010 grams.<\/li>\r\n \t
              3. Find the percentage of boxes with weight above 1,010 grams.<\/li>\r\n \t
              4. Determine the 40th percentile for the weight of the boxes of cereal.<\/li>\r\n \t
              5. A particular box of cereal is weighed, and it is determined that 5% of boxes are heavier than this particular box. What is the approximate weight of this box of cereal?<\/li>\r\n<\/ol>\r\n
                Show\/Hide Answer<\/summary>\r\n
                  \r\n \t
                1. [latex]P(X < 950)=[\/latex] [latex]P(Z < (950-1000)\/40)=[\/latex] [latex]P(Z < -1.25)=0.1056[\/latex]. That is 10.56%.<\/li>\r\n \t
                2. [latex]P(980 < X < 1010 )[\/latex] [latex]= P (-0.5 < Z < 0.25)[\/latex] [latex]= P(Z < 0.25) - P(Z < -0.5)[\/latex] [latex]= 0.5987 - 0.3085 = 0.2902[\/latex], that is, 29.02%.<\/li>\r\n \t
                3. [latex]P(X \\: > \\: 1010) = P(Z \\: > \\: (1010 - 100)\/40))[\/latex] [latex]= P(Z \\: > \\: 0.25)[\/latex] [latex]= 1 - P(Z < 0.25) = 1 - 0.5987 = 0.4013[\/latex], that is, 40.13%.<\/li>\r\n \t
                4. The 40th percentile of the standard normal distribution is [latex]z=-0.25[\/latex]. Therefore, [latex]x = 1000 + (-0.25) \\times 40 = 990[\/latex], so that the 40th percentile among boxes of cereal is 990 grams.<\/li>\r\n \t
                5. The 95th percentile of the standard normal distribution is [latex]z=1.645[\/latex]. Therefore, [latex]x = 1000 + (1.645) \\times 40 = 1065.8[\/latex], so that the 95th percentile among boxes of cereal is 1065.8 grams.<\/li>\r\n<\/ol>\r\n<\/details><\/div>\r\n<\/div>","rendered":"

                  Through standardization, we can solve problems related to any normal distribution [latex]N(\\mu, \\sigma)[\/latex]\u00a0using the standard normal table. The following diagram shows the procedure.<\/a><\/p>\n

                  \"Diagram
                  Figure 5.9<\/strong>: Diagram for Working with Any Normal Distribution [Image description (See Appendix D Figure 5.9)<\/a>]<\/figcaption><\/figure>\n
                  \n
                  \n

                  Example: Working With Any Normal Distribution Using the Standard Normal Table (Table II)<\/p>\n<\/header>\n

                  \n

                  Suppose the grades in a Statistics class are approximately normally distributed with a mean 70 and a standard deviation 10, i.e., [latex]X \\sim N(\\mu=70, \\sigma =10)[\/latex].<\/p>\n

                    \n
                  1. Find the percentage of students whose grades are below 60.<\/li>\n<\/ol>\n

                    [latex]P(X < 60) = P \\left( \\frac{X - \\mu}{\\sigma} < \\frac{60-\\mu}{\\sigma} \\right) = P\\left(Z < \\frac{60-70}{10}\\right) = P(Z < -1) = 0.1587.[\/latex]\n\n\n

                    15.87% of the students have a grade below 60.<\/p>\n

                      \n
                    1. Find the percentage of students whose grades are above 95.<\/li>\n<\/ol>\n

                      [latex]\\begin{align*} P(X \\: > \\: 95) &= P\\left(\\frac{X-\\mu}{\\sigma} \\: > \\: \\frac{95 - \\mu}{\\sigma}\\right) \\\\ \u00a0&= P\\left(Z \\: > \\: \u00a0\\frac{95-70}{10}\\right) \\\\ \u00a0&= P(Z \\: > \\: 2.5) \\\\ &= 1 - P(Z < 2.5) \\\\ &= 1 - 0.9938 = 0.0062.\u00a0 \\end{align*}[\/latex]\n\n\n

                      or [latex]P(X \\: > \\: \u00a095) = P(Z \\: > \\: 2.5) = P(Z < -2.5) = 0.0062[\/latex].<\/p>\n

                      Only 0.62% of students have a grade above 95.<\/p>\n

                        \n
                      1. Find the percentage of students whose grades are between 60 and 90.The [latex]z[\/latex]-score for 60 is [latex]z = \\frac{x - \\mu}{\\sigma} = \\frac{60-70}{10} = -1[\/latex]; the [latex]z[\/latex]-score for 90 is [latex]z = \\frac{x - \\mu}{\\sigma} = \\frac{90-70}{10} = 2[\/latex]. [latex]\\begin{align*} P(60 < X < 90) &= P(-1 < Z < 2) \\\\ &= P(Z < 2) - P(Z < -1) \\\\ \u00a0&= 0.9772 - 0.1597 \u00a0\\\\ &= 0.8185. \\end{align*}[\/latex]\n\n\n

                        81.85% of the students have a grade between 60 and 90.<\/p>\n<\/li>\n<\/ol>\n

                          \n
                        1. Suppose the bottom 5% of students fail. Find the minimum grade needed in order to pass the course.<\/li>\n<\/ol>\n

                          Find the grade x<\/em>\u00a0that bounds the bottom 5% of grades, i.e., [latex]P(X \\leq x) = 0.05[\/latex]. The [latex]z[\/latex]-score that captures the bottom 5% of the standard normal distribution is [latex]z=-1.645[\/latex].\u00a0Therefore, the corresponding x<\/em>\u00a0value is<\/p>\n

                          [latex]x = \\mu +z \\times \\sigma = 70 + (-1.645) \\times 10 = 70 - 16.45 = 53.55.[\/latex]<\/p>\n

                          The passing grade is 53.55.<\/p>\n

                            \n
                          1. Suppose the top 2% of students will get an A. Find the minimum grade needed in order to obtain an A.<\/li>\n<\/ol>\n

                            Find the grade x<\/em> that bounds the top 2% of grades, i.e., [latex]P(X \\ge x) = 0.02[\/latex]. The [latex]z[\/latex]-score that captures the top 2% of the standard normal distribution is [latex]z_{0.02}=2.05[\/latex]. Therefore, the corresponding x<\/em>\u00a0value is<\/p>\n

                            [latex]x = \\mu +z \\times \\sigma = 70 + (2.05) \\times 10 = 70 + 20.5 = 90.5.[\/latex]<\/p>\n

                            The cutoff is 90.5.<\/p>\n

                              \n
                            1. Find the quartiles of student grades.<\/li>\n<\/ol>\n

                              First, observe that the first, second and third quartiles of the standard normal distribution are [latex]z_1 = -0.67, z_2 = 0[\/latex] and [latex]z_3=0.67[\/latex],\u00a0 since [latex]P(Z < -0.67)\\approx 0.25, P(Z<0)=0.5[\/latex], and [latex]P(Z < 0.67)\\approx 0.75[\/latex]. Thus, the first, second and third quartiles of student grades are:<\/p>\n

                              [latex]Q_1 = \\mu + z_1 \\times \\sigma = 70 + (-0.67) \\times 10 = 63.3,[\/latex]<\/p>\n

                              [latex]Q_2 = \\mu + z_2 \\times \\sigma = 70 + (0) \\times 10 = 70,[\/latex]<\/p>\n

                              [latex]Q_3 = \\mu + z_3 \\times \\sigma = 70 + (0.67) \\times 10 = 76.7.[\/latex]<\/p>\n

                              Thus, the quartiles of student grades are:<\/p>\n

                              [latex]Q_1= 63.3, \\quad Q_2 = 70, \\quad Q_3 =76.7.[\/latex]<\/p>\n

                              Note: Recall that one of the properties of a symmetric distribution is that the mean and median are equal. So, we could have alternatively used the symmetry of a normal distribution to establish that [latex]Q_2 = \\mu =70[\/latex].<\/p>\n<\/div>\n<\/div>\n

                              \n
                              \n

                              Exercise: Work With Any Normal Distribution Using Standard Normal Table (Table II)<\/p>\n<\/header>\n

                              \n

                              Suppose the weight of boxes of a certain brand of cereal follows a normal distribution with a mean of 1,000 grams and a standard deviation of 40 grams.<\/p>\n

                                \n
                              1. A box is rejected by the quality control department if its weight is below 950 grams. What percentage of boxes will be rejected?<\/li>\n
                              2. Find the percentage of boxes with weight in between 980 grams and 1,010 grams.<\/li>\n
                              3. Find the percentage of boxes with weight above 1,010 grams.<\/li>\n
                              4. Determine the 40th percentile for the weight of the boxes of cereal.<\/li>\n
                              5. A particular box of cereal is weighed, and it is determined that 5% of boxes are heavier than this particular box. What is the approximate weight of this box of cereal?<\/li>\n<\/ol>\n
                                \nShow\/Hide Answer<\/summary>\n
                                  \n
                                1. [latex]P(X < 950)=[\/latex] [latex]P(Z < (950-1000)\/40)=[\/latex] [latex]P(Z < -1.25)=0.1056[\/latex]. That is 10.56%.<\/li>\n
                                2. [latex]P(980 < X < 1010 )[\/latex] [latex]= P (-0.5 < Z < 0.25)[\/latex] [latex]= P(Z < 0.25) - P(Z < -0.5)[\/latex] [latex]= 0.5987 - 0.3085 = 0.2902[\/latex], that is, 29.02%.<\/li>\n
                                3. [latex]P(X \\: > \\: 1010) = P(Z \\: > \\: (1010 - 100)\/40))[\/latex] [latex]= P(Z \\: > \\: 0.25)[\/latex] [latex]= 1 - P(Z < 0.25) = 1 - 0.5987 = 0.4013[\/latex], that is, 40.13%.<\/li>\n
                                4. The 40th percentile of the standard normal distribution is [latex]z=-0.25[\/latex]. Therefore, [latex]x = 1000 + (-0.25) \\times 40 = 990[\/latex], so that the 40th percentile among boxes of cereal is 990 grams.<\/li>\n
                                5. The 95th percentile of the standard normal distribution is [latex]z=1.645[\/latex]. Therefore, [latex]x = 1000 + (1.645) \\times 40 = 1065.8[\/latex], so that the 95th percentile among boxes of cereal is 1065.8 grams.<\/li>\n<\/ol>\n<\/details>\n<\/div>\n<\/div>\n","protected":false},"author":19,"menu_order":5,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-655","chapter","type-chapter","status-publish","hentry"],"part":588,"_links":{"self":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/655","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/users\/19"}],"version-history":[{"count":35,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/655\/revisions"}],"predecessor-version":[{"id":5291,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/655\/revisions\/5291"}],"part":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/parts\/588"}],"metadata":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/655\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/media?parent=655"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapter-type?post=655"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/contributor?post=655"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/license?post=655"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}