{"id":2650,"date":"2023-10-03T11:17:00","date_gmt":"2023-10-03T11:17:00","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2650"},"modified":"2023-10-03T11:27:05","modified_gmt":"2023-10-03T11:27:05","slug":"week-11-ss1-third-term-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-11-ss1-third-term-mathematics-notes\/","title":{"rendered":"Week 11 – SS1 Third Term Mathematics Notes"},"content":{"rendered":"

\u00a0
\n\u00a0WEEK ELEVEN\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
\n<\/strong>TOPIC: Mean Deviation, Variance and standard Deviation of Grouped Data use in solving practical problems related to real life situations
\n<\/strong>
\n\u00a0Mean Deviation of Grouped Data
\n<\/strong>Example 1
\nThe speeds of 40 cars in a certain road are tabulated as follows: <\/p>\n

\n\n\n\n\n
Speed (km\/h)<\/td>\n50 \u2013 54<\/td>\n55 \u2013 59<\/td>\n60 \u2013 64<\/td>\n65 \u2013 69<\/td>\n60 \u2013 74<\/td>\n75 \u2013 80<\/td>\n80 – 84<\/td>\n<\/tr>\n
Frequency<\/td>\n5<\/td>\n10<\/td>\n15<\/td>\n12<\/td>\n10<\/td>\n6<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

For this distribution, calculate <\/p>\n

    \n
  1. \n
    The mean\n<\/div>\n<\/li>\n
  2. \n
    The mean deviation\n<\/div>\n<\/li>\n<\/ol>\n

    \u00a0Solution
    \n<\/strong>The complete table of the distribution is shown below. <\/p>\n

    \n\n\n\n\n\n\n\n\n\n\n\n
    Class interval<\/strong><\/td>\nMid \u2013 value (xm<\/sub>)<\/strong><\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
    50 \u2013 54<\/td>\n52<\/td>\n5<\/td>\n260<\/td>\n13.17<\/td>\n65.85<\/td>\n<\/tr>\n
    55 \u2013 59<\/td>\n57<\/td>\n10<\/td>\n570<\/td>\n8.17<\/td>\n81.5<\/td>\n<\/tr>\n
    60 \u2013 64<\/td>\n62<\/td>\n15<\/td>\n930<\/td>\n3.17<\/td>\n47.55<\/td>\n<\/tr>\n
    65 \u2013 69<\/td>\n67<\/td>\n12<\/td>\n804<\/td>\n1.83<\/td>\n21.96<\/td>\n<\/tr>\n
    60 \u2013 74<\/td>\n72<\/td>\n10<\/td>\n720<\/td>\n6.83<\/td>\n68.3<\/td>\n<\/tr>\n
    75 \u2013 80<\/td>\n77<\/td>\n6<\/td>\n462<\/td>\n11.83<\/td>\n70.98<\/td>\n<\/tr>\n
    80 \u2013 84<\/td>\n82<\/td>\n2<\/td>\n164<\/td>\n16.83<\/td>\n33.66<\/td>\n<\/tr>\n
    Total <\/strong><\/td>\n\u00a0<\/td>\n<\/td>\n<\/td>\n\u00a0<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

    \u00a0<\/p>\n

      \n
    1. \n
      Mean, =\n<\/div>\n

      The mean is 65.2km\/h to 1 d.p.\n<\/li>\n

    2. \n
      Mean deviation =\n<\/div>\n

      The mean deviation is 6.5km\/h <\/p>\n

      \u00a0EVALUATION
      \n<\/strong><\/li>\n

    3. \n
      Calculate the mean and the mean deviation of the following:\n<\/div>\n
        \n
      1. \n
        8, 5, 12, 8, 13, 4, 9, 5, 4, 7\n<\/div>\n<\/li>\n
      2. \n
        9.25, 8.04, 12.08, 9.82, 10.05, 2.05, 8.25, 7.64, 7.02, 8.02\n<\/div>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

        \u00a0Variance and Standard Deviation of a Grouped Data
        \n<\/strong>Example 1
        \nThe table shows the time to the nearest hours of television watched by a group of students in a week. <\/p>\n

        \n\n\n\n\n
        Time<\/strong><\/td>\n1 \u2013 5<\/td>\n6 \u2013 10<\/td>\n11 \u2013 15<\/td>\n16 \u2013 20<\/td>\n21 \u2013 25<\/td>\n26 \u2013 30<\/td>\n31 \u2013 35<\/td>\n36 \u2013 40<\/td>\n<\/tr>\n
        Frequency<\/strong><\/td>\n2<\/td>\n5<\/td>\n8<\/td>\n10<\/td>\n14<\/td>\n6<\/td>\n4<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

        Calculate <\/p>\n

          \n
        1. \n
          The mean\n<\/div>\n<\/li>\n
        2. \n
          The variance\n<\/div>\n<\/li>\n
        3. \n
          The standard deviation\n<\/div>\n<\/li>\n<\/ol>\n

          \u00a0Solution
          \n<\/strong>Let xm<\/sub> represents the mid-value (or class mark) of the interval.<\/p>\n

            \n
          1. \n
            =\n<\/div>\n

            Now subtract 19.8 from each value in the 2nd<\/sup> column to obtain the results in the 5th<\/sup> column. Then complete the other two columns as shown in the table.\n<\/li>\n

          2. \n
            S2<\/sup> =\n<\/div>\n

            Variance = 64.8h to 3 s.f.\n<\/li>\n

          3. \n
            S = = 8.047h\n<\/div>\n

            Standard deviation is 8.05h to 3 s.f.\n<\/li>\n<\/ol>\n

            \u00a0Alternative method
            \n<\/strong><\/p>\n

            \n\n\n\n\n\n\n\n\n\n\n\n\n
            Interval<\/strong><\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
            1 \u2013 5 <\/td>\n3<\/td>\n2<\/td>\n6<\/td>\n9<\/td>\n18<\/td>\n<\/tr>\n
            6 \u2013 10<\/td>\n8<\/td>\n5<\/td>\n40<\/td>\n64<\/td>\n320<\/td>\n<\/tr>\n
            11 \u2013 15<\/td>\n13<\/td>\n8<\/td>\n104<\/td>\n169<\/td>\n1352<\/td>\n<\/tr>\n
            16 \u2013 20<\/td>\n18<\/td>\n10<\/td>\n180<\/td>\n324<\/td>\n3240<\/td>\n<\/tr>\n
            21 \u2013 25<\/td>\n23<\/td>\n14<\/td>\n322<\/td>\n529<\/td>\n7406<\/td>\n<\/tr>\n
            26 \u2013 30<\/td>\n28<\/td>\n6<\/td>\n168<\/td>\n784<\/td>\n4704<\/td>\n<\/tr>\n
            31 \u2013 35<\/td>\n33<\/td>\n4<\/td>\n132<\/td>\n1089<\/td>\n4356<\/td>\n<\/tr>\n
            36 \u2013 40<\/td>\n38<\/td>\n1<\/td>\n38<\/td>\n1444<\/td>\n1444<\/td>\n<\/tr>\n
            Total <\/strong><\/td>\n\u00a0<\/td>\n<\/td>\n<\/td>\n\u00a0<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

            \u00a0EVALUATION
            \n<\/strong><\/p>\n

              \n
            1. \n
              Calculate to 1 d.p the mean and standard deviation of the following numbers:\n<\/div>\n
                \n
              1. \n
                5, 7, 12, 10, 5, 15, 14, 9, 7, 8\n<\/div>\n<\/li>\n
              2. \n
                6.5, 8.5, 6.5, 8.4, 6.9, 2.5, 6.2, 5.5\n<\/div>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                \u00a0GENERAL EVALUATION
                \n<\/strong><\/p>\n

                  \n
                1. \n
                  The table bellows shows the age distributions of a group of people.\n<\/div>\n<\/li>\n<\/ol>\n
                  \n\n\n\n\n
                  Age (yrs)<\/strong><\/td>\n20 \u2013 29<\/td>\n30 \u2013 39<\/td>\n40 \u2013 49<\/td>\n50 \u2013 59<\/td>\n60 \u2013 69<\/td>\n70 \u2013 79<\/td>\n<\/tr>\n
                  Frequency<\/strong><\/td>\n3<\/td>\n5<\/td>\n10<\/td>\n13<\/td>\n7<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                  Calculate: <\/p>\n

                    \n
                  1. \n
                    The mean age\n<\/div>\n<\/li>\n
                  2. \n
                    The variance\n<\/div>\n<\/li>\n
                  3. \n
                    The standard deviation\n<\/div>\n<\/li>\n<\/ol>\n

                    \u00a0READING ASSIGNMENT
                    \n<\/strong>Essential Mathematics for Senior Secondary 1 pgs 237 – 248<\/p>\n

                    \u00a0WEEKEND ASSIGNMENT
                    \n<\/strong><\/p>\n

                      \n
                    1. \n
                      The lowest temperatures of a city in Asia for 10 consecutive days are recorded as: – 5o<\/sup>C, – 6o<\/sup>C, -5o<\/sup>C, 4o<\/sup>C, 0o<\/sup>C, 1o<\/sup>C, 2o<\/sup>C, 3o<\/sup>C, 4o<\/sup>C, 7o<\/sup>C. Find the mean deviation. A. 3.9 B. 4.0 C. 3.6 D. 6.4\n<\/div>\n<\/li>\n<\/ol>\n

                      Use the table below to answer question 2 to 4
                      \nA dice is thrown 100 times. The results are recorded as shown in the following table <\/p>\n

                      \n\n\n\n\n
                      Score<\/strong><\/td>\n1 <\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n6<\/td>\n<\/tr>\n
                      Frequency<\/strong><\/td>\n15<\/td>\n18<\/td>\n17<\/td>\n21<\/td>\n14<\/td>\n15<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                      Calculate:<\/p>\n

                        \n
                      1. \n
                        The mean score A. 4.0 B. 3.5 C. 1.0 D. 5.6\n<\/div>\n<\/li>\n
                      2. \n
                        The variance A. 2.7 B. 3.7 C. 2.1 D. 1\n<\/div>\n<\/li>\n
                      3. \n
                        The standard deviation A. 4 B. 5.1 C. 1.6 D. 7\n<\/div>\n<\/li>\n
                      4. \n
                        Find the variance of x, 2x, 3x, 4x, 5x, 6x, 7x, 8x, 9x and 10x. A. 8.25x2<\/sup>9x2<\/sup> B. 10x2<\/sup> 7.25x2<\/sup>\n\t\t\t\t<\/div>\n<\/li>\n<\/ol>\n

                        \u00a0THEORY
                        \n<\/strong><\/p>\n

                          \n
                        1. \n
                          The shoe sizes of a group of people are as follows:\n<\/div>\n<\/li>\n<\/ol>\n
                          \n\n\n\n\n
                          Shoe size<\/strong><\/td>\n5<\/td>\n6<\/td>\n7<\/td>\n8<\/td>\n9<\/td>\n10<\/td>\n11<\/td>\n12<\/td>\n13<\/td>\n<\/tr>\n
                          Frequency<\/strong><\/td>\n3<\/td>\n8<\/td>\n14<\/td>\n16<\/td>\n20<\/td>\n10<\/td>\n5<\/td>\n3<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                          For this distribution, calculate the mean deviation <\/p>\n

                            \n
                          1. \n
                            The table below show the age distributions of a group of people.\n<\/div>\n<\/li>\n<\/ol>\n
                            \n\n\n\n\n
                            Age (yrs)<\/strong><\/td>\n20 \u2013 29<\/td>\n30 \u2013 39<\/td>\n40 \u2013 49<\/td>\n50 \u2013 59<\/td>\n60 -69<\/td>\n70 \u2013 79 <\/td>\n<\/tr>\n
                            Frequency<\/strong><\/td>\n3<\/td>\n5<\/td>\n10<\/td>\n13<\/td>\n7<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                            Calculate (a) the mean age (b) the variance (c) the standard deviation <\/p>\n

                            \u00a0
                            \n\u00a0
                            \n\u00a0
                            \n\u00a0
                            \n\u00a0
                            \n\t\t\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"

                            \u00a0 \u00a0WEEK ELEVEN\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 TOPIC: Mean Deviation, Variance and standard Deviation of Grouped Data use in…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,219],"tags":[],"class_list":["post-2650","post","type-post","status-publish","format-standard","hentry","category-posts","category-third-term-ss1-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2650","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=2650"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2650\/revisions"}],"predecessor-version":[{"id":2651,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2650\/revisions\/2651"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2650"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2650"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2650"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}