{"id":2654,"date":"2023-10-03T11:18:27","date_gmt":"2023-10-03T11:18:27","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2654"},"modified":"2023-10-03T11:27:04","modified_gmt":"2023-10-03T11:27:04","slug":"week-9-ss1-third-term-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-9-ss1-third-term-mathematics-notes\/","title":{"rendered":"Week 9 – SS1 Third Term Mathematics Notes"},"content":{"rendered":"

\u00a0
\n\u00a0WEEK NINE\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
\n<\/strong>TOPIC: Calculation of Range, Median and Mode of Grouped data
\n<\/strong>
\n\u00a0RANGE:<\/strong> This is defined as the difference between the HIGHEST variable and the LEAST variable.<\/p>\n

\u00a0Example:<\/strong> Find the range of the following distribution: 2.2, 2.5, 2.2, 1.6, 1.8, 2.7,and 1.4
\nSolution:<\/strong> Range= Highest score \u2013 Lowest score
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Highest score = 2.8
\nLeast score = 1.4
\nRange = 2.8 \u2013 1.4 = 1.4
\nThe above example is ungrouped data; therefore, the range is as simple as that.
\nTo find the range from Grouped, just identify the highest (Upper) class interval and the Least (Lower) Class interval and find the difference.<\/p>\n

\u00a0Example 1<\/strong>: find the range of the distribution:
\n1-10 11-20 21-30 31- 40 and 41- 50
\nHighest = 50
\nLeast = 1
\nRange = 50 \u2013 1 = 49<\/p>\n

\u00a0\"\"\/THE MEAN: <\/strong>This is also known as Arithmetic mean, it is denoted with the symbol X.Simply put, arithmetic mean is also known as average.
\nFor simple data, Such as: EXAMPLE (1) 2.2, 2.5, 2.2, 1.6, 1.8, 2.7, and 1.4, to calculate the arithmetic mean,the required formula is the same as that of the average: e g
\n MEAN = SUM OF THE ALL VARIABLES\/SCORES
\n\t\t NUMBER VARIABLES\/SCORES<\/p>\n

\u00a02.2 + 2.5 + 2.2 + 1.6 + 1.8 + 2.7 + 1.4 = 14.4 = 2.06
\n 7 7
\nThe basic formular for the calculation of the arithmetic mean is given below:
\n X = \u2211Fx where,\u2211 (Sigma) means summation.
\n\"\"\/ \u2211F<\/p>\n

\u00a0Hence,Mean (X) = Sum of the product of the frequency and scores
\n Sum of the frequencies<\/p>\n

\u00a0EXAMPLE 2: <\/strong>The table below gives the scores of a group of students in a mathematics test<\/p>\n

\u00a0<\/p>\n

\n\n\n\n\n
SCORES<\/strong><\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n6<\/td>\n7<\/td>\n<\/tr>\n
Number of Students<\/strong><\/td>\n2<\/td>\n4<\/td>\n7<\/td>\n2<\/td>\n3<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Calculate the mean mark of the distribution:
\nSolution (Method,1)
\n\"\"\/Mean = \u2211Fx= ( 2X 2) + (3 X 4) + (4 X 7) + ( 6X 3) + ( 7X2)
\n \u2211F 2 + 4 + 7 + 2 + 3 + 2<\/p>\n

\u00a0 = 4 + 12 + 28 + 18 + 14
\n 20
\n = 86 = 4.3
\n 20
\n(Method 2): A simple frequency distribution may be constructed<\/p>\n

\n\n\n\n\n\n\n\n\n\n\n
SCORES<\/strong><\/td>\nFREQUENCY(f)<\/strong><\/td>\nFx<\/strong><\/td>\n<\/tr>\n
2<\/td>\n2<\/td>\n4<\/td>\n<\/tr>\n
3<\/td>\n4<\/td>\n12<\/td>\n<\/tr>\n
4<\/td>\n7<\/td>\n28<\/td>\n<\/tr>\n
5<\/td>\n2<\/td>\n10<\/td>\n<\/tr>\n
6<\/td>\n3<\/td>\n18<\/td>\n<\/tr>\n
7<\/td>\n2<\/td>\n14<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u2211f = 20 <\/td>\n\u2211fx= 86<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

\u00a0\u2211fx= 86 and \u2211 = 20
\ntherefore,Mean = 86 = 4.3
\n 20
\nARITHMETIC MEAN FROM GROUPED DATA:
\n<\/strong>To calculate the arithmetic mean from grouped data, a frequency table is necessary, only the Class intervals, frequencies, class marks(Mid Mark) and fx column is required.
\nEXAMPLE 3<\/strong>:The distributions of the waiting time for some students in a school is given as follows:<\/p>\n

\u00a0\"\"\/ Waiting Time (minuetes) Number of customers
\n<\/strong>\"\"\/
\n\t\t 1.5 \u2013 1.9 3
\n<\/strong> 2.0 \u2013 2.4 10
\n<\/strong> 2.5 \u2013 2.9 18
\n<\/strong> 3.0 \u2013 3.4 10
\n<\/strong> 3.5 \u2013 3.9 7
\n<\/strong> 4.0 \u2013 4.4 2
\n<\/strong>
\n\u00a0
\n\u00a0Calculate the mean time of the distribution:
\nSolution: Prepare a simple frequency distribution table for a grouped data:<\/p>\n

\n\n\n\n\n\n\n\n\n\n\n
Time intervals
\n (Minuetes)<\/td>\n		Mid Time (x)<\/td>\nNo of Students
\nFrequencies<\/td>\n		Fx<\/td>\n<\/tr>\n
1.5 \u2013 1.9<\/td>\n1.7<\/td>\n3<\/td>\n5.1<\/td>\n<\/tr>\n
2.0 \u2013 2.4<\/td>\n2.2<\/td>\n10<\/td>\n22.0<\/td>\n<\/tr>\n
2.5 \u2013 2.9<\/td>\n2.7<\/td>\n18<\/td>\n48.6<\/td>\n<\/tr>\n
3.0 \u2013 3.4<\/td>\n3.2<\/td>\n10<\/td>\n32.0<\/td>\n<\/tr>\n
3.5 \u2013 3.9<\/td>\n3.7<\/td>\n7<\/td>\n25.9<\/td>\n<\/tr>\n
4.0 \u2013 4.4<\/td>\n4.2<\/td>\n2<\/td>\n8.4<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\n\t\t\t\t\t\t\t\u2211 f = 50<\/strong><\/td>\n \u2211fx=142.0<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

\u00a0Mean Time (Average Time) = \u2211fx= 142.0 = 2.8 minutes
\n\u00a0\u00a0\u00a0\u00a0 \u2211f 50<\/p>\n

\u00a0THE MODE
\n<\/strong>The mode is the variable or score with the highest frequency. The variable with the highest occurrence or which appears most in an event is known as the MODE. <\/p>\n

\u00a0EXAMPLE:<\/strong> Determine the modal mark in the distribution table:<\/p>\n

\n\n\n\n\n
Marks <\/td>\n4<\/td>\n5<\/td>\n6<\/td>\n7<\/td>\n8<\/td>\n9<\/td>\n10<\/td>\n<\/tr>\n
Frequency<\/td>\n5<\/td>\n3<\/td>\n2<\/td>\n6<\/td>\n5<\/td>\n1<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

\u00a0Solution<\/strong>: Modal mark = 7 with the frequency of 6.
\nIt is possible to record more than one variable as the modal mark.When only one number appears most (as mode) it is said UNIMODAL.When two numbers appears as the mode it is said to be BIMODAL and when more than two numbers appear as mode it is said to be MULTIMODAL.<\/p>\n

\u00a0THE MEDIAN<\/strong>
\n\t\tMedian is the number(s) which appears at the middle.It is possible for two numbers to appear at the middle,especially when the total variable is even number,in such a case,the average of the two mid numbers,is calculated as the MEDIAN<\/strong>.it must be noted that before the median is picked or calculated,the variables or scores must be arranged in an order of magnitude.i.e,ascending or Descending Order of Magnitude. <\/p>\n

\u00a0EXAMPLE:<\/strong> Calculate the median of the distribution:
\n2, 6, 4, 5, 5, 8, 8, 6, 6, 5, 9, 9, 2, 7, 4, 6, 3, 5, 6, 2, 7, 2, 9, 8, 10,6
\n<\/strong>Steps in <\/strong>the variables in an order of magnitude
\n:2,2,2,2,3,4,4,5,5,5,5,6,6,6,6,6,6,7,7,8,8,8,9,9,9,10 calculating MEDIAN from ungrouped (even) variables
\n<\/strong>STEP (i) Rearrange
\nSTEP(ii),Divide total number by 2. i.e,26\/2=13.
\nSTEP (iii) Count 13 numbers from both left and right
\nSTEP (iv) subtract one from each, result is 12.Hence 12 numbers are then counted from both left and right as shown below: 2,2,2,2,3,4,4,5,5,5,5,6<\/strong>, 6,6 ,6,6,6,7,7,8,8,8,9,9,9,10
\n<\/strong>From the above, two numbers are at the centre (6, 6) therefore the average of these numbers is the<\/p>\n

\u00a0median= 6 + 6 = 12 = 6. Therefore,median = 6
\n2 2
\nEXAMPLE: 2
\n<\/strong>Find the median of the scores below:
\n2.0, 1.8, 3.9, 4.5, 2.6, 3.7, 5.0, 2.1 and 3.3
\nSolution:
\n<\/strong>Rearranging the scores: 1.8, 2.0 ,2.1, 2.6, 3.3, 3.7, 3.9, 4.5, 5.0
\nThere are nine scores in all; 9\/2= 4.5
\nCounting four numbers from both left and right 1.8, 2.0 ,2.1, 2.6,<\/strong> 3.3, 3.7, 3.9, 4.5, 5.0<\/strong>
\n\t\tMEDIAN = 3.3<\/p>\n

\u00a0MEDIAN FROM TABLES:
\n<\/strong>EXAMPLE 3: The table shows the marks scored by SSS 1 students in a mathematics test<\/p>\n

\n\n\n\n\n
MARK<\/td>\n4<\/td>\n5<\/td>\n6<\/td>\n7<\/td>\n8<\/td>\n9<\/td>\n10<\/td>\n<\/tr>\n
FREQUENCY<\/td>\n5<\/td>\n3<\/td>\n2<\/td>\n6<\/td>\n5<\/td>\n1<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Find the median
\nMake a table as follows:
\n<\/strong>Marks ( x )\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Frequency (f)
\n<\/strong>\"\"\/4 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a05
\n5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03
\n6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02
\n\"\"\/\"\"\/7 6
\n8 5
\n9 1
\n10 3<\/p>\n

\u00a0Position of Median = \u2211f + 1 = 25 + 1 = 26 = 13
\n 2 2 2<\/p>\n

\u00a0Counting down the frequency column as shown on the above table,the position of the median (i.e,13th<\/sup> position) occurs opposite 7.
\nThus the median mark = 7<\/p>\n

\u00a0EVALUATION:
\n<\/strong>The table gives the frequency distribution of marks obtained by a group of students in a test<\/p>\n

\n\n\n\n\n
Marks<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n6<\/td>\n7<\/td>\n8<\/td>\n<\/tr>\n
Frequency<\/td>\n5<\/td>\nX \u2013 1<\/td>\nx<\/td>\n9<\/td>\n4<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

If the mean mark is 5 (a) Calculate the value of x
\n(b)Find the (i) mode (ii) Median (iii) Range of the distribution<\/p>\n

\u00a0
\n\u00a0
\n\u00a0READING ASSIGNMENT
\n<\/strong>Essential Mathematics for Senior Secondary Schools 1 pg 336 \u2013 351 <\/p>\n

\u00a0WEEKEND ASSIGNMENTS
\n<\/strong>OBJECTIVES
\n<\/strong><\/p>\n

    \n
  1. Which of the following is the same as the arithmetic mean of a distribution?A.Mean deviation B. average C. Ordinary mean D. Percentage\n<\/li>\n
  2. A bundle of tally consists of ____________ strokes?A.12 B. 10 C. 5 D. 4\n<\/li>\n
  3. Frequency is defined as the\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..A.The number of times a variable occur in a distribution B. The number of bundles in a cell of tallies C. The highest occurrence scores D. The average score\n<\/li>\n
  4. The range of the distribution: -2,3,3,1,1.7,2.4 and 2.6 is _____ ? A. 4 B. 0.6 C. 4.6 D. 3.5\n<\/li>\n
  5. `Find the average age of the following distribution:1.23,2.32,1.17,2,3.11,2.11and2.12\n<\/li>\n<\/ol>\n

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

      \n
    1. \n
      A group of students were asked to state their year of birth,the results are as follows\n<\/div>\n

      1990 1992 1990 1989 1991 1990
      \n<\/strong>1990 1988 1990 1989 1989 1991
      \n<\/strong>1992 1992 1990 1989 1988 1990
      \n<\/strong>1991 1991 1990 1988 1992 1991
      \n<\/strong>1990 1990 1992 1991
      \n<\/strong><\/p>\n

        \n
      1. prepare a frequency table for this data\n<\/li>\n
      2. which year of birth has this highest frequency\n<\/li>\n
      3. what fraction and percentage of the student were born in 1990 and above\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

        \u00a0<\/p>\n

          \n
        1. \n
          The height in meters of student in sss1 class in a certain secondary school were given as follows\n<\/div>\n

          1.3 1.3 1.2 1.4 1.2 1.5 1.5 1.4 1.3 1.6
          \n<\/strong>1.6 1.5 1.3 1.6 1.3 1.4 1.5 1.3 1.2 1.1
          \n<\/strong>1.3 1.2 1.5 1.5 1.4 1.3 1.2 1.4 1.6 1.5
          \n<\/strong>1.4 1.5 1.2 1.1 1.6 1.5 1.5 1.5 1.5 1.4
          \n<\/strong>1.2 1.3 1.4 1.5 1.4 1.5 1.5 1.4 1.3 1.2
          \n<\/strong>1.5 1.5
          \n<\/strong><\/p>\n

            \n
          1. Prepare a frequency distribution table for this data\n<\/li>\n
          2. How many student are in sss1?\n<\/li>\n
          3. What is the different between the highest and lowest height in cm?\n<\/li>\n
          4. How many student are more than 1.3 m tall?\n<\/li>\n
          5. What percentage of the student are 1.3 m tall and less?\n<\/li>\n
          6. State whether the data is discrete or continuons\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

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

            \u00a0 \u00a0WEEK NINE\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 TOPIC: Calculation of Range, Median and Mode of Grouped data \u00a0RANGE: This…<\/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-2654","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\/2654","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=2654"}],"version-history":[{"count":2,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2654\/revisions"}],"predecessor-version":[{"id":2656,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2654\/revisions\/2656"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2654"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2654"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2654"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}