Form 3 Mathematics – STATISTICS msomimaktaba, November 13, 2018August 17, 2024 STATISTICSIs the study or the methods of collecting, summarizing and presenting data and interpreting the information.MEASURES OF CENTRAL TENDENCY1. Mean2. Median3. ModeMEAN “X”Is obtained by adding up all the data values then divide by the number of characters.I.e.i.e. =meanx1+x2+x3………. Sum of observationsN =number of observationExample 1. Find mean score from the following scores of biology test 10, 25, 45, 15,63 42,7 = = 29.57When the data is given with frequency or in grouped data; = or f= frequency ∑= summation2. Find the mean number of children per family from the following tableNo. of children [x]012345678No. of families [f]36781012842SolutionFinding the mean of the numbers in the table below;No. of children [x]No. of families [f]fx030166271438244104051260684874288216Total 60236 = = = 3.93Exercise1. Football club has the following number of goals scored against them 0,1,0, 2, 9,0 , 1, 2,1. What is the mean number of goals scored against them?Solution== = 1.77 2. In a class of 30 girls the mean mass was 50kg calculate the total mass of the class.Solution=50N =30= = 30 x 50 = x 30∑fx = 1500kgMEAN OF THE GROUPED DATA1. The table below shows a distribution of 100 students find the mean mark.Class intervalClass mark [x]Frequency [f]fx91-95930086-908818881-8583649876-80781078071-757315109566-706834231261-656322138656-60581058051-55532106N=100∑fx=6845 Mean = =The mean is 68.45 MEAN BY ASSUMED MEAN METHOD =A + Where,A = assumed meanD = difference between the class marks and the assumed mean d= x-AF= frequencyN= total frequencyFrom the above example use the data to find the mean by assumed mean method.Take the assumed mean as 58. Class intervalClass mark[x]F D=x-Afd91-9593035086-90881303081-858362515076-8078102020071-7573151522566-7068341034061-656322511056-6058100051-55532-5-10Total1001045 Total A=58= A + = 58 + = 68.45The mass of students were recorded as shown below in the following figure. Class mark[x]ffx6110610642012806730201070151050735365Total805315 = = = 66.4365= 66.44Exercise a)1. Show the distribution of the children’s age in a month. Calculate the mean age in months using assumed mean that is the formula; = A +Calculate the mean age in months using the formula; = A + Class markfrequency41-46335-40429-34923-281217-221811-16285-1026Solution:To calculate the mean age in months using an assumed mean that is the formula = A + Class mark[x]Frequency [f]D= x-Afd41-463309035-404249629-3491816223-28121214417-2218610811-1628005-1026-6-156Total100444 TotalLet A= 11-16Let A = 13.5=A + = 13.5 + =13.5+4.44=17.942. Calculate the mean age in months using the formula for mean calculation.Solution = ==17.943. A survey was of 200 children under 10 years to see how many visits they made to the clinic during the courses of the year. The results were recorded as shown in the table below. Number of visitsfrequency5166331478549311010114122130142151 Solution = Number of visits[x]Frequency [f]Fx516806331987473298544329312791010100114441222413001422815115Total 2001529 ===7.645Mean number of visits per child = 7.645 4. A histogram for 100 mathematics scores use the histogram to find the mean score Solution from =A+Let A = 37 =37+375/100 = 40.75 MEDIANMedian is a point that divides the data into two parts such that equal numbers of the data fall above and below that point.Computation of the median depends on whether the data is ODD or EVEN or there is duplication of data [i.e. data with frequency] MEDIAN OF ODD NUMBERS OF DATASTEP 1Arrange the numbers in ascending/descending order1,1, 2,2,5, 5STEP 2Pick the number which is between those numbers. If it is even find the average of the two middle numbers e.g:- 2+2= 4/2=2Median of numbers 2,3,9,11, 2, 2,2, 2, 3, 9, 11STEP 1Arrange the numbers in ascending order 2,2,2,2,2,3,3,9,9,11,11STEP Pick the number which is between those numbersMedian =3 Example1.1.Find the median of the following observations 1, 7, 4, 3, 8Solution 1, 3, 4, 7, 8Median =4Exercise 1) 1, 2, 5, 3. Find the median of the given data solution Step 1Arrange the data in ascending order1, 2, 3, 5Step 21, 2, 3, 5= = Median is 2.52) 1, 1, 3, 2, find the median given even numbers of dataStep1Arrange the data in ascending order1, 1, 2, 3Step 21, 1, 2, 3= =1.5 Median is 1.53) 5, 3, 1, 6, 8 find the median given odd number of dataStep 1Arrange the data in ascending order1, 3, 5, 6, 8,Step 21, 3, 5, 6, 8Median is 54) Obtain the media of the following 1, 1, 6, 9, 8, 5Step 1Arrange the data in ascending order1, 1, 5, 6, 8, 9Step 21, 1, 5, 6, 8, 95+6Median is 5.55) Obtain the media of the following 2, 3, 9, 7, 1.Step 1Arrange the data in ascending order1, 2, 3, 7, 9Step 21, 2, 3, 7, 9Median is 3MEDIAN FOR GROUPED DATA Example1. The following table shows the distribution of nails in [mm]. Calculate the median length. Length [mm]fCumulative frequency88-963397-10558106-114917115-1231229124-132534133-141438142-150240Total 40169 L=?Median class=115-123Median position= ===20.5L= lower limit – 0.5From 115-123L=114.5N = 40nb=17nw =12i= 9Exercise 1. The following is the distribution of marks obtained in a test given to 50 candidates Marks frequencyCumulative frequency11-201121-303431-40101441-50213551-6064161-7054671-8045050 Find the median mark?Solution L=?Median class= ?Median position= Median position = = = 25.5Median position =25.5Median class= 41-50L = lower limit- 0.5From 41-50L= 41-0.5L= 40.5N=50nb=14nw=21i = 10From the formula: The following figure represents the graph of frequency polygon of a certain data . To find the median distributionSolution a) Class markfrequencyCumulative frequency923310151811910331281346137753Total53 L =?M.p == =27L =114.5N= 53nb=18nw=15i= 9=114.5+ ()x 9=114.5+[8.5/15]9= 114.5+5.1= 119.6b)SolutionClass markfrequencyCumulative frequency1500205525152030123235104240042 L =?M.P =N+1 = 42+1= 21.52 2L = 25+30 = 55/2= 27.5 L =27.5N=42nb= 20nw=12i =5 Exercise1. The height in centimeters of 100 people was recorded as shown below.Height [cm]160165170175180185frequency2123224218 Find the median height?Solution Height in [cm]frequencyCumulative frequency1603316512151703247175247118021921858100total100 L =?Median position = = = 50.5L= ? L= 170 + 175 = 345/2= 172.5L= 172.5N =100nb=47nw=24i= 5 Figure 5.13 is a histogram representing test marks of 50 candidates find the median mark.SolutionL =?Median Point = = = = 25.5 Class markfrequencyCumulative frequency15.51125.53435.5101445.5213555.564165.554675.5450Total 50 L = ?L = 40.5N =50nb =14nw=21i=10Figure 5.14 shows the frequency histogram for daily wages in TSHS of 70 people find the wagesSolution Wages in [TSHS}frequencyCumulative frequency5588651018751634851549951059105564Total 64 Median position = = = 32.5L= 75+85 = 160/2 L= 80N= 64nb = 34nw= 15i=10Figure 5.15 is a frequency Polygon for masses in kilogram’s of 80 students find the median mass. Solution Mass in kgfrequencyCumulative frequency470470052161651203662145061126272870776768248087080Total 80 L =?Median position = = = = 40.5L = 62+ 61 =123/2 L = 61.5N =80nb=36nw=14i = 5Median = 61.5 + ( ) x 5 \Median = 62.93. MODEMode is the value of data which occurs most frequently [data with the highest frequency].Data may have only one mode, more than one mode or no mode at all.Example Find the mode from the following datai) 3, 5, 7, 3, 2, 10, 8, 2, 7, 2Mode is =2ii) 2, 1, 2, 5, 3, 1, 1, 4, 2, 7.Mode is=1 and 2MODE FOR GROUPED DATA Figure 5.19 shows a histogram for heights of little children in centimeters calculate the mode of these heightsSolutionHeight in cmfrequency8210851188149110949976Total 60 Modal class =88L = = 86.5L= 86.5t1= 14-11 = 3t2= 14-10 = 4i= 3M = L + M = 86.5 + M = 87.79The mode is 87.79 ALL NOTES FOR ALL SUBJECTS QUICK LINKS:AGRICULTURE O LEVEL PURE MATHEMATICS A LEVELBAM NOTES A LEVELBASIC MATH O LEVELBIOLOGY O/A LEVELBOOK KEEPING O LEVELCHEMISTRY O/A LEVELCIVICS O LEVELCOMPUTER(ICT) O/A LEVELECONOMICS A LEVELENGLISH O/A LEVELCOMMERCE O/A LEVELACCOUNTING A LEVELGENERAL STUDIES NOTESGEOGRAPGY O/A LEVELHISTORY O/A LEVELKISWAHILI O/A LEVELPHYSICS O/A LEVELMOCK EXAMINATION PAPERSNECTA PAST PAPERS Basic Mathematics Study Notes Form 3 Basic Mathematics Study Notes FORM 3MATHEMATICSPost navigationPrevious postNext postRelated Posts Basic Mathematics Study Notes Form 4 Mathematics – VECTORS November 13, 2018August 17, 2024VECTORS Displacement and Position Vectors. Displacement Examples of vector quantities are displacement, Velocity, acceleration, force, momentum, electric field and magnetic field. Quantities which have magnitude only scalars, for example distance, speed, Pleasure, time and temperature. Sometimes a single. Small letter with a bar below like a or a like a… Read More Basic Mathematics Study Notes Form 2 Mathematics – TRIGONOMETRY November 13, 2018August 17, 2024TRIGONOMETRY Trigonometric ratio Introduction: TRI – is the Greek word which means three. -Trigonometry is the branch of mathematics which deals with measurement. -A Trigonometric ratio consists of three parts that is – Hypotenuse, Adjacent and opposite. Consider the diagram below which is the right angled triangle For example… Read More Basic Mathematics Study Notes Form 2 Mathematics – CONGRUENCE OF SIMPLE POLYGON November 13, 2018August 17, 2024 CONGRUENCE OF SIMPLE POLYGON The triangles above are drawn such that CB= ZY AC=XZ B=YX Corresponding sides in the triangles are those sides which are opposite to the equal angles i.e. If the corresponding sides are equal i.e. In general, polygons are congruent if corresponding sides and corresponding… Read More Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment *Name * Email * Website Save my name, email, and website in this browser for the next time I comment. Δ
Basic Mathematics Study Notes Form 4 Mathematics – VECTORS November 13, 2018August 17, 2024VECTORS Displacement and Position Vectors. Displacement Examples of vector quantities are displacement, Velocity, acceleration, force, momentum, electric field and magnetic field. Quantities which have magnitude only scalars, for example distance, speed, Pleasure, time and temperature. Sometimes a single. Small letter with a bar below like a or a like a… Read More
Basic Mathematics Study Notes Form 2 Mathematics – TRIGONOMETRY November 13, 2018August 17, 2024TRIGONOMETRY Trigonometric ratio Introduction: TRI – is the Greek word which means three. -Trigonometry is the branch of mathematics which deals with measurement. -A Trigonometric ratio consists of three parts that is – Hypotenuse, Adjacent and opposite. Consider the diagram below which is the right angled triangle For example… Read More
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