Form 2 Mathematics – STATISTICS msomimaktaba, November 13, 2018August 17, 2024 STATISTICSDefinition-is a branch of mathematics dealing with the study of method of collecting,organizing, analyzing, presenting and interpreting numerical details to reach conclusions.Frequency distributionIs a number of times each data pointExample 11. Make a frequency table from the following data from the followings data of ages 10 students14, 15, 16. 14, 17, 15, 16, 13, 2. In mathematics test the following marks were obtained;48 , 47,42, 67, 73, 50, 76 ,47, 44, 44, 57, 58, 54, 45, 58, 56 , 66, 67, 45, 43, 71, 48, 64, 52, 42, 54, 62, 32, 49, 34, 35, 46, 89, 37, 47, 54, 45, 60, 64, 44,.If the class size of the class interval is 8 group the works starting with the interval 32-39 and draw the frequency distribution table.solutionMarkFrequency88-95180-87072-79 264-71656-63648-55840-471332-394n=40 ExampleFrom example below find the class mark of the class interval 88 – 95 and 80 – 87class mark: and and = 91.5 and 83.5Class limit ; example in class interval 88 – 9588 is the class lower limit95 is the class upper limitCLASS REAL LIMITSclass lower real limit – is the number obtained by subtracting 0.5 from a class lower limit e.g. 88-0.5= 87.5class upper real limit obtained by adding 0.5 to the upper class limit eg, 95 + 0.5 = 95.5class size- is the value obtained by the difference between the upper real limit and the lower real limitexample .from class interval 88-95 and 31-35find the class sizesolution:Lower class real limit = 88 – 0.5 = 87.5Upper class real limit = 95+ 0.5 = 95.5Class size = 95.5 – 87.5 = 8Lower class real limit = 31 – 0.5 = 30.5 Upper class real limit = 35+ 0.5 = 35.5Class size = 35.5– 30.5= 5Exercise 1:(1).In biology class test the following marks when obtained;54,54,40,55,54,43,73,34, 75, 47, 35, 45,73,46,31,43,47,35,35,60,67,51,44,48,55,45,50,37,51,36By grouping the marks in class interval 20-29 ,30-39, 40-49, etc construct the the frequency Solution: DISTRIBUTION TABLEMarksFrequency(f)20 -29030 –39740 -491050 -59860 -69270 -793N = 30(2) The following data represent the masses of 10 people in kg. Construct the frequency distribution table for these people30 25 35 28 38 40 25 25 40 24Solution: (3). The following is a set of marks on a geography examination presents the frequency distribution table with class intervals, real limit, class marks, interval size starting with the interval 8-15 at the bottomSolution:class intervalReal limits class marksintervalf88-95 87.5-95.5 91.5 8 380-8779-87.5 83.5 8 372-7971.5-79.575.5 8 664-71 63.5-71.5 67.5 8 356-63 55.5-63.5 59.5 8 648-55 47.5-55.5 51.5 8 440-47 39.5-47.5 43.5 8 732-39 31.5-39.5 35.5 8 424-31 23.5-31.5 27.5 8 816-23 15.5-23.5 19.5 8 28-15 7.5-15.5 11.5 8 4 n=50(4). Fill in the blank columnsDistribution of 100 math s examination score class interval real limit class marksintervalf95-99 94.5-99.5 97 5 390-94 89.5-94.5 92 5 785-89 84.5-89.5 87 5 980-84 79.5-84.5 82 5 1375-79 74.5-79.5 77 5 2070-74 69.5-74.5 72 5 2365-69 64.5-69.5 67 5 1760-64 59.5-64.5 62 5 8N=100Note:Class real limits are also known as class boundariesGRAPHS OF FREQUENCY DISTRIBUTIONS:HISTOGRAMSHistograms of frequency distribution are rectangular figures plotted with class marks against frequency . The width of the histogram equal to the class size.Example:1. Draw a histogram of 100 mathematics examination scores in the table belowclass intervalclass markfrequency95-9997390-9492785-8987980-84821375-79772070-74722365-69671760-646283. . 2. Use the following distribution table below to draw a histogramageFrequency131144152162171 SolutionFREQUENCY POLYGONIs the line graph of class frequency plotted against class marksSteps ;1. 1. Add one interval below the lowest interval and one above the highest interval and assign them as zero frequency.2. 2. Plot a point and join them by straight linesExample1. Draw a frequency polygon from the following data.c-intervalc-markfrequency100-104102095-9997390-9492785-8987980-84821375-79772070-74722365-69671760-6462855-59570 EXERCISE1. The following table shows female death between 0 and 34 years to the nearest numbers represent this information byA) HistogramB) Frequency polygonExpected death of female per 100 womenages F(death risks)age0-434025-995710-14551215-19601720-24952225-291102730-341203235-3912537N=1000Solution:A) HistogramB) FREQUENCY POLYGON2. Table below show the distribution of marks obtained by 110 students in two different monthly tests. Draw the frequency polygon on the same chartmarksfrequencymarksfrequency21-30 421-30 231-40 731-40 1241-50 1041-50 1551-60 551-60 461-70 361-70 371-80 171-80 4N=40N=40 CUMULATIVE FREQUENCY CURVE (ORGIVE) – Cumulative frequency is the sum of all the frequency less than or equal to a given mark or class interval– To calculate the cumulative frequency start with the smaller upper real limit– Add the frequency of the smallest interval to the next interval downwards or up wards depending on whether the data is arranged in descending or ascending orderNote: The last entry in the cumulative frequency is always equal to the total number of observations– Plot upper real limit against class marks.– Join adjacent points by a free hand.EXAMPLES1. Draw an orgivefor the scores data below.scoref70-741665-691260-641455-591050-54845-491840-44635-39430-342N=90 Solution; THE CUMULATIVE FREQUENCY DISTRIBUTION,Scorefrequencycumulative frequencyless than 34.522less than39.546less than 44.5612less than 49.51830less than 54.5838less than 59.51048less than 64.51462less than 69.512less than 74.51690N =902. Motor vehicle company tested 100 cars to see how far they could travel on 10 litres of petrol. Draw the cumulative frequency curve for this companydistance in km100-109110-119120-129130-139140-149numbers of car515253520solutiondistance in km fcum. Fless than 109.555less than 119.51520less than 129.52545less than 139.53580less than 149.520100N=1003. Platform in each square metre of a lawn were counted and recorded as follows. Draw an orgive for the platformno.of plat formsfc. Frequency0101018182725353044345537 N = 37REVISION EXERCISE1. 1. The ages of the 22 players in a football match were recorded in the following17 18 15 16 16 16 18 15 18 15 15 18 18 15 16 17 15 16 17 15 15 16 15 18 15Express the data in a frequency table. Solution:AGESFREQUENCY1510165172185N = 222. 2. The examination marks of 45 students are,65 58 71 62 64 35 72 32 64 46 59 82 73 76 64 63 75 71 61 36 64 80 61 64 76 64 60 68 48 35 92 73 46 24 35 43 30 50 70 40 46 64 24 28A)Make a frequency distribution using class interval 21-30, 31-40, 41-50, Solution:c-intervalfrequency21-30431-40641-50651-60461-701471-80981-902n=45B) Draw cumulative frequency curveSolution:3. 3. Two plot A and B were treated with different families. The frequency number of potatoes on on samples of 100 plants on each plot are shown belowno.of potatoes38131823283338plot A12628275832plot B172830143620Draw a histogram for plot B. Plot B4. In a certain examination the result were as follows;3 student got marks between 0and 105 students got marks between 10 and 156 students got marks between 20 and 404 students got marks between 30 and 402 students got marks between 40 and 50Construct a histogram 5] final score of history examination were recorded as shown in table belowscorefrequencyc- mark50-5415255-5725760-64116265-69106770-74137275-79127780-84218285-8968790-9499295-99497A) What is the size of class intervals?Solution:5 is the size of class intervals .B)Draw a histogram to represent the scores . 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