4. STATISTICS. 1. Introduction. 2. Data. 2.1 Fundamental Characteristics of Data. 2.2 Types of Data.

Transcription

1 4. STATISTICS 1. Introducton In varous felds, we need nformaton n the form of numercal fgures called data. These data may relate to the marks obtaned by the pupls of a class n a certan examnaton; the weghts, heghts, ages, etc., of pupls n a class; the monthly wages earned by workers n a factory; the populaton of a town or the profts of a company durng last few years, etc. Evaluaton of such data helps analysts study the varous growth patterns and formulate future targets or polces or derve certan nferences. Statstcs: It s the scence whch deals wth the collecton, presentaton, analyss and nterpretaton of numercal data. In sngular form, statstcs s taken as a subject. And, n plural form, statstcs means data.. Data The word data means a set of gven facts n numercal fgures..1 Fundamental Characterstcs of Data () Numercal facts alone form data. Qualtatve characterstcs, lke honesty, poverty, etc., whch cannot be measured numercally do not form data. () Data are aggregate of facts. A sngle observaton does not form data. () Data collected for a defnte purpose may not be sutable for another purpose.. Types of Data () Prmary Data: The data collected by the nvestgator hmself wth a defnte plan n mnd are known as prmary data. () Secondary Data: The data collected by someone, other than the nvestgator, are known as secondary data.

2 4. Statstcs 3. Varable A quantty whch can take dfferent values s called a varable. Ex: Heght, Age and Weght of pupls n a class are three varables. If we denote them by x, y and z respectvely, then values of x gve the heghts of the pupls; the values of y gve the ages of the pupls and the values of z gve the weghts of the pupls. 3.1 Contnuous and Dscrete Varables Varables are of types. Contnuous Varable: A varable whch can take any numercal value wthn a certan range s called a contnuous varable.. Ex. () Wages of workers n a factory () Heghts of chldren n a class () Weghts of persons n a group etc. Dscontnuous (or dscrete) Varable: A varable whch cannot take all possble values between two gven values, s called a dscontnuous or dscrete varable. Ex. () Number of members n a famly () Number of workers n a factory Such varables cannot take any value between 1 and, and 3, etc. 3. Important Terms Range: The dfference between the maxmum and mnmum values of a varable s called ts range. Varate: A partcular value of a varable s called varate. Presentaton of Data: Puttng the data n condensed form n the form of a table, s known as presentaton of data. Frequency: The number of tmes an observaton occurs s called ts frequency. Frequency Dstrbuton: The tabular arrangement of data showng the frequency of each observaton s called ts frequency dstrbuton. 4. Raw or Ungrouped Data The data obtaned n orgnal form are called raw data or ungrouped data. Ex. The marks obtaned by 5 students n a class n a certan examnaton are gven below: 5, 8, 37, 16, 45, 40, 9, 1, 4, 40, 5, 14, 16, 16, 0, 10, 36, 33, 4, 5, 35, 11, 30, 45,48. Ths s the raw data. Array: An arrangement of raw data n ascendng or descendng order of magntude s called an array. Arrangng the marks of 5 students n ascendng order, we get the followng array. 8, 10, 11, 1, 14, 16, 16, 16, 0, 4, 5, 5, 5, 9, 30, 33, 35, 36, 37, 40, 40, 4, 45, 45, 48.

3 Foundaton for Mathematcs To Prepare Frequency Dstrbuton Table for Raw Data Usng Tally Marks We take each observaton from the data, one at a tme, and ndcate the frequency (the number of tmes the observaton has occurred n the data) by small lnes, called tally marks. For convenence, we wrte tally marks n bunches of fve, the ffth one crossng the fourth dagonally. In the table so formed, the sum of all the frequences s equal to the total number of observatons n the gven data. Ex. The sale of shoes of varous szes at a shop, on a partcular day s gven below: The above data s clearly raw data. From ths data, we may construct a frequency table, as gven below: Frequency Table Sze Tally Marks Frequency 4 II IIII 4 7 IIII Total Mean of Ungrouped Data The average of numbers n arthmetc s known as the Arthmetc Mean or smply the mean of these numbers n statstcs. Sumof observatons Mean Numberof observatons The mean of n observatons x 1, x,..., x n s gven by (x1 x x 3... x n) x Mean, x n n where the symbol å, called sgma stands for the summaton of the terms.

4 4.4 Statstcs 4.3 Mean for an Ungrouped Frequency Dstrbuton Let n observatons consst of values x 1, x,..., x n of a varable x, occurrng wth frequences f 1, f,..., f n respectvely. Then, the mean of these observatons s gven by: (f1x1fx... fnx n) fx Mean, x (f f... f ) f 1 n Illustraton 1: The ages of 40 students n a class are gven below: Fnd the mean age of the class. Age (n years) Number of students We prepare the table as gven below: Age (n years) (x ) Number of students f f x f = 40 f x = 585 Mean age = fx 585 f 40 = years. 4.4 Medan of Ungrouped Data Medan: After arrangng the gven data n an ascendng or a descendng order of magntude, the value of the mddle-most observaton s called the medan of the data. Method for Fndng the Medan of an Ungrouped Data Arrange the gven data n an ncreasng or decreasng order of magntude. Let the total number of observatons be n. () If n s odd, then medan = n +1 th value of observaton.

5 Foundaton for Mathematcs 4.5 1n n () If n s even, then medan = th observaton 1thobservaton Illustraton : The weghts (n kg) of 10 chldren are: 4, 50, 36, 45, 33, 31, 5, 39, 46, 3 Fnd the medan weght. Arrangng the weghts n ascendng order, we have: 31, 3, 33, 36, 39, 4, 45, 46, 50, 5 Here, n = 10, whch s even Medan weght = th term 1thterm 1 81 = {39 + 4} kg = kg = 40.5 kg Hence, medan weght = 40.5 kg = 1 {5th term + 6th term} PLANCESS CONCEPTS Defnton Applcablty Relevance to the data set How to calculate Mean The mean s the arthmetc average of a set of numbers, or dstrbuton. The mean s used for normal dstrbutons. The mean s not a robust tool snce t s largely nfluenced by outlers. A mean s computed by addng up all the values and dvdng that score by the number of values. Medan The medan s descrbed as the numerc value separatng the hgher half of a sample, a populaton, or a probablty dstrbuton, from the lower half. The medan s generally used for skewed dstrbutons. The medan s better suted for skewed dstrbutons to derve at central tendency snce t s much more robust and sensble. The Medan s the number found at the exact mddle of the set of values. A medan can be computed by lstng all numbers n ascendng order and then locatng the number n the centre of that dstrbuton. Vabhav Gupta Gold Medalst, INPhO

6 4.6 Statstcs 5. Grouped Data To put the data n a more condensed form, we make groups of sutable sze, and menton the frequency of each group. Such a table s called a grouped frequency dstrbuton table. Class-Interval: Each group nto whch the raw data s condensed, s called a class-nterval. Each class s bounded by two fgures, whch are called class lmts. The fgure on the left sde of a class s called ts lower lmt and that on ts rght s called ts upper lmt. 5.1 Types of Grouped Frequency Dstrbuton. Exclusve Form (or Contnuous Interval Form): A frequency dstrbuton n whch the upper lmt of each class s excluded and lower lmt s ncluded, s called an exclusve form.. Ex. Suppose the marks obtaned by some students n an examnaton are gven. We may consder the classes 0 10, 10 0 etc. In class 0 10, we nclude 0 and exclude 10. In class 10-0, we nclude 10 and exclude 0. Inclusve Form (or Dscontnuous Interval Form): A frequency dstrbuton n whch each upper lmt as well as lower lmt s ncluded, s called an nclusve form. Thus, we have classes of the form 0 10, 11 0, 1 30 etc. In 0 10, both 0 and 10 are ncluded. 5. Important Terms Related to Grouped Data 5..1 Class Boundares or True Upper and True Lower Lmts () In the exclusve form, the upper and lower lmts of a class are respectvely known as the true upper lmt and true lower lmt. () In the nclusve form, the number mdway between the upper lmt of a class and lower lmt of the subsequent class gves the true upper lmt of the class and the true lower lmt of the subsequent class.. Thus, n the above table of nclusve form, we have: True upper lmt of class 1 10 s = 10.5, and, true lower lmt of class 11-0 s Smlarly, true upper lmt of class 11 0 s = 0.5, and, true lower lmt of class 1 30 s Class Sze The dfference between the true upper lmt and the true lower lmt of a class s called ts class sze. Trueupper lmt Truelower lmt 5..3 Class Mark of a Class = The dfference between any two successve class marks gves the class sze.

7 Foundaton for Mathematcs Method of Formng Classes of a Data. Determne the maxmum and mnmum values of the varate occurrng n the data.. Decde upon the number of classes to be formed.. Fnd the range,.e., the dfference between the maxmum value and the mnmum value. Dvde the range by the number of classes to be formed to get the class-sze. v. Be sure that there must be classes havng mnmum and maxmum values occurrng n the data. v. By countng, we obtan the frequency of each class. 5.4 Rule to Convert Dscontnuous (or Inclusve) form to Contnuous (or Exclusve) Form In a dscontnuous nterval or nclusve form, we have : Adjustment factor = 1 [(Lower lmt of one class Upper lmt of prevous class)] Thus, f the classes are 1 10, 11 0, etc., then adjustment factor = 1 (11 10) = 0.5. To convert data gven n dscontnuous form to the contnuous form, we subtract the adjustment factor from each lower lmt and add the adjustment factor to each upper lmt to get the true lmts. Illustraton 3: Convert the followng frequency dstrbuton from dscontnuous to contnuous form: Marks (Class-ntervals) Frequency Adjustment factor = 1 (11 10) = 0.5. Subtract 0 5 from each lower lmt and add 0 5 to each upper lmt. Then, the requred table n contnuous form may be prepared as under: Marks (before adjustment) Marks (after adjustment) Frequency

8 4.8 Statstcs Marks (before adjustment) Marks (after adjustment) Frequency Total 5 6. Cumulatve Frequency of a Class-Interval The sum of the frequences of all the prevous classes and that partcular class, s called the cumulatve frequency of the class. 6.1 Cumulatve Frequency Table A table whch shows the cumulatve frequences over varous classes s called a cumulatve frequency dstrbuton table. Illustraton 4: Followng are the ages (n years) of 360 patents, gettng medcal treatment n a hosptal. Age (n years) Number of Patents Construct the cumulatve frequency table for the above data. The cumulatve frequency table for the above data s gven below. Class nterval Frequency Cumulatve Frequency Ths table may be presented n less than form, as under. Age (n years) Number of Patents Frequency Less than Less than Less than Less than Less than Less than

9 Foundaton for Mathematcs Graphcal Representaton of Statstcal Data The tabular representaton of data s an deal way of presentng them n a systematc manner. When these numercal fgures are represented pctorally or graphcally, they become more notceable and easly ntellgble, leavng a more lastng effect on the mnd of the observer. Wth the help of these pctures or graphs, data can be compared easly. There are varous types of graphs. In ths chapter, we shall be dealng wth the followng graphs: 1. Bar Graphs. Hstogram 3. Frequency Polygon 7.1 Bar Graph (or Column Graph or Bar Chart) A bar graph s a pctoral representaton of numercal data n the form of rectangles (or bars) of equal wdth and varyng heghts. These rectangles are drawn ether vertcally or horzontally. The heght of a bar represents the frequency of the correspondng observaton. The gap between two bars s kept the same. Illustraton 5: The followng table shows the number of students partcpatng n varous games n a school. Crcket Tenns Football Badmnton Draw a bar graph to represent the above data. Take the games along x-axs and the number of students along Y-axs. Along y-axs, take the scale 1 cm = 6 students. The bar-graph may, thus, be drawn as shown alongsde. 7. Hstogram A hstogram s a graphcal representaton of a frequency dstrbuton n an exclusve form n the form of rectangles wth class ntervals as bases and the correspondng frequences as heghts, there beng no gap between any two successve rectangles.

10 4.10 Statstcs 7..1 Method of Drawng a Hstogram Step 1: If the gven frequency dstrbuton s n nclusve form, convert t nto an exclusve form. Step : Takng sutable scales, mark the class-ntervals along x-axs and frequences along y-axs. Note that the scales chosen for both the axes need not be the same. Step 3: Construct rectangles wth class-ntervals as bases and the correspondng frequences as heghts. Illustraton 6: Draw a hstogram to represent the followng data: Class nterval Frequency Draw rectangles wth bases 30 36, 36 4, 4 48, and and heghts 15, 5, 0, 30 and 10 respectvely. Note: Snce the scale on x-axs starts at 30, we make a knd n the begnnng. 7.3 Frequency Polygon Let x 1, x,..x n be the class marks (.e., md ponts) of the gven frequency dstrbuton and let f 1, f,... f n be the correspondng frequences. We plot the ponts (x 1, f 1 ), (x, f )... (x n, f n ) on a graph paper and jon these ponts by lne segments. We complete the dagram n the form of a polygon by takng two more classes (called magned classes), one at the begnnng and the other at the end, each wth frequency zero. Ths polygon s known as the frequency polygon of the gven frequency dstrbuton. Illustraton 7: Draw the frequency polygon representng the followng frequency dstrbuton. Class nterval Frequency

11 Foundaton for Mathematcs 4.11 Though the gven frequency table s n nclusve form, yet we fnd that class marks n case of nclusve and exclusve forms are the same. We take the magned classes 5 9 at the begnnng and at the end, each wth frequency zero. Thus, we have: Class nterval Class Mark Frequency Now plot the ponts (7, 0), (3, 1), (37, 16), (4, 0), (47, 8), (5, 10), (57, 4) and (6, 0) and jon them successvely to obtan the requred frequency polygon, as shown below: 7.4. Hstogram and Frequency Polygon on the Same Graph When a hstogram and a frequency polygon are to be drawn on the same graph, we frst draw the hstogram wth the gven data. We then jon the md-ponts of the tops of adjacent rectangles by lne segments to obtan the frequency polygon. Illustraton 8: The followng table gves the number of doctors workng n government hosptals n a cty n varous age groups. Draw a hstogram and frequency polygon for the gven data. Age (n years) Number of doctors Step-1: Draw rectangles wth bases 0 5, 5 30, 30 35, and and heghts 40, 60, 50, 0 and 10 respectvely. Snce the scale on x-axs starts at 0, we make a knk n the begnnng. Thus, we obtan the requred hstogram. Step-: Mark the md-pont of the top of each rectangle of the hstogram. Step 3: Mark the md-ponts of class-ntervals 15-0 and on x-axs.

12 4.1 Statstcs Step-4: Jon the consecutve md-ponts by lne segments to obtan the requred frequency polygon. 8. Mean of Grouped Data To fnd mean of grouped Data we have three methods () Drect Method () Assumed Mean Method () Step Devaton Method 8.1 Drect Method STEP-1: Fnd the class mark (x ) of each class usng, x = STEP-: Calculate f x for each STEP-3: Use the formula: mean, x = n 1 n 1 fx f, lower lmt Upper lmt 8. Shortcut Method or Assumed Mean Method In ths case, to calculate the mean, we follow the followng steps: lower lmt Upper lmt STEP-1: Fnd the class mark x of each class usng, x = STEP-: Choose a sutable value of x n the mddle as the assumed mean and denote t by a. STEP-3: Fnd d = x a for each STEP-4: Fnd f d for each STEP-5: Fnd N = f STEP-6: fd Calculate the mean, ( x ) by usng the formula x = a +. N

13 Foundaton for Mathematcs Step-Devaton Method Sometmes, the values of x and f are so large that the calculaton of mean by assumed mean method becomes qute nconvenent. In ths case, we follow the followng steps: lower lmt Upper lmt STEP-1: Fnd the class mark x of each class by usng x = STEP-: Choose a sutable values of x n the mddle as the assumed mean and denote t by a. STEP-3: Fnd h = (upper lmt lower lmt) for each class. STEP-4: x a Fnd u = h for each class. STEP-5: Fnd f u for each. fu STEP-6: Calculate, the mean by usng the formula x = a + h, where N = f N. Illustraton 9: Fnd the mean of the followng data: Class Interval Frequency We may prepare the table as gven below: Class nterval Frequency (f ) Class mark (x ) f x f =40 f x =856 Mean, x = n 1 n 1 fx f = = 1.4 Illustraton 10: Fnd the mssng frequences f 1 and f n the table gven below, t s beng gven that the mean of the gven frequency dstrbuton s 50. Class Total Frequency 17 f 1 3 f 19 10

14 4.14 Statstcs We may prepare the table as gven below: Class Frequency (f ) Class mark (x ) f x f f f 70 70f Mean, x = f =68 +f 1 +f f x = f 1 +70f fx f 70f f = 1 68 f1 f Gven, mean = f1 70f 50 = 68 f f 1 0f 1 0f = 80 f 1 f = 4...() And f = 68 + f 1 + f 10 = 68 + f 1 + f [ f 1 = 10] f 1 + f = 5...() Addng (1) and (), we get f 1 = 56 f 1 = 8 f = f f = f f Hence, the mssng frequences f 1 and f are 8 and 4 respectvely. Illustraton 11: Fnd the mean marks from the followng data: Marks No. of Students Below Below Below Below Below Below Below Below Below Below We may prepare the table as gven below: Marks No. of students Class Interval f Class mark (x ) f x Below Below

15 Foundaton for Mathematcs 4.15 Marks No. of students Class Interval f Class mark (x ) f x Below Below Below Below Below Below Below Below Mean = x = fx = 4140 N 85 = Medan of a Grouped Data N=85 f x =4140 Medan: It s a measure of central tendency whch gves the value of the mddle most observaton n the data. In a grouped data, t s not possble to fnd the mddle observaton by lookng at the cumulatve frequences as the mddle observaton wll be some value n a class nterval. It s, therefore, necessary to fnd the value nsde a class that dvdes the whole dstrbuton nto two halves. Medan Class: The class whose cumulatve frequency s greater than N s called the medan class. To calculate the medan of a grouped data, we follow the followng steps: STEP-1: Prepare the cumulatve frequency table correspondng to the gven frequency dstrbuton and obtan N = f. STEP-: Fnd N STEP-3: Look at the cumulatve frequency just greater than and fnd the correspondng class STEP-4: (Medan class). N C c Use the formula Medan, M = l + h f Where, l = Lower lmt of medan class. f = Frequency of the medan class. c = Cumulatve frequency of the class precedng the medan class.

16 4.16 Statstcs h = Sze of the medan class. n = f. Illustraton 1: The length of 40 leaves of a plant are measured correct to the nearest mllmeter, and the data obtaned s represented n the followng table. Fnd the medan length of the leaves. Length (n mm) No. of leaves The gven seres s n nclusve form. We may prepare the table n exclusve form and prepare the cumulatve frequency table as gven below: Length (n mm) No. of leaves (f ) Cumulatve frequency N=40 Here, N = 40, N = 0 The cumulatve frequency just greater than 0 s 9 and the correspondng class s So, the medan class s l = 144.5, N = 40, c = 17, f = 1 and h = 9 N C c Therefore, medan = l + (0 17) h = = = f 1 1 = Hence, medan length of leaves s mm. 10. Mode of a Grouped Data Mode: Mode s that value among the observatons whch occurs most often.e. the value of the observaton havng the maxmum frequency.

17 Foundaton for Mathematcs 4.17 In a grouped frequency dstrbuton, t s not possble to determne the mode by lookng at the frequences. Modal Class: The class of a frequency dstrbuton havng maxmum frequency s called modal class of a frequency dstrbuton. The mode s a value nsde the modal class and s calculated by usng the formula. f 1 f 0 Mode = l + h f1 f0 f Where l = Lower lmt of the modal class. h = Sze of class nterval f 1 = Frequency of modal class f 0 = Frequency of the class precedng the modal class f = Frequency of the class succeedng the modal class. Illustraton 13: Gven below s the frequency dstrbuton of the heghts of players n a school. Heghts (n cm) No. of students Fnd the average heght of maxmum number of students. The gven seres s n nclusve form. We prepare the table n exclusve form, as gven below: Heghts (n cm) No. of students We have to fnd the mode of the data. Here, the class has maxmum frequency, so t s the modal class. Illustraton 14: The mode of the followng seres s 36. Fnd the mssng frequency f n t. Class Frequency 8 10 f Snce the mode s 36, so the modal class wll be l = 30, h = 10, f 1 = 16, f 0 = f and f = 1 f 1 f 0 Therefore, mode = l + h f1 f0 f 16 f (16 f) 36 = = f 1 (0 f) 10 6f = f 4f = 40 f = 10 Hence, the value of the mssng frequency f s 10.

18 4.18 Statstcs 11. Graphcal Representaton of Cumulatve Frequency Dstrbuton 11.1 Cumulatve Frequency Polygon Curve (Ogve) Cumulatve frequency s of two types and correspondng to these, the ogve s also of two types. (A) Less Than Seres (B) More Than Seres (a) Less Than Seres To construct a cumulatve frequency polygon and an ogve, we follow these steps: STEP-1: Mark the upper class lmt along x-axs and the correspondng cumulatve frequences along y-axs. STEP-: Plot these ponts successvely by lne segments. We get a polygon, called cumulatve frequency polygon. STEP-3: Plot these ponts successvely by smooth curves, we get a curve called cumulatve frequency curve or an ogve. (b) MORE THAN SERIES To construct a cumulatve frequency polygon and an ogve, we follow these steps: STEP-1 : Mark the lower class lmts along x-axs and the correspondng cumulatve frequences along y-axs. STEP- : Plot these ponts successvely by lne segments, we get a polygon, called cumulatve frequency polygon. STEP-3 : Plot these ponts successvely by smooth curves, we get a curve, called cumulatve frequency curve or an ogve. 11. Applcaton of an Ogve Ogve can be used to fnd the medan of a frequency dstrbuton. To fnd the medan, we follow these steps. Method I STEP-1: Draw anyone of the two types of frequency curves on the graph paper. N STEP-: Compute (N f ) and mark the correspondng ponts on the y-axs. STEP-3: Draw a lne parallel to x-axs from the pont marked n step, cuttng the cumulatve frequency curve at a pont P. STEP-4: Draw perpendcular PM from P on the x-axs. The x - coordnate of pont M gves the medan. Method -II STEP-1: Draw less than type and more than type cumulatve frequency curves on the graph paper.

19 Foundaton for Mathematcs 4.19 STEP-: Mark the pont of ntersectng (P) of the two curves drawn n step 1. STEP-3: Draw perpendcular PM from P on the x-axs. The x- coordnate of pont M gves the medan. Illustraton 15: The followng table gves producton yeld per hectare of wheat of 100 farms of a vllage. Producton yeld (n kg/ha) No. of farms Change the dstrbuton to more than type dstrbuton and draw ts ogve. From the gven table, we may prepare more than type cumulatve frequency dstrbuton table, as gven below: Producton more than (n kg/ha) Cumulatve frequency Now, plot the ponts (50,100), (55,98), (60,90), (65,78), (70,54) and (75,16). Jon these ponts by a freehand to get an ogve of more than type. PLANCESS CONCEPTS Emprcal Relatonshp between Mean, Medan and Mode Followng the relaton : Mode = 3 Medan Mean Medan = Mode + (Mean Mode) 3 Mean = Mode + 3 (Medan Mode) Symmetrc Dstrbuton: A dstrbuton s sad to be symmetrc f Mean = Medan = Mode Uday Kran G KVPY Fellow

20 4.0 Statstcs Illustraton 16: Fnd the mean of the followng dstrbuton. Class Number of Students Class (f ) Number of Students (x ) Md-value Product = f x = = = = = = = = 10 Total N = Mean = M = 1 nn fx where 8 fx = 199 and n = M = = 19.9 Mean of the requred dstrbuton = M = f x 199

21 Foundaton for Mathematcs 4.1 SUMMARY We lve n the age of nformaton. There s data all around us but t s meanngful only f we try to make sense out of t. Ths s exactly what makes Statstcs so mportant. It s a tool we need to make sense of data. And for that, t s mportant to know what knd of data t s and what nformaton can be extracted out of t. Ths s exactly what we have learnt n ths chapter. So try to apply what we have learnt n your day to day stuatons. You could try to statstcally analyze the next crcket match you watch!

22 4. Statstcs SOLVED EXAMPLES Example 1: The arthmetc mean of the followng frequency dstrbuton s 5.5. Fnd the value of p. Classes Frequency p Classes Frequency (f ) Md-value (x ) Product = Freq. M.Value f x = = = p 70 p 70 = 70 p = 1890 Total N = 95 + p 5 M = Mean of the dstrbuton = nn fx = ( p) = p p 55 p = = p = 4375 p = 5 Hence, the value of p s 5. 1 f x p

23 Foundaton for Mathematcs 4.3 Example : Calculate the average heght of each student n a class from the followng frequency dstrbuton. Heght (n cm) No. of Students Here, the class nterval (.e. heght nterval) s same (= 10) and md-values of t wll be of very large values rangng from 145 to 195. So, to avod lengthy calculaton. So devaton method s used to fnd the mean. Heght n cm () Md-values Step Devaton No. of students = f fd s (x) d s = x A =A f = 5 = N fd s = = 7 Here, A = assumed mean = 165 (arbtrarly chosen near mddle of the gven class ntervals) Usng the formula, fds Mean = A + = = cm. N 5 Hence the average heght of each student of the class s cm. Example 3: The daly wages (n rupees) of 100 workers n a factory are gven below: Daly wages (n Rs.) No. of workers

24 4.4 Statstcs Fnd the medan wage of a worker for the above date. Daly wages (n Rs.) No. of workers Cumulatve Frequency N = 100 (even) th th N N Medan = observaton 1 observaton = th 50 observaton 51 observaton th = Medan wage of a workers n the factory s Rs Example 4: Fnd the medan of the followng dstrbuton = Marks Frequency Let us construct the cumulatve frequency table whch s gven below:

25 Foundaton for Mathematcs 4.5 Marks Frequency (f) Cumulatve Frequency cf cf = f = 0 65 Medan Class N Here, N = total frequency = 100, calculate n 100 = = 50 Cumulatve frequency just greater than 50 s 65, the span of the group correspondng to 65 s Therefore, s the medan class. l = the lower lmt of the medan class = 500 cf = the cumulatve frequency of the class proceedng the medan class = 45 f = the frequency of the medan class ( ) = 0 h = the wdth of the medan class = 100 Usng the formula Nn Medan = l cf h = f 0 5 = = = 55 0 Hence, the medan of the gven dstrbuton s 55 marks. Example 5: Fnd the mssng frequency n the followng dstrbuton table (by usng medan formula) f N = 5 and medan = 36 Class nterval Frequency

26 4.6 Statstcs Class nterval Frequency 0 30 x We prepare the cumulatve frequency table as under: Class Interval Frequency (f ) Cumulatve Frequency x cf = 15 + x f = x Medan Class x x x Total N = 5 It s gven that the medan 36 les n the nterval l = lower lmt of the medan class = 30 cf = cumulatve frequency of the medan just before the medan class = 15 + x h = the wdth of the class nterval = 10 f = frequency of the medan class = 10 N = total frequency = 5 nn 5 30 n Medan = l cf h f cf 15 x f 10 h 10

27 Foundaton for Mathematcs (15 x) Medan = = (30 + 6) (15 + x) 36 = 56 (15 + x) [Q Medan = (Gven)] 15 + x = = x = 0 x = 0 15 = 5 Mssng frequency = x = 5 (Answer) Example 6: The followng table shows the marks obtaned by 100 students of class X n a school durng a partcular academc sesson. Fnd the mode of ths dstrbuton. Marks less than Number of students Snce the gven dstrbuton s cumulatve frequency dstrbuton, we frst convert t nto smple frequency dstrbuton. Marks less than Number of students = = = f 0 = = f 1 = = 0 modal class f = = = = 08

28 4.8 Statstcs Here the maxmum frequency s 0, and the class correspondng to ths frequency s So the modal class s h = class sze or wdth of the nterval = 10 l = lower lmt of the modal class = 40 f 1 = frequency of the modal class = 0 f 0 = frequency of the class proceedng the modal class = 1 f = frequency of the class succeedng the modal class = 11 Usng the formula, f1 f 0 Mode = l + f1 f0 f h Mode = = Modal class = = = Hence, the mode of the dstrbuton s marks. Example 7: Fnd the medan, from the followng data: 5, 7, 9, 11, 13, 15, 17. The data s already s ascendng order. Snce, n = 7 (odd) th n 1 Medan = term = 4th term = 11. Example 8: Calculate the medan for the followng dstrbuton class: Class Frequency th N 55 th () Frst we fnd value.e. 7.5 whch les n 0-30 Here l 0, N 7.5,c.f. 15,f 0,h medan0 10 Medan th Class Frequency c.f

29 Foundaton for Mathematcs 4.9 Class Frequency c.f Example 9: A famly requres the commodtes lsted n the table below for regular use. The mportance (weghts) attached to each commodty s also gven. Fnd the dfference n the mean and the weghted mean prce per kg. Commodty Weght (kg) Prce /kg (Rs) Rce Wheat 3.75 Pulses 8.50 Vegetable Ols For calculatng weghted mean and mean prce per kg, we prepare the followng table: Commodty Weght (w) (kg) Prce /kg (n Rs)(x) Product (wx) Rce Wheat Pulses Vegetable Ols Total Mean 10.55; weght mean 8.34 kg 5 5 Dfference Dfferencen prce per kgrs.1 Example 10: Fnd the mean of the followng dstrbuton: Class nterval Frequency

30 4.30 Statstcs Class nterval Frequency (f ) Mdpont (x ) x A u h fu (A) Total 90 6 Mean fu 6 5 A h f 90 9 Example 11: Construct a hstogram from the dstrbuton of total marks obtaned by 65 students of X class n the fnal examnaton. Marks (md-ponts) Number of females Lower and upper class lmts: Snce the dfference between the second and frst md-pont s =10 h h 10 5 So, lower and upper lmts of the frst class are and e. 145 and 155 respectvely. Frst class nterval s Usng the same procedure, we get the classes of other md-ponts us under:

31 Foundaton for Mathematcs 4.31 Class frequency Example 1: The number of students admtted n dfferent facultes of a college are Faculty Scence Arts Commerce Law Educaton Total No. of students Construct the pe chart for the followng dstrbuton. The followng table gves the share of each faculty as a component of Faculty No. of students Share as a component of Scence Arts Commerce Law Educaton Example 13: Fnd the mean for the followng dstrbuton: Marks Frequency Items Md values x No. of students f f x

32 4.3 Statstcs Items Md values x No. of students f f x f 40 f x 1430 fx x f 40 4 Example 14: Fnd the mean followng dstrbuton wth step-devaton method: class Frequency Calss x f x 7.5 u 5 fu (A) f 40 fu 3 x A h x f 40 Example 15: Fnd the mode of the followng dstrbuton: fu 3 Daly wages No. of workers Daly wages No. of workers Daly wages No. of workers

33 Foundaton for Mathematcs 4.33 Daly wages No. of workers Daly wages No. of workers * Modal class frequency s l 4.5, f1 0, f0 1, f 15, h Mode 4.5 6, Mode 46. Example 16: The followng frequency dstrbuton gves the monthly consumpton of electrcty of 68 consumers of a localty. Fnd the mean, medan and mode of the data. Monthly consumpton (n unts) Number of consumers Class nterval f Cumulatve frequency x u x 75 5 fu

34 4.34 Statstcs Class nterval f Cumulatve frequency x u x 75 5 fu (A) f 68 fu 7 Mean x A h f 68 Nn s the medan class. n c.f 34 Medan = L h f 0 Maxmum frequency s s the modal class. fu 7 Mode = L 0 f f 0 13 h f1 f0 f

35 Foundaton for Mathematcs 4.35 EXERCISE 1 For School Examnatons Fll n the Blanks Drectons: Complete the followng statements wth an approprate word/term to be flled n the blank space(s). Q.1. The algebrac sum of the devatons from arthmetc mean s always. Q.. Percentle dvdes the number of tems nto equal parts. Q.3. The mean- devaton from the medan s than that measured from any other value. Q.4. The dfference between the maxmum and the mnmum observatons n data s called the of the data. Q.5. The md-pont of a class nterval s called ts. Q.6. Value of the mddle-most observaton (s) s called. True / False Drectons: Read the followng statements and wrte your answer as true or false. n cf Q.7. The medan for grouped data s formed by usng the formula, Medan = L h f True False Q.8. The medan of grouped data wth unequal class szes cannot be calculated. True False Q.9. The modal value s the value of the varate whch dvdes the total frequency nto two equal parts. True False Q.10. (Medan -Mean)=Mode-Mean. True False Q.11. Mean may or may not be the approprate measure of measure of central tendency. True False Match the Followng Columns Drectons: Each queston contans statements gven n two columns whch have to be matched. Statements (A, B, C, D) n column I have to be matched wth statements (p, q, r, s) n Column II Column II gves data for descrpton gven n column I, match them correctly.

36 4.36 Statstcs Q.1. Column I Column II (A) the drect method (p) fu x a h f (B) step devaton method (C) mode fl f0 (q) l f f f 1 0 n cf (r) l h f h (D) medan (s) fx x f Q.13. The table shows a frequency dstrbuton of the lfe tme of 400 rado tubes tested at a company. Lfe tme (hours) Number of tubes Total 400 Column I Column II (A) upper lmt of the ffth class (p) 100 (B) Lower lmt of the eght class (q) 949.5

37 Foundaton for Mathematcs 4.37 Column I Column II (C) class marks of the seventh class (r) 1000 (D) class nterval sze (s) 799 Very Short Answer Questons Drectons: Gve answer n one word or one sentence. Q.14. In a school 85 boys and 35 grls appeared n a publc examnaton. The mean marks of boys was found to be 40%, whereas the mean marks of grls was 60% Determne the average marks percentage of the school. Q.15. The mean of 1 observatons s 14. By an error one observaton s regstered as 4 s read of -4. Fnd. Fnd the actual mean. Q.16. The mean weght of 0 students s 5 kg and the mean weght of another 10 students s 40 kg. Fnd the mean weght of the 30 students. Short Answer Questons Drectons: Gve answer n two to three sentences. Q.17. The mean of 50 numbers s 40. It was found that three numbers 3, 9 and 0 were taken as 8, 9 and. Fnd the correct mean. Q.18. Gven n (x 3n) 84 and n (x n) 144, fnd n and the mean. I I Q.19. The mean of n observaton x 1,x,x 3,...x n s x. If each observaton s multpled by p, prove that the mean of the new observatons s px Q.0. The mean wages of 1000 workers n a factory was (INR) 600. The mean wages of 00 workers of the nght was (INR) Fnd the mean wages of the remanng 800 workers of the day shft. Long Answer Questons Drectons: Gve answer n four to fve sentences. Q.1. There are sx numbers. Combnatons of 3 numbers are selected and ther mean was calculated. The resultng means were, 4, 6,.36, 38, 40. What s the average of the orgnal sx numbers? Q.. If the mean of the followng table s 30, fnd the mssng frequences. Class nterval Total Frequency 10 a b 8 60

38 4.38 Statstcs Q.3. The mean of the followng frequency table s 50. But the frequences f 1 and f n class 0-40 and are mssng. Fnd the mssng frequences. Class Total Frequency 17 f 3 1 f Q.4. A survey regardng the heghts (n cm) of 51 grls of class X of a school was conducted and the followng data was obtaned: Heght (n cm) Number of grls Less than Less than Less than Less than Less than Less than Fnd the medan heght and nterpret the result.

39 Foundaton for Mathematcs 4.39 EXERCISE For Compettve Examnatons Multple Choce Questons Drectons: Ths secton contans multple choce questons. Each queston has 4 choces (a), (b), (c) and (d) out of whch ONLY ONE s correct. Q.1. The mean of dscrete observatons y 1,y,...yns gven by Q.. Q.3. Q.4. Q.5. Q.6. (a) (c) n l n n l n y yf If the mean of the numbers 7 +x, 31+x,89+x,107+x,156+x s 8, then the mean of 130+x, 16+x, 68+x, 50+x,1+x s (a) 75 (b) 157 (c) 8 (d) 80 The number of observatons n a group s 40. If the average of frst 10 s 4.5 and that of the remanng 30 s 3.5, then the average of the whole group s (a) 1/5 (b) 15/4 (c) 4 (d) 8 In a class of 100 students there are 70 boys whose average marks n a subject are 75. If the average marks of the complete class are 7, then the average marks of the grls (a) 73 (b) 65 (c) 68 (d) 74 The numbers 3, 5, 7, and 9 have respectve frequences x-, x+, x-3 and x+3. If the arthmetc mean s 6.5 then the value of x s (a) 3 (b) 4 (c) 5 (d) 6 If the sets of data had means of 15,.5 and 4 based on 6, 4, and 5 observatons respectvely, then the mean of these three sets combned s (a) 0.0 (b) 0.5 (c).5 (d) 4.0 (b) (d) n l n n l l n l y yf f

40 4.40 Statstcs Q.7. Consder the followng statements (1) Mode can be computed from hstogram () Medum s not ndependent of change of scale (3) Varance s ndependent of change of orgn and scale. Whch of these s /are correct? (a) (1), () and (3) (b) Only () (c) Only (1) and () (d) Only (1) More than One Correct Drectons: Ths secton contans multple choce questons. Each queston has 4 choces (a), (b), (c) and (d) out of whch ONLY OR MORE may be correct. Q.8. Whch of the followng s/are correct? upper class lmt lower class lmts (a) Class mark= f1 f 0 (b) Mode l h f1 f0 f (c) medan =mode + mean th n 1 (d) If n s odd, then medan = term where, n s the number of terms. Passage Based Questons Drectons: Study the gven passage (s) and answer the followng questons. Q.9. Passage I The followng table gves the weekly wages of workers n an factory: Weekly wages Md-value (x ) No. of (f ) workers f 80 fx f x 550 Cumulatve frequency

41 Foundaton for Mathematcs The mean s.. (a) 70 (b) 68 (c) 71 (d) 69 The modal class s (a) (b) (c) (d) none of these The number of workers gettng weekly wages, below (INR) 80 s (a) 50 (b) 70 (c) 60 (d) 80 Q.10. Passage II The marks of 0 students n a test were as follows: 5, 6, 8, 9, 10, 11, 11, 1, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 0. The mean (a) 14 (b) 13 (c) 1 (d) 15. The medan s (a) 13.5 (b) 1.5 (c) 14.5 (d) The modal s (a) 0 (b) 10 (c) 15 (d) 5 Asserton and Reason Drectons: Each of these questons contans an Asserton followed by Reason. Read them carefully and answer the questons on the bass of followng optons. You have to select the one that best descrbes the two statements. a) If both asserton and reason are correct and reason s the correct explanaton of asserton. b) If both asserton and reason are correct, but reason s not the correct explanaton of asserton. c) If asserton s correct but reason s ncorrect. d) If asserton s ncorrect but reason s correct. Q.11. Asserton: If the value of mode and mean s 60 and 66 respectvely, then the value of medan s 64. Reason: Mean =1/ (mode + mean)

42 4.4 Statstcs Subjectve Questons Drectons: Answer the followng questons. Q.1. Durng the medcal check-up of 35 students of a class, ther weght were recorded as follows: Weght (n kg) Less than 38 0 Less than 40 3 Less than 4 5 Less than 44 9 Less than Less than 48 8 Less than 50 3 Less than 5 35 Number of students Draw a less than type gve for the gven data. Hence obtan the medan weght from the graph and verfy the result by usng the formula. Q.13. The followng table shows the ages of the patents admtted n a hosptal durng a year: Ages (n year) Number of patents Fnd the mean of the data gven above. Compare and nterpret the two measure of central tendency. Q.14. Gve below s a frequency dstrbuton wth medan 46. In ths dstrbuton, some of the frequences are mssng: Determne the mssng frequences. Marks ? ? Total 9 No. of students

43 Foundaton for Mathematcs 4.43 Q.15. The adjonng pe chart the marks scored n an examnaton by a student n Englsh, Hnd Scence, socal Scence and Mathematcs. If the total marks obtaned by the student were 540, answer the followng questons. (a) In whch subject scored 105 marks? (b) How many more marks were obtaned by the student n Mathematcs than n Hnd? (c) Examne whether the sum of marks obtaned n socal Scence and mathematcs s more than that n Scence and Hnd.

44 4.44 Statstcs SOLUTIONS EXERCISE 1 For School Examnatons Fll n the Blanks 1. Zero. Hundred 3. Less 4. Range 5. Class-mark 6. Medan True / False 7. True 8. False 9. False 10. False 11. True Match the Followng Columns 1. (A) s (B) p (C) q (D) r 13. (A) s (B) r (C) q (D) p Very Short Answer Questons 14. n 1.x1 n.x 5500 xc n n kg Short Answer Questons 17. Correct mean = 39 n 18. x 3n x x...x 3n 84 1 n 1 1 n S 3n 84...(1) S x x...x 1 n x n x x x...x n n S n 144 From (1) and (), we get S=10, n=1 Mean S n 1 x1 x x 3...xn 19. Accordng to queston, x...(1) n x1p xp x3p... xnp New observaton are x1p,xp,x3p,...xnp ; New mean n

45 Foundaton for Mathematcs 4.45 p x x x... x n 1 3 n p.x [From (1)] 0. Total wages of 1000 worker= Total wages of 00 nght shft workers Total wages of 800 day shft workers Mean wages of the day shft workers (INR) Long Answer Questons Let the sx numbers be x 1,x,x 3,x 4,x 5,x 6. Number of combnatons of 3 C The mean of these 0 combnatons are, 4, 6,..36, 38, 40 Sum of all these means In the above sum, each number from the orgnal sx s repeated 10 tmes. 10 x1x x3 x4 x5 x x1x x3 x4 x5 x x1x x3 x4 x5 x , 4 3. Let the assumed mean be A=50 and h=0. Calculatng of mean x Class Frequency f Md-values x 50 u 0 fu f f f 70 1 f N f 68 f f 1 fu 4 f f 1 We have, N f 10 (Gven )

46 4.46 Statstcs 68 f1 f f1 f5...() Now, mean f1 f A h fu N 0 4f1 f 4f1 f f f 0 f f 4...() 1 1 Solvng equatons () and (), we get f 1 =8 and f =4. 4. To calculate the medan heght, we need to fnd the class ntervals and ther correspondng table wth the gven cumulatve frequences s gven below: Class ntervals Frequency Cumulatve frequency N Here, N=51. So, n Ths observaton les n the class Then, l (the blower lmt)=145, c.f (the cumulatve frequency of the class precedng )=11, f(the frequency of the medan class )=18, h (the class sze)=5. Usng the formula, Medan Nn c.f l h f Ths means that the heght of 50% grl s less than the heght cm and 50% are taller than the heght cm.

47 Foundaton for Mathematcs 4.47 EXERCISE For Compettve Examnatons Multple Choce Questons 1. (a). (a) Gven 7 x 31 x 89 x 107 x 156 x x x 0 Requred mean s 130 x 16 x 68 x 50 x 1 x x x x (b) 4. (b) Let the average marks of the grl students be x, then x (Number of grls ).e., x x (c) 6. (a) 7. (d) More than One Correct 8. (a), (b), (c) and (d) Passage Based Questons 9. Passage I. (d) Mean fx 550 Rs. 69 f 80. (b) Modal class: We know that class of maxmum frequency s called the modal class.e., s the modal class.. (c) Number of workers gettng weekly wages below (INR) 80 accordng to table =60 workers.

48 4.48 Statstcs 10. Passage II Arrange n ascendng order. (b) Mean=( )/ (a) Here, n =0 whch s an even number th th th th n n 0 0 term 1 term term 1 term Medan th th 10 term 11 term (c) In the data, 15 occurs the maxmum tmes.e., 3 tmes Mode =15 Asserton and Reason 11. (c) Medan= 1 (mode 3 mean) 3 1 (60 66) 64 3 Subjectve Questons 1. Use table cumulatve frequency for each less than the upper lmt. We mark the upper lmts od the class on X-axs and correspondng cumulatve frequency at Y-axs. We can draw the ogve by plottng the ponts: (38, 0),(40, 3), (4, 5), (44, 9), (40, 35), (46, 14), (48, 8), (50, 3) and (5, 35). N Here n Located the pont on the gve whose ordnate s The X-coordnate of ths pont wll be the medan. Ths fnd the medan by formula, we should convert less than type n contnues class nterval as gven below: Weght (n kg) Class frequency Cumulatve frequency Less than

49 Foundaton for Mathematcs 4.49 N Here N =35. So n 17.5x Medan class s , c.f. 8, f 14, h From formula, medan= l Nn cf h Medan 46 f Age (n year ) Number of patents (f ) Md- values (x ) f x Total Mean = fx years. f 80 Here, maxmum class frequency s 3, and the class correspondng to ths frequency s So, the modal class s Now, l 35, h 10, f1 3, f0 1, f 14

50 4.50 Statstcs 14. f Mode = l 1 f 0 0 h year. f1 f0 f Mode =36.8 year, Mean =35.37 years. Maxmum number of patents admtted n the hosptal are of the age 36.8 years (approx.), whle on an average the age of a patent admtted to the hosptal s years. Marks No. of students c.f ? 4+x x 50+60? 107+x+y x+y x+y Total 9 Let the mssng frequency correspondng to and class be x and y respectvely. N Here, n=total number of students= 9 n , and Medan 46 Medan class =40-50 and l 40, c.f. 4x,f 65,h 10 N n c.f. We know, Medan = l h x 10 f x or x 33.5 x 34 ( Number of students cannot be n fracton) Now f x y 150 9; x y (1) Puttng the value of x n (1), we get 34+y=79 ; y=79-34=45 x=34, y=45 Hence, the mssng frequency are 34 and 45 correspondng to class and class respectvely (a) For the total marks 540, central angle = For 105 marks, central angle = Snce, central angle of the representng Hnd s 70 0, hence the students scored 105 marks n Hnd.

51 Foundaton for Mathematcs 4.51 (b) The dfferent of central angles of the scores = = Correspondng dfference of marks= Hence the students scored 30, marks more n mathematcs than that n Hnd (c) The sum of central angles of the sectors representng Socal Scence and Mathematcs= =155 0 Snce > 150 0, so the sum of marks obtaned n Socal scence and mathematcs s more than that n Scence and Hnd.