Data Handling

DATA HANDLING TYPES OF DATA CONTENTS  Data (i) Primary data (ii) Secondary data  Types of Data (iii) Raw (ungrouped) data  Some Terms  Frequency Distribution Table  Central Tendency & Its Types  Arithmetic Mean of Tabulated Data  Median  Mode  Bar Graph   Probability 2. Sachin made highest runs in today's match Secondary data : The secondary data is not collected by observer. It is obtained from published or unpublished sources. Note: The Primary data is more reliable than the secondary data as the information collected by observer is more accurate in comparison to information collected by another person (other than observer). For example : 5, 7, 9, 11, 11, 15, 16, 19, 21, 22, 23, 24, 26, 29, 30. The collection of facts which are expressed numerically with the specific purpose is called data. or appropriate information and facts called data. Some people died in accident of two cars Primary data : The data which is collected by the observer is called Primary data. Grouped data : If representation of data are with some group, is called grouped data. DATA 1. (iv) Grouped data Raw data : The data which is collected for specific purpose and put as it is (without any arrangement) is called raw data. Each entry (number) in raw data is known as observation.   Double Bar Graph For example, 3 Three people died in accident of two cars. Sachin made 190 runs in today's match In grouped form : 0 – 10, 10 – 20, 20 – 30, etc. Each group is called class. SOME TERMS Array : To understand the data in better way, we arrange the data either in ascending order or descending order which is called an array. Range : The difference between the highest and lowest values of the given data is called range. For example, 0, 3, 8, 10, 10, 12, 18, 19, 19, 20 So, Range = 20 – 0 = 20 Frequency : The number of times a particular observation occurs is called its frequency. It is denoted by (fi). Ex. 2 A die is a cube where six faces are marked with numbers (or dots) from 1 to 6 one number on each face. The score obtained in 25 throws are 5, 4, 3, 2, 1, 1, 2, 5, 4, 6, 6, 6, 3, 2, 1, 4, 3, 2, 1, 5, 5, 6, 2, 1 and 3. Prepare a frequency table for the above scores. Sol. Arranging the data in ascending order as follows : Tally Marks : The way of representation of frequency numbers is called tally marks. One Six Two Seven Three Eight four Nine five Ten ,4,4 1 ,1, 1,1 ,1, 2 ,2 ,2, 2 ,2 3 ,3 ,3 ,3 4  5 times 5 ,5 ,5 ,5 6 ,6 ,6 ,6 4 times 1 EXAMPLES  The number of children in 20 families of a locality are given below : 2, 2, 3, 2, 1, 3, 3, 2, 2, 1, 2, 2, 3, 1, 2, 1, 1, 3, 2, 2 Represent this information in a frequency distribution table. Arranging the data in ascending order, we have 10 times Number of Children 1 Number of Families 5 2 5 + 5 = 10 3 5 Total 20 Frequency 5 5 3 4 4 3 5 4 6 4 Total 25 CENTRAL TENDENCY AND ITS TYPES Average is a number that represents or shows the central tendency of the given data. Types of central tendency 5 times Tally Marks Tally Marks 2 1 ,1, 1,1 ,1,2 ,2 ,2 ,2 ,2 ,2 ,2, 2,2 ,2,3 , 3, 3,3 ,3,   5 times 3 times 4 times Scores If we rearrange above data in the tabular form, showing the frequency of each observation, the tabular form of representation of the data is called frequency distribution and the table is called frequency table. Sol. 4 times The frequency distribution table is as under : FREQUENCY DISTRIBUTION TABLE Ex. 1 5 times Arithmetic mean or Mean Arithmetic mean : Mean = Median Mode Sum of the given observations (x) Number of the given observations (n) Ex.3 EXAMPLES  The maximum temperature recorded in a city during the last six months of a year are given below. Find the mean of the maximum temperature. ARITHMETIC MEAN OF TABULATED DATA Let x1, x2, x3, ...., xn are n observations and f1, f2, f3, ....., fn are their frequencies respectively. July Aug. Sep. Oct. Nov. Dec. Month Maximum 42 40 40 37 29 22 Temp. (in ºC) Sol. = Sol. Ex.5 Sol. Frequency (fi) xi × f i x2 f2 x2f2 x1 Total temp. = 42 + 40 + 40 + 37 + 29 + 22 = 210 Number of months = 6  Arithmetic mean = Ex.4 Observation (xi)  Mean = Thus, average or mean temperature = 35ºC. The runs scored by two batsman in their recent five matches are given as below : 81 75 71 69 74 Batsman-I 12 81 5 105 117 Batsman-II Who performed better ? Arithmetic mean of the scores of Batsman-I 81  75  71  69  74 370 = = = 74 5 5 Arithmetic mean of the scores of Batsman-II 12  81  5  105  117 320 = = = 64 5 5 The mean of the scores of Batsman-I is more than the mean of the scores of Batsman-II. So, the performance of Batsman-I is better. Note : The average or mean is helpful for making comparisons.  fn xnfn fi = N Total 210 = 35 6 x1f1  xn Sum of temperature Number of months The average or mean weight of 50 students of a class is 32 kg. Find the total weight of the students. Sum of all the items Mean = Number of items or Mean × Number of items = Sum of all the items For the given data, we have Number of item (students) = 50 Average or Mean weight = 32 kg From the formula we have Sum of all weights = (32 × 50) kg = 1600 kg f1 (f1x1  f 2 x 2  f 3 x 3  ...  f n x n ) (f1  f 2  f 3  ...  f n ) xifi ( (called sigma) shows addition.)  (f x ) n or x  i ( N) Here N = Ex.6 i i 1 f n i 1 i EXAMPLE  The ages (in years) of 50 players are given below : Age (in years)  Sol.  Number of players Find mean age. 24 14 25 15 26 3 Age in years (xi) No. of players (fi) 24 14 26 3 25 15 27 28 10 8 xi × f i 336 375 78 27 10 270 Total fi = 50 xiyi = 1283 28 Mean = 8 (x i f i ) 1283 = f i 50 Mean = 25.66 224 8 8    th term    1 th term 2 2  =   2 MEDIAN The another method for finding central tendency called median. = Definition : The median is that value of the given data, which divides it into exactly two parts.  Method : Arrange the data in ascending or descending order. Let the total number of observations be n. Case I : When n is odd  n 1  Median =   th term.  2  Ex.7 Sol. = Ex.9 Find the median of the following data : Sol. On arranging the data in ascending order, we get 67, 93, 94, 95, 97, 98, 100 Middle number is 4th  Median = 4th number = 95 First arranging in ascending order, we get Working rule to find median. Steps : 27, 28, 34, 35, 45, 46, 56 (i) Arrange the data descending order. Here, n = 7(which is odd)  7 1  Median =   th term  2  Median = Median = 35 Sol. 30, 42, 24, 37, 21, 25, 38, 31. Find median. Arranging the weights in ascending order, we get 21, 24, 25, 30, 31, 37, 38, 42 n = 8 (even) n n    th term    1 th term 2 2   Median =   2 or (iii) If it is odd, observe one middle value and write it as median and if it is even, observe the two middle values and find median by the following formula : = 35 (4th term) The weights of 8 students (in kg) are : in ascending (ii) Count the number of items. 8 =   th term = 4th term 2 Ex.8 97, 93, 67, 100, 95, 98, 94 Number of observations = 7 (odd) Find median of the data 34, 46, 56, 27, 28, 35, 45. n n    th term    1 th term 2 2   Median =   2 30  31 61 = 2 2 Median weight = 30.5 kg. EXAMPLES  Case II : When n is even 4th term  5th term 2 Sum of the middle values 2 Ex.10 Given the mean of 5 numbers is 25. If the four numbers are 16, 26, 20, 32, find the fifth number. Sol. Let the missing number be x. Then sum of the numbers = 16 + 26 +20 +32 + x = 94 + x NowMean = Sum of the numbers 5  5 × Mean = Sum of the numbers i.e., 5 × 25 = 94 + x or 125 = 94 + x  x = 125 – 94 = 31 Thus, the fifth number is 31. Ex.11 A data is given as below in which one number was missed by the observer : 18, 16, 37, 40, 28, 20, ....., 35 Sol. If the median of the data is 26, find the missing number. On arranging the given numbers in ascending order, we get EXAMPLES  Ex.13 Find mean, median and mode of the following data. Choose the best average(s) to represent the data. 3, 11, 7, 5, 17, 11 Sol. = 3 + 11 + 7 + 5 + 17 + 11 = 54 16, 18, 20, 28, 35, 37, 40 Mean = There are 8 numbers including the missing number.  Mode = 11 Also for 8 numbers, the middle numbers will be 4th and 5th. Thus, the missing number must be at 4th place. Median : Step 1 : On arranging the data in ascending order, we get 4th  5th Median = 2 3, 5, 7, 11, 11, 17 Mis sin g number  28 or 26 = 2 Step 2 : Two middle values are 7 and 11. or 52 = Missing number + 28  Median = or 26 × 2 = Missing number + 28 or Missing number = 52 – 28 = 24 The first five multiples of 6 are : 6, 12, 18, 24, 30 which are in ascending order. n = 5 (odd)  n 1 Median value =   th value  2   5 1 rd =  th value = 3 value  2  Median value = 18 MODE This is also used for central tendency. Definition : The observation which has maximum frequency is called the mode. 7  11 =9 2 We find that mean and median are the same and are in the centre of the data. The mode is not an appropriate central value for this data. Ex.12 Find the median of first five multiples of 6. Sol. Sum of the numbers 54 = =9 Number of items 6 Mode : 11 as it occurs two times. But median is 26 which is less than 28. Now, Mean : Sum of the numbers Ex.14 Find the mode of numbers Sol. 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 6, 8, 8 Number 2 3 4 5 6 8 Tally Marks Frequency 4 3 2 3 2 2 Number 2 has highest frequency (4), so mode is 2. Ex.15 Find the mode of the following data : 2, 5, 4, 7, 5, 5, 8, 12, 8, 9 Sol. Number Tally Marks Frequency 2 1 4 1 5 3 7 1 8 2 9 1 12 1 Mode : Marks 5 9 10 12 15 16 19 The number 5 has the highest frequency 3. Hence mode of given numbers is 5. 20 23 Ex.16 The following table shows the marks obtained by 40 students of class VII : Marks obtained 30 No. of Students Sol. Find the mode. 3 25 2 23 4 19 11 17 13 14 7 As the marks obtained by maximum students is 17 so the mode of the given data is 17. Ex.17 The scores in mathematics test (out of 30) of 15 students are as follows : 20, 19, 12, 25, 24, 23, 20, 25, 20, 16, 9, 20, 15, 10, 5 Find mean, median and mode. Are they equal? Sol. As we know that Mean = Sum of all observations Total number of observations 20  19  12  25  24  23  20  25  20  16  9  20  15  10  5 = 15 263 Mean = 17.54 15 Now arranging the given marks in ascending order, we get 5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25 n = 15 (odd)  n 1 Median =   th term  2   15  1  th =  th term = 8 term  2  Median marks = 20 = 24 25 Tally Marks Frequency 1 1 1 1 1 1 1 4 1 1 2 Number 20 has the highest frequency i.e., 4, so, mode marks = 20. BAR GRAPH A bar graph is a pictorial representation of the numerical data by a number of bars (rectangles) of uniform width erected horizontally or vertically with equal spacing between them. Steps : (i) On a graph paper draw horizontal line and vertical line OX and OY respectively, which are called X-axis and Y-axis respectively. (ii) Marks points at equal space (intervals) along the X-axis. Below these points write the names of the data whose values are to be plotted. (iii) Assume (choose) a suitable scale. On that scale find the heights of the bars for the given numerical values. (iv) Mark these heights parallel to Y-axis from the point taken in step (ii) (v) Draw bars of equal width for heights marked in step (iv) on X-axis. The bars should be centred on the points marked on X-axis. These bars show the given numerical data. Ex.18 Raju recorded the temperature at 2 p.m. for a week as follows : Days Temperature (ºC) 30 25 35 30 20 38 28 Draw a bar graph for the above data. We construct the bar graph using the following steps : (i) Take a graph paper and draw perpendicular lines OX and OY. two 50 40 30 20 10 O (iv) Along OY, choose an appropriate scale and mark the temperature in Celsius. Here, 1 large division = 10ºC (v) Calculate the heights of various bars as follows: (a) Height of bar for Monday = Thu Fri Sat Sun X Y 100 90 80 70 60 50 40 30 20 10 O Mon Tue Wed Thu Days of the week Fri Sat X (iii) On which day was the highest attendance recorded ? 1 × 35 = 3.5 large divisions. 10 1 × 30 = 3 large divisions. 10 (e) Height of bar for Friday = Wed Days 1 × 25 = 2.5 large divisions. 10 (d) Height of bar for Thursday = Tue (i) What does the bar graph represent ? (c) Height of bar for Wednesday = Mon = 2.8 large divisions. 1 × 30 = 3 large divisions. 10 (b) Height of bar for Tuesday = 1 × 28 10 Ex.19 Answer the following questions for the given bar graph. (ii) Along OX mark "Names of the days" and along OY "Temperature in ºC". (iii) Along OX, choose suitable width for each bar. = Y Number of students Present in the class Sol. Mon Tues Wed Thu Fri Sat Sun (g) Height of bar for Sunday Temperature (in ºC) EXAMPLES  1 × 20 10 = 2 large divisions (f) Height of bar for Saturday 1 = × 38 10 = 3.8 large divisions. (ii) On which day minimum number of students attended the class ? (iv) If all students attended the class on Wednesday, how many students are there in the class ? Sol. (i) Along vertical line, the students who were present on days of a weak are shown. So the bar graph represents the attendance of a class during the days of a week. (ii) The bar for Saturday is the shortest, therefore, the minimum attendance was on Saturday. (iii) The bar for Wednesday is the highest, therefore, the maximum/highest attendance was on Wednesday. (iv) Take a scale/ruler and put it parallel to the line OX at the highest point of the bar for Wednesday. We find that the scale touches the OY line at 100 marks. So the total number of students in the class are 100. DOUBLE BAR GRAPH = A double bar graph is a bar graph that represent two sets of data on the same graph. The two bars corresponding to each category are drawn next to each other. = EXAMPLES  = Ex.20 The number of two types of drinks sold on days of a week are given below. Construct a horizontal double bar graph for the data. Monday Strawberry Drinks Chocolate Drinks 80 73 147 Tuesday Wednesday 80 90 Saturday 190 154 75 Sunday 165 81 Cities 140 Ambala Sun Sat Fri Thu Wed Tue Mon O Chennai Here also, a zig-zag is marked along OX which shows that marking is skipped and begins with 50. Scale 1 cm = 10 drinks Shaded bars correspond to strawberry drinks 140 165 154 75 69 81 Maximum 27ºC 36ºC 26ºC 37ºC 28ºC Jaipur 29ºC 39ºC Bhopal 25ºC 28ºC 26ºC 38ºC 41ºC 35ºC Construct a vertical double bar graph. Also answer the following questions on the basis of the bar graph : 90 80 75 73 80 Minimum 24ºC Delhi 190 Temperature Bangalore Jammu (i) Which city has the greatest difference in the minimum and maximum temperatures on the given day ? 128 147 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 Y 735 = 105 chocolate drinks 7 Ex.21 The minimum and maximum temperature of different cities on a particular day are given below : Also find the averages in both types of drinks and tell for which drink the sale is more. Sol. 128  73  69  90  81  154  140 7 Therefore, the average sale of strawberry drinks is more than the average sale of chocolate drinks. 69 Thursday Friday = 128 75 812 = 116 strawberry drinks 7 Average (mean) sale of chocolate drinks  Days 147  80  75  80  75  190  165 7 Number of drinks sold Now, Average (mean) sale of strawberry drinks (ii) Which is the hottest city and which is the coldest city ? X (iii) Name the city which has the least difference between the minimum and maximum temperatures. (iv) Name the two cities where the maximum temperature of one was less than the minimum temperature of the other. Sol. (i) Difference in temperatures for various cities : Ambala Ex.22 Sales of English and Hindi books in the years 1995, 1996, 1997 and 1998 are given below : (37 – 26)ºC = 11ºC Chennai (36 – 27)ºC = 9ºC (28 – 24)ºC = 4ºC Jaipur (39 – 29)ºC = 10ºC (38 – 28)ºC = 10ºC Jammu (35 – 25)ºC = 10ºC Sol. 41ºC Scale : 0.5 cm = 1ºC Shaded bars correspond to minimum temperature (ii) Can you say that the demand for English books rose faster ? We construct the bar graph using the following steps : 39ºC two 35ºC (c) We choose a suitable scale to determine the heights of bars. Here, we choose the scale as 1 large division is equal to 100. For year 1995, 26ºC Height of bar for English 25ºC Bhopal Jammu Jaipur Delhi 650 (b) Along OX, mark the years and along OY, mark the subjects. 29ºC 28ºC 28ºC Bangalore 600 620 (d) Calculate the height of the bar for the different years for both subjects. 24ºC Chennai 525 450 (a) Take a graph paper and draw perpendicular lines OX and OY. 38ºC 36ºC 37ºC 27ºC 26ºC Ambala Temperature (in ºC)  42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 500 400 (i) In which year was the difference in the sale of the two language books least ?  Greatest difference is in Jammu. Y Hindi 350 Draw a double bar graph and answer the following questions : (41 – 26)ºC = 15ºC Bhopal 1995 1996 1997 1998 English Bangalore Delhi Years Cities (ii) Hottest city is the city which has the highest maximum temperature i.e., Jammu. Coldest city is the city which has the lowest minimum temperature i.e., Bangalore (iii) City with least difference in minimum and maximum temperatures is Bangalore. (iv) The maximum temperature of Bangalore is less than the minimum temperature of Jaipur. X = 350 100 = 3.5 large divisions Height of bar for Hindi = 500 = 5.0 large divisions. 100 For year 1996, Height of bar for English = 400 = 4 large divisions 100 Height of bar for Hindi = 525 = 5.25 large divisions 100 For year 1997, Basic terms Height of bar for English = Experiments : The activities which we perform to achieve something. Eg. Throw a die, Toss a coin etc. 450 100 = 4.5 large divisions Random Experiment : The activities or experiment whose possible outcomes are known to us but it is not possible to predict the outcome when the experiment is performed. The activities of tossing a coin or throwing a die are random experiments. Height of bar for Hindi = 600 100 = 6.0 large divisions For year 1998, Outcome : When a random experiment is performed, it ends into some happening. Such a happening is called an outcome of the experiment. Height of bar for English = 620 100 For example, let us list out the outcomes of following experiments. = 6.2 large divisions Height of bar for Hindi = 650 100 = 6.5 large divisions Y 700 English Hindi Subject  600 500 400 300 200 100 O 1995 1996 year 1997 1998 X (i) In year 1998, the difference in the sale of the two language books is least. (ii) Yes, the demand for English books rose faster. PROBABILITY Definition : The measure of the chance of happening something is called 'probability'. Measure of Probability : Probability of a sure or certain happening = 1 Probability of an impossible situation = 0 Probability of an event that may or may not happen = between 0 and 1. Experiment Outcomes Tossing a coin Head or tail Throwing a die 1 or 2 or 3 or 4 or 5 or 6 dots on upper face. Two coins tossed together Two heads or two tails or one head and one tail Watching traffic on a busy road There may be an accident or all will go smooth Equally Likely Outcomes : In a random experiment when all the outcomes have equal chance to occur or they have equal probability then the outcomes are called equally likely. For example : (i) Getting head or tail on tossing a coin are equally likely outcomes. So each have half-half chance to occur. or Probability of head = Probability of tail = 1 . 2 (ii) Getting 1, 2, 3, 4, 5 or 6 dots on top of a die are equally likely outcomes. Each outcome has 1 equal chance or probability that is one-sixth or . 6