The Leading Maths - 9

17.1 Collection of Data and Frequency Table At the end of this topic, the students will be able to: ¾ collect the data and prepare frequency table from the given ungrouped data. Learning Objectives Introduction The word Statistics ordinarily means information expressed in terms of numbers. Such numerical information are known as data. Data are collected and compiled. They may be discrete like individuals or continuous like time. They can be grouped or organized systematically. Such data may be presented in the form of numerical tables, graphs, charts or various types of pictures or diagrams. All these help in drawing certain conclusions. As a science, Statistics deals with the techniques of collecting, presenting, analyzing and interpreting the data. Collections, Classification and Representation of Data Age of a person, his height, his weight or the like can be specified numerically or as data. But, we cannot do so with his religion, his nationality or the like. Data may be collected from the original place of inquiry or sources. Such data are primary and others are secondary. Data thus obtained are usually crude, ungrouped and unsystematic. They become more usable when grouped or classified. Grouped or classified data look more comfortable when presented in the form of tables, graphs, charts, and diagrams. Let's now see how this can be done: (a) Data Collection i. Primary Data The marks of 20 students in Statistics are 35, 59, 65, 43, 78, 59, 37, 78, 71, 75, 35, 78, 65, 78, 71 43, 59, 78, 35, 78. This is also called raw data. ii. Secondary Data The marks of the first ten students in Statistics from the school record are 78, 78, 78, 78, 78, 78,75, 71, 71, 65, 65, 59, 59, 59, 43, 43, 37, 35, 35, 35. We use a letter such as x to denote any one of such data or information. CHAPTER 17 CLASSIFICATION AND REPRESENTATION OF DATA Statistics and Probability Statistics and Probability 350 Allied The Leading Mathematics-9 Classification and Representation of Data 351

(b) Series Representation In each of the above cases, the data are crude and unsystematic. They can be arranged i. Ascending (increasing) order of magnitude as 35, 35, 35, 37, 43, 43, 59, 59, 59, 65, 65, 71, 71, 75, 78, 78, 78, 78, 78, 78; ii. Descending (decreasing) order of magnitude as 78, 78, 78, 78, 78, 78,75, 71, 71, 65, 65, 59, 59, 59, 43, 43, 37, 35, 35, 35. In the above series, the smallest value is 35 and the largest value is 78. They are called the lower and upper limits respectively. The data in which the values are arranged as individual is called individual data. We also note that some of them occur and recur, i.e., occur frequently or can be grouped together as: 35 occurs 3 time, 37 occurs 1 times, 43 occurs 2 times, 59 occurs 3 times, 65 occurs 2 times, 71 occurs 2 times, 75 occurs 1 times, 78 occurs 6 times. The number of times each (The number of repeated same value) of the data (or values) occurs is called the frequency of the datum or value or item. iii. Tabular Representation We can put the data in the form of a simple table. In such tables, repeated data may be grouped together. In other words, frequency of a particular item or value can be written down. Such a table is known as frequency table or frequency distribution table. The following is one such table: Method 1: Frequency Distribution of Marks of 20 Students Marks of Students (x) Tally Marks Number of Students (Frequency f ) 35 37 43 59 65 71 75 78 /// / // /// // // / //// / 3 1 2 3 2 2 1 6 Total 20 This table has individual value for frequency. Therefore, it is called frequency table of ungrouped data. It is also called discrete data. Statistics and Probability Statistics and Probability 352 Allied The Leading Mathematics-9 Classification and Representation of Data 353

Method 2: Frequency Distribution of Marks of 20 Students Marks of Students (x) Tally Marks Number of Students (Frequency f ) 30-40 40-50 50-60 60-70 70-80 //// // /// // //// //// 4 2 3 2 9 Total 20 This table has the given values in the form of classes. It has equal class interval in each class. Therefore, it is called frequency table of grouped data. This data is also called a continuous data. In each class, the values are taken from the lower limit to the value just less than upper limit. That means, in the class 30-40, it contains the values from 30 to 39. In the table, a) In the first row, we have the column headings i. Marks of the Students (x), ii. Tally Marks, and iii. Number of Students (or frequency, f ) b) In the first column, we have the marks(variables) written in increasing (or ascending) order. In the second column, we drew as many tally marks (/) as the number of times a value or item occurred. While drawing tally marks for five, we drew four parallel tally marks and one crossing them diagonally. In the last column, we wrote the number of tally marks of each age group. c) In the third row of the first column, we wrote the word Total, left the second blank and then wrote the total number or sum of the frequencies in the last one. In the table, the frequencies at each stage is added up (or accumulated) and a table so formed is called a cumulative frequency distribution table. Statistics and Probability Statistics and Probability 352 Allied The Leading Mathematics-9 Classification and Representation of Data 353

a) Cumulative frequency distribution of ungrouped data of marks of 20 students Marks of students (x) No. of students (frequency f ) Cumulative frequency (cf ) 35 37 43 59 65 71 75 78 3 1 2 3 2 2 1 6 3 3 + 1 = 4 4 + 2 = 6 6 + 3 = 9 9 + 2 = 11 11 + 2 = 13 13 + 1 = 14 14 + 6 = 20 Total 20 B. Cumulative frequency distribution of grouped data of marks of 20 students Marks of students (x) Number of students (frequency f ) Cumulative frequency (cf ) 30-40 40-50 50-60 60-70 70-80 4 2 3 2 9 4 4 + 2 = 6 6 + 3 = 9 9 + 2 = 11 11 + 9 = 20 Total 20 PRACTICE 17.1 Read Think Understand Do Keeping Skill Sharp 1. (i) Given the length of some pencils (in cm) : 12, 8, 6, 11, 9, 10, 12, 14, 12, 8. (a) Define frequency of an item. (b) Make the frequency table. (c) Find the sum of frequencies. (d) Which length of the pencil is equal to the sum of frequencies? (ii) Given the marks of 20 students out of 25 : 21, 23, 19, 17, 12, 15, 15, 17, 17, 19, 23, 23, 21, 23, 25, 25, 21, 19, 19,19 (a) What value corresponds to tally marks ? (b) Construct the frequency table by using tally bars. (c) Find the sum of frequencies. (d) Why the sum of frequencies is not equal to 25? Give reason. Statistics and Probability Statistics and Probability 354 Allied The Leading Mathematics-9 Classification and Representation of Data 355

2. (i) Weight of school' bag of 25 students (in kg) are given below : 5.3, 6.2, 5.6, 6.2, 5.2, 4.8, 6.3, 5.2, 5.3, 5.6, 5.8, 5.3, 5.6, 6.2, 6.3, 5.3, 6.4, 6.2, 5.2, 5.8, 6.3, 5.7, 5.6, 6.3, 5.4. (a) Define cumulative frequency. (b) Construct the cumulative frequency table of the data. (c) Of which weight of bag is found most? (d) Find the difference of number of the heaviest bag and number of the lightest bag. (ii) The marks obtained by 30 students in class test of statistics are given as below: 34, 44, 54, 25, 56, 65, 45, 34, 29, 56, 67, 76, 47, 78, 67, 52, 71, 68, 59, 20, 45, 37, 73, 45, 25, 28, 56, 34, 56, 73. (a) Define grouped data. (b) Make the frequency table taking class interval 10 of the following data. (c) Which group has the highest number of students? (d) Find the difference between total number of students and total frequency. Why the answer is like that? 3. (i) The ages of teachers in a government school are as follows: 45, 44, 56, 45, 45, 56, 45, 46, 53, 46, 56, 54, 54, 57, 54, 49, 59, 48, 47, 59, 53, 51, 50, 40, 55, 52, 41, 59, 49, 47 (a) Define the class interval in frequency distribution table. (b) Construct the cumulative frequency distribution table for the data taking the class interval of 4. (c) How many teachers are there in the youngest group? (d) If the retirement age limit is 60, how many teachers will be retired in coming 5 years? (ii) Here the number of family members are given which was asked by team Sagarmatha of grade nine in project work of Chapter "Sets": 3, 4, 5, 4, 3, 6, 4, 6, 7, 3, 5, 4, 7, 5, 2, 4, 2, 5, 4, 3, 6, 2, 4, 6, 7, 4, 3, 5, 6. (a) Define cumulative frequency. (b) Construct the cumulative frequency distribution table for the data. (c) How much families did the team visited? (d) How many families has most number of family members? (iii) The score secured by some students in IQ test. 75, 85, 84, 83, 88, 90, 89, 70, 98, 97, 90, 86, 79, 70, 78, 93, 82, 81, 71. (a) Determine the best class interval for the cumulative frequency distribution table. (b) Construct the cumulative frequency distribution table with the same class interval as you determined. (c) Which group has less IQ ? (d) By how much percent the highest IQ is more than the least IQ? Statistics and Probability Statistics and Probability 354 Allied The Leading Mathematics-9 Classification and Representation of Data 355

4. (i) The following table represents the frequency distribution of marks obtained by some students of class 9 in the First terminal examination, 2080 Obtained Marks 25 36 45 48 53 55 67 78 89 No. of Students 1 3 2 7 6 8 4 5 3 (a) What is the type of data series given above? (b) Construct the cumulative frequency distribution table for the data. (c) How many students scored the highest marks? (d) Find the difference between the marks obtained by highest number of students and least number of students. (ii) Ashopkeeper weighed some of available books in his store and found the result as follows: Weight of Books (in gm) 245 432 342 356 432 453 453 342 No. of Students 5 7 6 5 4 2 9 1 (a) Which alphabet is used to present the number of students? (b) Construct the cumulative frequency distribution table for the given data. (c) How many books did the shopkeeper weigh altogether? (d) What is the difference between the heaviest and the lightest book? 5. (i) The health post is distributing the nutrition supplements "Baal Vita" to the babies under 15 months. The preliminary reports show the number of children by their ages months is as follows: Age of Children (in month.) 2-4 4-6 6-8 8-10 10-12 12-14 No. of Children 16 15 18 17 14 12 (a) What type of data presentation is it? (b) Construct the cumulative frequency table from the following data: (c) If each children are to be given 5 packets of Bal Vita, find the total number of packets needed. (d) Howmanymore packets need are ifthe children above 10months are given 10 packets each? (ii) 19 students of taekwondo class are found to be the weight of as the following: Weight of Students (in kg) 25-30 30-35 35-40 40-45 45-50 No. of Students 5 3 4 2 5 (a) What is the class interval in the given data presentation? (b) Construct the cumulative frequency distribution table for the data. (c) How many students are of more than 40 kgs? (d) How many students need to increase their weight if they want to select in 40kg weight championship? Consult with your teacher. Answers Statistics and Probability Statistics and Probability 356 Allied The Leading Mathematics-9 Classification and Representation of Data 357

17.2 Histogram At the end of this topic, the students will be able to: ¾ construct the histogram from the given data. Learning Objectives Introduction A histogram is especially important in presenting statistical data diagrammatically when data are given in the form of frequency distributions and the classes are of different sizes or magnitude. In such case, the area of the rectangle represents the frequency of the data. When the data are presented in discrete form or the class size are equal then the height of the rectangles of equal width represents the frequency of the respective score or the class. Histogram for Discrete Series When the data are presented in the discrete form, we draw rectangles of equal width with height representing the frequency of the individual score. Did you notice here, in the histogram, ™ rectangles of equalwidth are used. ™ rectangles are joined together. ™ height of the rectangles or vertical scale are representing the frequencies. ™ horizontal scale is used for the variables. Example-1 Look at the histogram given in the figure and answer the questions that follows: (a) What is the total number of workers? (b) Find the number of workers who earn less than 200. (c) Find the number of workers who earn more than 200. Solution: Here, (a) Number of workers in total = 100 + 150 + 300 + 225 + 175 + 75 = 1025 (b) Number of workers earning less than 200 = 100 + 150+ 300 = 550 (c) Number of workers earning more than 200 = 175 + 125 = 300 2 0 Points Scored Frequency Not Histogram 50 55 60 65 70 4 6 8 10 2 0 Points Scored Frequency Histogram 50 55 60 65 70 4 6 8 10 50 50 100 150 200 250 300 Wages Rs. 100 150 200 250 300 Statistics and Probability Statistics and Probability 356 Allied The Leading Mathematics-9 Classification and Representation of Data 357

Example-2 10 coins were tossed simultaneously and the occurrence of the heads were recorded in the form of a table. No. of heads 0 1 2 3 4 5 6 7 8 9 10 Frequency 5 10 25 35 50 70 90 80 60 30 10 Represent this information in a histogram. Solution: Drawing the histogram according to the given data, Example-3 In an examination of 100 marks in mathematics, 300 students scored the following marks. Marks 10–20 20–30 30–40 40–50 50–60 60–70 70–80 No. of Students 20 30 45 50 60 40 30 Represent this in the form of a histogram. Solution: Here, the classes are of equal size, we use the rectangles of equal width and represent the frequency by the height of the rectangles. PRACTICE 17.2 Keeping Skill Sharp 1. Read the given histogram and answer the questions given below it. (a) Which axis should the marks be shown to draw a histogram? (b) How many students are there in all? (c) Which age group has maximum students? (d) Find the difference between the number of students above 6 years and the number of students below 7 years. 2. A veterinarian recorded the weights of animals in a histogram. Answer the following questions: (a) Which axis should the frequencies be shown to draw a histogram? (b) Find the total number of animals recorded? (c) How many animals weight between 5 kg and 10 kg? (d) Find the difference between the heaviest group and the lightest group. 20 0 1 2 3 4 5 6 7 8 9 10 30 40 50 60 70 80 90 100 No. of heads Frequency 10 0 Ages No. of students 5 6 7 8 9 20 30 40 50 70 60 5 0 Weight (pounds) Animal Weights No. of animals 0-4.9 5-9.9 10.14.9 15.19.9 20-24.9 10 15 20 10 5 0 10 20 30 40 50 60 70 80 20 30 40 50 60 70 Marks No. of students Statistics and Probability Statistics and Probability 358 Allied The Leading Mathematics-9 Classification and Representation of Data 359

3. A group of students in an entrance exam of Engineering college scored the following marks out of 75 full marks. Marks 20 30 40 50 60 70 Frequency 5 18 32 45 20 3 (a) Which axes should the marks and frequencies be shown to draw a histogram? (b) Construct the histogram for the data given below. (c) What is the marks secured by most of the students? (d) If the pass marksis 35, how many students will have to be back from engineering college? 4. The Birthing Centre recorded the height of mothers as follows: Height (cm) 120 125 130 140 145 150 Frequency 8 15 27 30 25 10 (a) Define histogram. (b) Construct the histogram for the given data. (c) What is the total number of mothers? (d) If 135 cm is the ideal height of mother then how many ideal mothers are there? 5. The marks scored by standardized test in mathematics grade nine is as below: Marks 0-20 20-40 40-60 60-80 80-100 Frequency 50 150 300 200 80 (a) What data series type is given above? (b) Construct histograms to represent the data given below. (c) Find the number of students who attended the test. (d) If below 35 marks is considered as Non Graded then how many of them have not achieved any grade? 6. The 140 families are asked about the daily wages and the result obtained as follows: Wages (Rs.) 150-200 200-250 250-300 300-350 350-400 Frequency 5 18 32 45 20 (a) Which graphical presentation is the best for the data given? (b) Represent the data in histogram. (c) If the family with daily income below Rs. 200 is considered as absolute poverty line, find the number of family below the absolute poverty line. (d) How much more money should they earn daily to rise above the absolute poverty line? Consult with your teacher. Answers Statistics and Probability Statistics and Probability 358 Allied The Leading Mathematics-9 Classification and Representation of Data 359

17.3 Frequency Polygon At the end of this topic, the students will be able to: ¾ construct the cumulative frequency curve (ogive) from the given data. Learning Objectives Observe the histogram of the given data shown alongside. Marks (x) 0-20 20-40 40-60 60-80 80-100 No. of Students (y) 6 4 10 16 14 Plot the midpoints A, B, C, D and E on the top of each pillars of the adjoining histogram. Also, find the midpoints of 0-(-20) and (100- 120) and plot the points P and Q. Now, join the points at P, A, B, C, D, E and Q by using scale and pencil and remove the bars of histogram. What do you see ? PABCDEQ is a polygon obtained by the frequency table. This is called frequency polygon. A frequency polygon is a line graph of class frequency plotted against the midpoints of the class of the data. It can be obtained by connecting the midpoints of the tops of the rectangles in the histogram or drawn as a line graph. It is the visualization tool that is used to understand the shape of a distribution. Also, we can directly draw the frequency polygon from the given data by taking the midpoints of the classes on the x-axis and their corresponding frequencies in the y-axis. Frequency polygon is another way to show the information in a frequency table. Frequency polygon of a given frequency distribution can be drawn in two ways. 1) By using histogram 2) Without using histogram Steps of constructing frequency polygon; To draw frequency polygons, first we need to draw histogram and then follow the below steps: 1. Choose the class interval and mark the values on the horizontal axes 2. Mark the mid value of each interval on the horizontal axes. 3. Mark the frequency of the class on the vertical axes. 4. Corresponding to the frequency of each class interval, mark a point at the height in the middle of the class interval. 5. Connect these points using the line segment. 6. The obtained representation is a frequency polygon. Let's consider an example to understand this in a better way. P A B C E Q D 0 P A B C E Q -20 Marks (x) No. of Students (y) 2 4 6 8 20 40 60 80 100 120 10 12 14 16 18 D Statistics and Probability Statistics and Probability 360 Allied The Leading Mathematics-9 Classification and Representation of Data 361

Example-1 In a batch of 400 students, the height of students is given in the following table. Represent it through a frequency polygon. Height in cm 140-150 150-160 160-170 170-180 No. of Students 74 163 135 28 Solution: Following steps are to be followed to construct a histogram from the given data: 1. Construct the midpoints table as below; Height in cm 130-140 140-150 150-160 160-170 170-180 180-190 Midpoints 135 145 155 165 175 185 No. of Students 0 74 163 135 28 0 2. The heights are represented on the horizontal axes on a suitable scale as shown. 3. The number of students is represented on the vertical axes on a suitable scale as shown. 4. Now, plot the points for the corresponding values of midpoints and no. of students as below. "OR" The rectangular bars of widths equal to the class size and the length of the bars corresponding to a frequency of the class interval is drawn. ABCDEF represents the given data graphically in form of frequency polygon as, A B OR C E F 135 140 145 150 155 160 165 170 175 180 185 Height in cm (x) No. of Students (y) 20 40 80 100 120 140 160 180 200 D PRACTICE 17.3 Keeping Skill Sharp 1. A cargo company is preparing to deliver the boxes. Different boxes must deliver in different van according to their weight. The boxes are separated as following: Weight in kg 10-20 20-30 30-40 40-50 50-60 No. of boxes 5 10 18 12 8 (a) With which value should we mark on y-axis corresponding to frequency to construct the frequency polygon? Statistics and Probability Statistics and Probability 360 Allied The Leading Mathematics-9 Classification and Representation of Data 361

(b) Present the given data into frequency polygon. (c) Which group has the highest number of boxes? (d) If Rs. 50 is charged for each box above the weight of 40 kg in toll checkpoint, how much money should be paid as toll charge? 2. A farmer recorded the height of the plants in his her farm as follows: Height in cm 0-5 5-10 10-15 15-20 20-25 No. of Students 5 10 18 12 8 (a) In which axis the height of plants need to mark to construct the frequency polygon? (b) Present the given data into frequency polygon. (c) How many plants are there in tallest plants group? (d) If each plant costs Rs. 15 for the plants of height 10 cm and above, how much money will the farmer get? 3. The height of students of grade nine of a particular school is as given in table: Height in cm 100-110 110-120 120-130 130-140 140-150 150-160 No. of Students 4 16 20 24 10 5 (a) In which axis the Number of Students need to mark to construct the frequency polygon? (b) Present the following data into frequency polygon. (c) How many total students are there altogether? (d) If only students above the height of 120 cm can swim in pool, how much money should be paid at Rs. 300 per each ticket? 4. In an exam of in full marks 50, the marks obtained by the students of class nine are as follows: Marks 2.5-7.5 7.5-12.5 12.5-17.5 17.5-22.5 22.5-27.5 27.5-32.5 No. of Students 7 5 12 8 4 2 (a) What is another name for frequency polygon? (b) Present the following data into frequency polygon. (c) How many total students are there altogether? (d) If the marks below 35% is considered as non-graded, find the number of students whose certificate will not be issued? Consult with your teacher. Answers Statistics and Probability Statistics and Probability 362 Allied The Leading Mathematics-9 Classification and Representation of Data 363

17.4 Cumulative Frequency Curve (Ogive) At the end of this topic, the students will be able to: ¾ construct the cumulative frequency curve (ogive) from the given data. Learning Objectives The curve obtained by plotting the cumulative frequencies against the class size is called the ogive. There are two types of ogive Less than type ogive. More than type ogive. Example-1 Table below shows the wages of employees in a company. Wages (Rs.) 200-300 300-400 400-500 500-600 600-700 700-800 800-900 900-1000 No. of Workers 100 150 250 300 200 140 60 20 (a) Draw less than type from the above data. (b) Draw more than type cumulative frequency curve or the ogive. Solution: (a) Constructing a less than type frequency table, we get Wages (Rs.) 200-300 (< 300) 200-400 (< 400) 200-500 (< 500) 200-600 (< 600) 200-700 (< 700) 200-800 (< 800) 200-900 (< 900) 200-1000 (< 1000) No. of Workers 100 150 250 300 200 140 60 20 Cumulative (F) 100 250 500 800 1000 1040 1200 1220 Here, N = 1220 Now, plotting frequencies against wages, we get the following less than type ogive, (b) Constructing a more than type cumulative frequency table, we get Wages (Rs.) 200-300 (> 300) 200-400 (> 400) 200-500 (> 500) 200-600 (> 600) 200-700 (> 700) 200-800 (> 800) 200-900 (> 900) 200-1000 (> 1000) No. of Workers 100 150 250 300 200 140 60 20 Cumulative (F) 1220 1120 970 720 420 220 80 20 200 400 600 800 1000 1200 1400 Wages Rs. Less than ogive No. of workers 200 300400 500 600 700 800 900 1000 Statistics and Probability Statistics and Probability 362 Allied The Leading Mathematics-9 Classification and Representation of Data 363

Plotting the cumulative frequencies against the wages, we get the following ogive (more than type ogive), Note: We combine the both less than and greater or more than cumulative frequency curves (ogives) in the same graph as follows: PRACTICE 17.4 Keeping Skill Sharp 1. The given ogive shows the records of the marks of students, (a) What is cumulative frequency? (b) What is the total number of students? (b) How many students are there who obtained less than 75 mark? (c) What is the number of students who obtained the marks between 15 to 35? 2. Study the given cumulative frequency curve and answer the following questions: (a) What does the adjoining graph represent? (b) What isthe number ofstudents who secured 70 to 80 marks? (c) How many students secured the marks more than 80? (d) If the pass mark is 40, how many students passed the test? 3. Study the given cumulative frequency curve and answer the following questions: (a) What does the adjoining graph represent? (b) What isthe number ofstudents who secured the mark 65? (c) Find the number of students who secured the marks between 5 and 45? (d) If the pass mark is 25, how many students failed the exam? 200 400 600 800 1000 1200 1400 Wages Rs. More that ogive No. of workers 200 300400 500 600 700 800 900 1000 200 400 600 800 1000 1200 1400 Wages Rs. No. of workers 200 300400 500 600 700 800 900 1000 More than ogive Less than ogive 5 0 10 0 20 30 40 50 60 70 10 15 20 25 30 Marks Cumulative frequency 5 0 40 0 50 60 70 80 90 100 10 15 20 25 30 Test marks Cumulative frequency Marks Cumulative Frequency 0 5 10 15 20 25 30 35 40 45 50 55 60 5 15 25 35 45 5565 75 85 95 105 X Y Scale x-axis: 1 cm = 10 units y-axis: 1 cm = 10 units Statistics and Probability Statistics and Probability 364 Allied The Leading Mathematics-9 Classification and Representation of Data 365

4. The given ogive shows the record of weights of some girls. Answer the following questions: (a) What does the adjoining graph represent? (b) Find the number of girls whose weight is between 20 kg and 40 kg. (c) How many girls are less than 20kg? (d) What it the total number of girls surveyed? 5. Here is given a data of some survey done by grade nine students on the topic of pocket money in class 8: Pocket Money (Rs.) 0-10 10-20 20-30 30-40 40-50 No of students 5 11 18 10 7 (a) What is less than ogive? (b) Construct the less than type cumulative curve (ogive) of the following data: (c) How many students are there in class 8? (d) If cost of a burger is Rs. 30, how many students cannot afford a burger? 6. The number of students of a government school who created a google account at following age for the first time are given below. Age in years 10-11 11-12 12-13 13-14 14-15 Frequency 15 27 35 23 10 (a) What is the first limit to construct the less than ogive for the above data? (b) Construct the less than type cumulative curve (ogive) of the given data: (c) How many students are involved in survey? (d) If the age limit is 12 years to create google account, how many of them used fake date of birth? 7. A hotel recorded the number of costumers for 40 days. Number of costumer 20-40 40-60 60-80 80-100 100-120 Number of days 4 8 6 12 10 (a) What is more than ogive? Define it. (b) Construct the more than type cumulative curve (ogive) of the above data: (c) How many days the hotel served more than 100 costumers? (d) If the cook of the hotel get bonus of Rs. 1500 per day when the number of costumer is more than 80, how much bonus did he get in those 40 days? 8. The administration of a fun park recorded the ages of the visitors in a certain day as follows: Age in years 5-10 10-15 15-20 20-25 25-30 30-35 Frequency 5 10 20 15 25 15 (a) What is the first limit to construct the more than ogive for the above table? Marks An Ogive More than cf 0 0 10 20 30 40 50 60 70 80 90 100 5 10 15 20 25 30 35 40 45 Statistics and Probability Statistics and Probability 364 Allied The Leading Mathematics-9 Classification and Representation of Data 365

(b) Construct the more than type cumulative curve (ogive) of the given data: (c) Which age group has maximum number of visitors? (d) If the rates of tickets are as follows: for below 15 –free and for more than 15 years Rs. 20 per person, find the total income of that day. 9. The marks obtained by some students are as following: Marks 0-10 10-20 20-30 30-40 40-50 No. of Students 5 11 18 10 7 (a) Define ogive. (b) Prepare the more than cf and less than cf table from the given data: (c) Construct the less than and more than type cumulative curve (ogives) of the following data in the same graph: (d) Determine the median marks of the students from the graph. [md=25.28] 10. Amedical recorded the data of ages of people who were infected by viral infection last week. Age in years 4-8 8-12 12-16 16-20 20-24 24-28 Frequency 3 4 7 3 5 6 (a) What is the last limit to construct the less than and the more than ogive for the above table? (b) Prepare the more than cf and less than cf table from the given data: (c) Construct the less than and more than type cumulative curve (ogives) of the following data in the same graph: (d) Determine the median age of people suffering from the viral infection from the graph. [md=17] Consult with your teacher. Answers Project Work Collect the obtained marks of your class friends in the unit test of Mathematics. Represent it in the continuous data and then draw the line graph, pie chart, histogram and less than ogive. Statistics and Probability Statistics and Probability 366 Allied The Leading Mathematics-9 Classification and Representation of Data PB

18.1 Arithmetic Mean At the end of this topic, the students will be able to: ¾ calculate the arithmetic mean of the ungrouped data and solve the related problems. Learning Objectives Raw data are almost non-informative. Classified data presented in the form of a frequency distribution are less informative. Data presented in the form of diagrams or charts provide a visual comparison of a certain portion of data with the whole observation or information. But, in life-situation we need short, concise and easily understandable conclusive information. For this purpose, we can take the help of Elementary Mathematics. The simple operations of addition and division help calculate certain numerical value (or statistic) or values (or statistics) from given numerical values (or data or statistics). In many cases, such numerical value (or statistic) or values can be used to describe or summarize the whole situation. In general, the name ‘average’is given to such a number. There are many kinds of averages. We shall be mainly concerned with three types or 3M’s, i.e., mean, median and mode. Arithmetic Mean for Individual Data Let's begin with the question: What do we mean by the word ‘mean’? Ordinarily speaking, a mean is a numerical value (or statistic) lying between the largest and the smallest observations. In other words, the term ‘mean’ means a particular type of average of all the values of the variable recorded in the raw data. There are several methods of finding an average of given data. One of the such ways is to sum up all the values (or items) and divide the sum by the total number of values (or items or observations). This average (or value) is ordinarily called a mean. More specifically, it is called an Arithmetical Mean; and is defined mathematically in the following ways: The arithmetic mean (AM) of 1) 2 values a and b of a variable x is the value (or number) a + b 2 , 2) 3 values a, b and c of a variable x is the value (or number) a + b + c 3 , … … … … … … … … … … … … … … … … … … … … … … … … … … 3) n values a,b,c,…, of a variable x is the value (or number) a + b + c + ...... n terms n It is denoted by x (read bar x or x bar). The standard ways of writing the values of the variables, their sum and the average are as follows: CHAPTER 18 MEASUREMENT OF CENTRAL TENDENCY Statistics and Probability Statistics and Probability PB Allied The Leading Mathematics-9 Measurement of Central tendency 367

The n values of a variable x are denoted by x1, x2, x3, .... xn, (x1 is read x sub 1, x2 is read x sub 2 & so on) and their sum by x1 + x2 + ... + xn = n i = 1 xi , where the symbol n i = 1 xi , is read ‘summation or sigma i = 1 to i = n’. Finally, the arithmetic mean is defined by the formula, x = x1 + x2 + ..... + xn n = n i = 1 xi n = x n Example-1 The following are the marks obtained by ten students of a class. Calculate the mean (or average marks): 40, 21, 27, 30, 22, 25, 18, 12, 10, 15 Solution: Here, the given data is 40, 21, 27, 30, 22, 25, 18, 12, 10, 15. Number of observation (n) = 10 ∑x = 40 + 21 + 27 + 30 + 22 + 25 + 18 + 12 + 10 + 15 = 220 Now, Mean (x) = ∑x n = 220 10 = 22. Example-2 The mean of the data x + 3, 2x – 1, 4x – 3, 4x + 1 is 22. Find the value of x. Solution: Here, the given data is x + 3, 2x – 1, 4x – 3, 4x + 1. Mean, (x) = 22. Number of observation (n) = 4 Sum of all items, ∑x = x + 3 + 2x – 1 + 4x – 3 + 4x + 1 = 11x Now, we know that Mean (x) = ∑x n or, 22 = 11x 4 or, 88 = 11x or, x = 8. Arithmetic Mean from Frequency Distribution Table Calculation of arithmetic mean from a given set of data by considering each individual item or value separately is a simple and direct process. But when each of the data is repeated a number of times or the given data are condensed into a group or class in a frequency distribution table, the mean is usually calculated in the following way: Suppose x1, x2, ..., xn are the different values of a variable and their frequencies are f1, f2, ..., fn respectively, then the mean x will be given by x = i = n i = n i = 1 i = 1 fi fi xi = ∑fx ∑f = ∑fx N . In case each of the value of a variable is repeated several times, the values of x can be calculated directly. i = n i = 1 fi xi = f1x1 + f1x2 + .... + fnxn and i = n i = 1 fi = N Statistics and Probability Statistics and Probability 368 Allied The Leading Mathematics-9 Measurement of Central tendency 369

Example-1 The following are the marks secured by 40 students of a class. Compute the arithmetic mean. Marks (x) 20 25 30 35 40 No. of Student (f) 10 5 8 12 5 Solution: Here, Marks (x) 20 25 30 35 40 Total No. of Student (f) 10 5 8 12 5 ∑f = N = 40 fx 200 125 240 420 200 ∑fx = 1185 Now, we know that, the required AM is, x = ∑fx N = 1185 40 = 29.625. Example-2 If the mean of the given data is 7.3, find the value of p. Age of Children 5 6 7 8 9 No. of Children 4 6 12 p 8 Solution: Here, mean of the given data (x) = 7.3 Age of Children 5 6 7 8 9 Total No. of Children 4 6 12 p 8 ∑f = N = 30 + p fx 20 36 84 8p 72 ∑fx = 212 + 8p Now, we know that, the required AM is, or, x = ∑fx N or, 7.3 = 212 + 8p 30 + p = 29.625 or, 219 + 7.3p = 212 + 8p or, 219 – 212 = 8p – 7.3p or, 7 = 0.7p or, 7 0.7 = p ∴ p = 10. PRACTICE 18.1 Keeping Skill Sharp 1. (a) Define arithmetic mean. (b) Write the formula to find the mean of an individual data. (c) Write the formula to calculate the mean of a discrete series. Statistics and Probability Statistics and Probability 368 Allied The Leading Mathematics-9 Measurement of Central tendency 369

(d) In a data, the sum of N observations is ∑fx. What is the mean of the data? 2. (a) In an individual data, the sum of x of 26 observations is 390. What is its mean? (b) If the sum of wages of 35 labours is Rs. 20125, what is their average wages? (c) In a discrete data, the sum of fx of 36 observations is 1620. What is the mean? (d) In a discrete data, the number of observations is 30 and ∑fx is 750. What is its AM? 3. Find the mean of each of the following sets of scores: (a) 2, 3, 2, 4, 6 (b) 1, 2, 2, 4, 5, 5, 7, 6, 3, 4 (c) 1, 0, 1, 0, 1, 0 (d) 1, 3, 5, 7, 5, 3, 1 4. (a) If the mean of 1, 3, 5, 7 and x is 10, find the value of x. (b) If the mean of 10, 30, 40, 50, 60, p and 90 is 50, find the value of p. 5. (a) If the mean of an individual data is 14 in which N = 24 + a and ∑x = 360 + 2a, find the value of a. (b) If the mean of a discrete data is 56 in which N = 32 + k and ∑fx = 2567 + 5k, find the value of k. 6. (a) If the following table gives the marks obtained by 50 students, find the mean. Marks (x) 20 25 30 35 40 No. of Students (f) 10 6 10 15 9 (b) If the following table gives the marks obtained by 50 students, find the mean. Marks (x) 14 15 16 17 18 No. of Students (f) 8 9 12 16 12 7. (a) If the arithmetic mean of the given data is 7, find the value of p. x 5 6 7 8 9 f 6 4 3 6 p (b) Find the value of k, if the arithmetic mean of the given data is 29.625. Marks (x) 20 25 30 35 40 No. of Students (f) 10 5 k 12 5 3. (a) 3.4 (b) 3.9 (c) 0.5 (d) 3.57 4. (a) 34 (b) 70 5. (a) 2 (b) 15.2 6. (a) 30.7 (b) 16.26 7. (a) 5 (b) 8 Answers Statistics and Probability Statistics and Probability 370 Allied The Leading Mathematics-9 Measurement of Central tendency 371

18.2 Median At the end of this topic, the students will be able to: ¾ calculate the median of the ungrouped data and solve the related problems. Learning Objectives We are familiar with the process of ranking the values or items in ascending or descending order of magnitude. While doing so, we may get a value lying at the middle. In other words, the number of values or items less than a certain value is equal to the number of values or items greater than that. For instance, in the following values arranged in ascending order of magnitude, 2, 3, 5, 7, 11, the value or item 5 lies at the middle. That is, the number of values or items (2) below 5 is the same as the number of values or items (2) above 5. Median for Individual Data A median is a value or item, which divides the total number of values or items, or observations into two equal halves. In case, the number of values or items (n) is odd, the N + 1 2 th value or item gives the value of the median. If the number of terms is even, there will be no single middle term, but two middle terms. In such a case, the average of the two middle terms is taken as the median value. For instance, there is no single middle term in the following series of values, 1, 3, 5, 7, 9, 11. Here, we have two middle values 5 and 7. The median in this case is taken to be 5 + 7 2 = 6. Example-1 Find the median in each of the following cases: (a) 2, 1, 3, 7, 4 (b) 1, 2, 3, 4, 6, 7 (c) 1, 1, 1, 2, 2, 3, 3, 4, 5, 5 Solutions: (a) Arranging the data in ascending order, we have 1, 2, 3, 4, 7. No. of items (N) = 5 Here, the values are distinct and odd in number. So, we have only one median, the 5 + 1 2 = 3rd value, i.e., 3. ∴ Median (Md) = 3 (b) The given data 1, 2, 3, 4, 6, 7 is in ascending order. N = 6 which is even in number Median (Md) = N + 1 2 = 6 + 1 2 = 7 2 = 3.5th item = 3 + 4 2 = 7 2 = 3.5. Statistics and Probability Statistics and Probability 370 Allied The Leading Mathematics-9 Measurement of Central tendency 371

(c) Here the values 1, 1, 1, 2, 2, 3, 3, 4, 5, 5 are in ascending order and 10 in number. Now we consider the 5th ( i.e. 10 2 = 5) item the left. It is 2. But the 5th item from the right is 3. We therefore may take any number lying between 2 and 3 as the median. Usually, we take the average of the two. Thus, the median value is 2 + 3 2 = 5 2 = 2.5. Example-2 x + 1, 2x – 1, x + 7 and 3x + 4 are in ascending order. If its median is 12, find the value of x. Solution: Here, the given ascending ordered data is x + 1, 2x – 1, x + 7, 3x + 4. The number of items (N) = 4 The median of the given data (Md) = 12, x = ? Now, we have Md = The position of N + 1 2 th item or, 12 = The position of 4 + 1 2 th item or, 12 = The position of 2.5th item Median for Discrete Data The median ofthe discrete seriesisthe value of N + 1 2 th item, when data are presented is ascending order of size in cumulative frequency table. In this case, the use of cumulative frequency (cf) distribution facilitates the calculations. The steps involved are: (i) Prepare the ‘less than’ table. (ii) Find the value of N + 1 2 th item. (iii) See the cf just greater than the value of N + 1 2 . (iv) The corresponding value of the variable gives the median. Example-3 Determine the median weight from the following data of 46 persons: Weights (in lbs) 150 100 140 130 120 160 No. of persons 2 3 16 14 10 1 Solution: First of all, arranging the variable in ascending order, we get the following array: 100, 120, 130, 140, 150, 160. We now prepare the following cumulative frequency table in ascending order of variables. Weights (in lbs) 100 120 130 140 150 160 Total No. of persons 3 10 14 16 2 1 46 Cumulative Frequency 3 13 27 43 45 46 or, 12 = 2x – 1 + x + 7 2 or, 24 = 3x + 6 or, 24 – 6 = 3x or, 18 3 = x ∴ x = 6 Statistics and Probability Statistics and Probability 372 Allied The Leading Mathematics-9 Measurement of Central tendency 373

Here, the total number of (persons) terms (N) = 46 Now, we have Median (Md) = The value of N + 1 2 th item = The value of 46 + 1 2 th item = The value of 23.5th item In the above cftable, the cfjust greaterthan 23.5 is 27 and the value of x corresponding to 27 is 130. Hence, the median of the given data is 130. PRACTICE 18.2 Keeping Skill Sharp 1. (a) Define median. (b) In which percent does the median divide the data? (c) Which is the median term of a data having N observations? (d) If a data has 12 observations, what is the position of the median in it? 2. Write the median in each of the following cases: (a) 3, 2, 7, 1, 4 (b) 10, 8, 6, 5, 4, 1, 2 3. Find the median in each of the following cases: (a) 3, 4, 7, 7, 7, 1, 2, 2, 7, 1, 9 (b) 9, 9, 9, 8, 8, 8, 8, 4, 4, 1, 1, 1, 1, 1, 6, 8, 5 4. (a) What is the value of x if the data 40, 50, 3x + 4 2 , 80, 90 are in ascending order and the median is 14? (b) If the data x – 1, x + 2, 2x – 1, 2x + 3, 3x + 2 are in ascending order and the median is 17, what is the value of x? (c) If the data x, 2x, 3x + 1, 4x – 1, 6x, 7x – 1 are in ascending order and its median is 17, find the value of x. 5. Find the median from the data in each of the following tables: (a) (b) (c) (d) 6. (a) Determine the median size of shoes from the following data of 51 pairs: Size 4 5 6 7 8 9 No. of pairs 8 14 k 10 3 1 (b) Calculate the median from the data given below: Marks 28 24 32 35 48 45 Total No. of Students 24 23 28 a 59 54 225 4. (a) 8 (b) 9 (c) 10 5. (a) 60 (b) 65 (c) 30 (d) 25 6. (a) 6 (b) 45 Answers x 50 60 70 80 90 99 f 15 11 14 5 3 1 Wages 45 55 65 75 85 No. of workers 24 23 31 12 23 Marks 10 20 30 40 50 No. of students 1 4 8 6 2 x 5 15 25 35 45 55 f 7 8 9 2 1 4 Statistics and Probability Statistics and Probability 372 Allied The Leading Mathematics-9 Measurement of Central tendency 373

18.3 Quartiles At the end of this topic, the students will be able to: ¾ calculate the quartiles of the ungrouped data and solve the related problems. Learning Objectives When a data are arranged in ascending order of magnitude the three values that split a distribution into four equal parts are known as quartiles. In order of magnitude usually represented by Q1, Q2 and Q3 are called the first, second and third quartiles. Q1 divides the data at 25% of the data, Q2 divides the data at 50% of the data and Q3 divides the data at 75% of the data. Q1, Q2 and Q3 are also called lower quartile, middle quartile and upper quartile respectively. Note by definition, the second quartile Q2 is just the median since it divides the area under the frequency curve into two equal parts. Schematically we can demonstrate it in the number line as follows: Q1 Q2 Q3 0% 25% 50% 75% 100% As in the median, we workout the quartiles by using the formulae. Quartiles for Individual Data Rewrite the items or observations of the given data in increasing order and then calculate the following values given below: First Quartile (Q1) = The position of N + 1 2 th item Second Quartile (Q2) = The position of = 2(N + 1) 4 th = N + 1 2 th item = Median (Md) Third Quartile (Q3) = The position of 3(N + 1) 4 th item Find the item for the above value. The item is the required quartile. It is noted that if the values of N + 1 4 th item and 3(N + 1) 4 th item are in the decimal form, i.e., n.25 and n.75 where n is a whole number, then Q1 = the position of whole number part item + 0.25 or 0.75 times the difference between that part and its forward part/item. i.e., Q1 = The position of nth item + 0.25 or 0.75 × {(n + 1)th item – nth item}. Similarly, Q3 = The position of nth item + 0.25 or 0.75× { (n + 1)th item – nth item}. Example-1 Find the first quartile for the data given below: 20, 18, 16, 22, 15, 27, 30, 21, 13, 15, 17, 33, 45, 19, 14 Solution: Arranging the data in order of size in ascending order, we get 13, 14, 15, 15, 16, 17, 18, 19, 20, 21, 22, 27, 30, 33, 45 => N = 15 Statistics and Probability Statistics and Probability 374 Allied The Leading Mathematics-9 Measurement of Central tendency 375

Now, we have Q1 = Value of N + 1 4 th item = Value of 15 + 1 4 th item = Value of 4th item = 15. Example-2 Find the third quartile of the given data: 45, 53, 34, 67, 53, 58, 61, 67 Solution: Arranging the data in order of size in ascending order, we get 34, 45, 53, 53, 58, 61, 67, 67 => N = 8 Now, we have Q3 = The value of 3(N + 1) 4 th item = value of 3(8 + 1) 4 th item = value of 6.75th item = The value of 6th item + 0.75× { (6+1)th item – 6th item} = 61 + 0.75 × (67 – 61) = 61 + 0.75 × 6 = 61 + 4.5 = 65.5. Quartiles for Discrete Data In the discrete data/series, we can calculate the first quartile (Q1), the second quartile (Q2) and third quartile (Q3) same in the individual data/series as follows: Q1 = value of N + 1 4 th term Q2 = value of = 2(N + 1) 4 th = N + 1 2 th term = Median (Md) Q3 = value of 3(N + 1) 4 th term It is noted that the values of the term are compared just greater in the cumulative frequency (cf) of the data. The corresponding variable of the cf is the required quartile. Example-3 Find the first and the third quartiles from the data given below: x 5 10 15 20 25 30 f 8 11 17 20 18 6 Solution: Making cf table x 5 10 15 20 25 30 Total f 8 11 17 20 18 6 80 cf 8 19 36 56 74 80 Here, N = 80 Now, we have Q1 = value of N + 1 4 th item = value of 80 + 1 4 th item = value of 20.25th item = 15. Q3 = value of 3(N + 1) 4 th item = value of 3(80 + 1) 4 th item = value of 60.75th item = 25. Statistics and Probability Statistics and Probability 374 Allied The Leading Mathematics-9 Measurement of Central tendency 375

PRACTICE 18.3 Keeping Skill Sharp 1. (a) Define quartiles. (b) What is the first quartile? (c) What is the third quartile? (d) What is the formula to calculate the third quartile of a data with N items? 2. (a) If a data has 7 items, what is the position of Q1 in it? (b) If a data has 11 observations, what is the position of the third quartile in it? 3. (a) Write the first quartile in each of the following data: (i) 1, 1, 2, 3, 4, 7, 9 (ii) 12, 10, 5, 4, 4, 3, 1 (b) Write the third quartile of the following data: (i) 12, 14, 15, 16, 17, 18, 21 (ii) 9, 9, 9, 8, 8, 8, 8, 4, 4, 3, 2 4. Find the first quartile from the data given below: (a) 18, 15, 5, 17, 20, 21, 23. (b) 15, 27, 18, 13, 12, 16, 22, 21, 25 5. Find the third quartiles from the data given below: (a) 12, 13, 10, 15, 8, 17, 6, 12, 14, 19, 25 (b) 16, 32, 10, 25, 22, 28, 34 6. (a) If the first quartile of the data given below is 41, find the value of x: 23, 2x – 3, 4x – 5, 29, 30, 34, 38. (b) If the third quartile of the data given below is 21, find the value of y. 12, 15, 17, 2y + 3, 3y – 1, 24, 27 7. Find the first quartile from the data given below: (a) (b) 8. Find the third quartile from the data given below: (a) (b) 9. Find the quartiles (Q1 and Q3) from the data given below: (a) (b) 4. (a) 15 (b) 14 5. (a) 17 (b) 32 6. (a) 22 (b) 9 7. (a) 20 (b) 35 8. (a) 15 (b) 48 9. (a) 50, 70 (b) 55, 75 Answers x 10 20 30 40 50 f 7 8 12 5 6 Weight (kg.) 32 35 40 42 50 No. of Students 5 15 18 13 10 Age (yrs.) 12 13 14 15 16 Frequency 6 12 19 11 6 Weight (kg.) 38 42 44 48 50 No. of Students 3 4 8 6 2 x 50 60 70 80 90 99 f 15 11 14 5 3 1 Wages 45 55 65 75 85 No. of workers 24 23 31 12 23 Statistics and Probability Statistics and Probability 376 Allied The Leading Mathematics-9 Measurement of Central tendency 377

18.4 Mode At the end of this topic, the students will be able to: ¾ find the mode of the given data and solve the related problems. Learning Objectives In our everyday life, we often talk of the commonest size of a shirt or the most common height of the Nepali citizen, the most common size of a pair of shoes and so on. Such a value of a variable (or variate) is called the mode of the variate values. Mode is defined as that value of a variable (or variate) which repeats maximum number of times or occurs most frequently. For instance, the mode of the data 1, 2, 2, 3, 4, 4, 4, 5, 6, 6 is 4, since it occurs the maximum number of times. Mode can often be found by inspection in the case of individual or discrete series. Mode in Individual Data In an individual data, the most repeated item or term or observation is the mode of the given data. If the most repeated observations are more than one, we can choose any one or both of them. Such types of data are called multimodal data. For example, the given list of data 1, 1, 2, 4, 4, the mode is not unique - the data may be said to be bimodal, while a set with more than two modes may be described as multimodal. Example-1 Determine the mode from the given data: 34, 35, 46, 34, 23, 34, 46, 35, 36, 34, 35, 34, 46, 34, 23 Solution: From the given data, 23 is repeated 2 times. 34 is repeated 6 times. 35 is repeated 3 times. 36 is repeated once. 46 is repeated 3 times. Here, 34 is the most repeated item. Therefore, the mode of the given data is also 34. Example-2 The mean of the given data is 5. Find its mode. 5, x + 2, 2x – 1, 9, 2, 2, 5, 7 Solution: The given data is 5, x + 2, 2x – 1, 9, 2, 2, 5, 7. No. of terms (N) = 8 Sum of 9 terms, ∑x = 5 + x + 2 + 2x – 1 + 9 + 2 + 2 + 5 + 7 = 31 + 3x. Mean (x) = 5, Mode (Mo) = ? Now, we know hat x = ∑x N or, 5 = 31 + 3x 8 or, 40 = 31 + 3x or, 40 – 31 = 3x or, 9 = 3x or, x = 3 Statistics and Probability Statistics and Probability 376 Allied The Leading Mathematics-9 Measurement of Central tendency 377

Substituting the value of x in the given data, we get 5, x + 2, 2x – 1, 9, 2, 2, 5, 7 = 5, 3 + 2, 2 × 3 – 1, 9, 2, 2, 5, 7 = 5, 5, 5, 9, 2, 2, 5, 7 In this data, 5 is the most repeated number. Therefore, the mode of the given data is also 5. Mode in Discrete Data A variable with the greatest frequency in a discrete data is called mode of the given discrete data. Example-3 Determine the mode of the given data: Marks 45 65 76 89 90 95 No. of Students 2 8 6 3 7 2 Solution: In the given data, 8 is the highest frequency. So, its corresponding variable 65 is the mode of the given data. Relation between Mean, Median and Mode The relation between mean, median and mode are given below: Mode = 3 Median – 2 Mean. Example-4 Establish the relation Mode = 3 Median – 2 Mean for the data given below: 1, 3, 3, 3, 4, 5, 2, 1, 5 Solution: Here, the given data is 1, 3, 3, 3, 4, 5, 2, 1, 5. For Mode: The most repeated number is 3. So, 3 is its mode. i.e., Mode (Mo) = 3 For Median: Ascending order of the given data = 1, 1, 2, 3, 3, 3, 4, 5, 5 No. of terms (N) = 9 ∴ Median (Md) = The value of N + 1 2 th term = The value of 9 + 1 2 th term = The value of 10 2 th term = The value of 5th term = 3 For Mean: No. of terms (N) = 9 Sum of 9 terms, ∑x = 1 + 3 + 3 + 3 + 4 + 5 + 2 + 1 + 5 = 27 ∴ Mean (x) = ∑x N = 27 9 = 3 Now, Mode = 3 Median – 2 Mean or, 3 = 3 × 3 – 2 × 3 or, 3 = 9 – 6 or, 3 = 3. Hence, it is proved. Statistics and Probability Statistics and Probability 378 Allied The Leading Mathematics-9 Measurement of Central tendency 379

PRACTICE 18.4 Keeping Skill Sharp 1. (a) Define mode. (b) How is the mode of an individual data found? (c) How is the mode of a discrete data found? 2. (a) What is the mode of the data: 3, 3, 4, 3, 4, 6, 7? (b) What is the mode of the data: 1, 1, 1, 1, 1, 0, 0, 0? 3. Find the mode for the following data: (a) 34, 35, 26, 35, 34, 38, 34, 39, 34, 39, 23, 31, 30, 33, 34, 35, 34 (b) 56, 65, 75, 56, 67 56, 43, 46, 37, 83, 56, 56, 34, 29, 45, 56, 67, 75, 83 4. (a) The mean of the given data 1, x – 3, 2x + 1, 3x + 5, 2, 2, 5, 7 is 7. Find its mode. (b) If the mean of the given data x, x + 3, 2x + 15, 3x – 1, 13, 3x + 10 is 10, find its mode. (c) The median of the given data 12, 15, 2x + 2, 2x + 3, 3x – 3, 18, 25, 27, 30 in ascending order is 18. Find its mode. (d) The data 25, 30, 35, 40, x + 10, 2x – 20, 55, 60, 65, 70 in ascending order. If its median is 40, find its mode. 5. (a) Find the mode for the following data: (b) Determine the modal weight of children: 6. Find the mean, median and mode in the given data and verify Mode = 3 Median – 2 Mean for each data: (a) 3, 4, 4, 4, 5, 2, 3, 2, 5, 6, 3, 4, 5, 3, 2, 6, 3, 4, 4, 5, 2, 1, 3, 3, 2, 3 (b) 15, 10, 12, 19, 10, 11, 18, 16, 15, 14, 15, 15, 14, 13, 15, 15, 16, 18, 13, 12 7. Find the mean, median and mode for the following data and prove that Mode = 3 Median – 2 Mean for each data: (a) (b) 3. (a) 34 (b) 56 4. (a) 2 (b) 5 (c) 18 (d) 40 5. (a) 50 (b) 17 6. (a) 3.5, 3, 2 (b) 15, 15, 15 7. (a) 61, 61, 61 (b) 18, 18, 18 Answers Project Work Record the weight (in gram) of your educational goods (books, copies, pens, pencils, sharpener, eraser, ruler, compass, etc) and represent it in the discrete data. Also, calculate the arithmetic mean, median, quartiles and mode of the data. Marks 50 60 70 80 90 99 No. of stud. 15 11 14 5 3 1 Size 14 15 16 17 18 19 No. of pairs 7 5 4 9 7 8 Heights (inc.) 59 61 62 63 64 No. of persons 12 13 7 7 1 Length (in cm) 12 15 18 21 24 No. of copies 3 4 9 8 1 Statistics and Probability Statistics and Probability 378 Allied The Leading Mathematics-9 Measurement of Central tendency 379

19.1 Introduction to Probability At the end of this topic, the students will be able to: ¾ define the basic terms of the probability. Learning Objectives An apple, falling naturally from a tree or dropped by someone, is known to hit the ground. Each such phenomenon (natural or manmade) can be considered as an experiment. In this experiment, the event of the apple hitting the ground is sure (or certain) to occur (or happen). But, we arrive at a different situation if we go the other way. Let's begin by considering an apple hitting the ground. We may ask the question: ‘Where did the apple come from?’ A simple answer would be ‘It might have fallen from an apple tree or been dropped by someone’. It is not sure. We thus experience an element of uncertainty. To express this state of uncertainty, we often use the word ‘possible’. In terms of this word, our answer would look like ‘It is possible that the apple has fallen from a tree or is dropped by someone’. In everyday life, we deal with more uncertainties than certainty. A branch of mathematics uses a measure of uncertainty to ascertain certainty. This branch is the Theory of Probability. It originated long ago in the game of chance or gambling. Pascal and Format developed a scientific basis for the Theory of Probability. Let's make the above naive introduction and complete by considering the following example. Consider a coin. A coin is known to have two faces: head and tail. Suppose the coin is tossed once, i.e., an experiment (or operation) of tossing the coin is performed. A head or tail may appear. This implies the existence of two possible outcomes. The two outcomes when taken together are then exhaustive, i.e., leave no possible outcome unconsidered. Here head can appear only once in one trial, and the same is the case with the tail. In other words, there exists only one outcome in favour of (or favourable to) head and one in favour of (or favourable to) tail. But the occurrence of the head (or tail) guarantees the non-occurrence of the tail (or head). That is, only one of the two events can happen. This situation is described by saying that one is the complement of the other. The fact that the number of outcomes favourable to the occurrence of head bears a 1:1 ratio with that of the occurrence of tail may be expressed by the statement: ‘Head is as equally likely to appear as the tail’. Another way of expressing this situation is to relate the number of outcomes favourable to the appearance of head (or tail) to the total number of possible outcomes. It is obviously 1:2 in each case. CHAPTER 19 PROBABILITY Statistics and Probability Statistics and Probability 380 Allied The Leading Mathematics-9 Probability 381

Random Experiment and Sample Space In science, technology and engineering, if an experiment is repeated any number of times in identical conditions; the outcome is the same every time. The theory of probability does not deal with such experiments. It is associated with those experiments or phenomena, whose outcomes are well defined but uncertain. For instances: a) A particular day that may be cloudy or sunny and the crop of a certain year that may be rich or poor are some examples of natural situations or phenomena. b) Tossing a coin whose outcome may be a head or a tail and throwing of a die that may show up a face with 1 or 2 or 3 or 4 or 5 or 6 dots marked on it are some experiments conducted by human being. An experiment ofthis kind in which the outcomes arewell defined but uncertain is roughly called a random experiment. In probability, we shall be mainly concerned with the set of outcomes of a random experiment or experiments. The set of all possible outcomes in a random experiment is called a sample space and it is denoted by S. For example, if a coin is tossed, there are two possible outcomes: head and tail. Usually, head is denoted by H and tail by T. In this experiment, the sample space is the set {H, T} or {T, H}. In the same way, if a fair die is thrown, there are six possible outcomes in a trial. They can be indicated by the list of six numbers 1, 2, 3, 4, 5, 6. The sample space in this case is the set {1, 2, 3, 4, 5, 6}. Each outcome is called an element of the sample space. In an experiment, if we get the desired outcome, we call it a success; otherwise we call it a failure. If we do an experiment one or more times, we may get one or more successes or failures. The totality of successes (or failures) constitutes what is known as an event. An event may therefore be realized as a subset of a sample space. Basic Terminology We begin by recalling the simple experiment of tossing a coin. The two simple events in this case are appearing of head (H) and appearing of tail (T). A game is set up as soon as we ask questions such as ‘ Is H the face appeared? Is T the face appeared?’ The answer would be ‘Yes’ or ‘No’. If the answer is ‘Yes’, we say that the event defined by the question has happened or occurred. We also say that we have a ‘success’. Otherwise, we say that we have a failure. Note that if the occurrence of head is regarded as a success, the occurrence of tail is a failure and vice versa. We explain below some terms associated with such experiments: (a) Sure and impossible events In the simple game of tossing a coin, the sample space is S = {H, T}. In each trial, either head or tail will appear. That is, one of the events represented by the sample space S = {H, T} is sure to occur. This is an example of a sure event. But, in no trial the both head and tail will show up at the same time. This is therefore an impossible event. Statistics and Probability Statistics and Probability 380 Allied The Leading Mathematics-9 Probability 381

(b) Elementary event and compound event If, instead of one coin, we toss two coins at the same time, the sample space will be S = {HT, TH, HH, TT}, The space is exhaustive (i.e., no possible is left unconsidered). An event, in which both the heads (i.e., the event {HH}) appear in only one way, is a simple event. In the same way, the event {TT}, (i.e., appearing of both the tails at the same time) is a simple or elementary event. But the event in which at least one head would appear (i.e., the event {HT, TH, HH} may occur more than one way. An event like this is called a compound or mixed or composite event. (c) Equally likely events In tossing a coin, neither head nor tail can be expected to occur in preference to the other. Two such events are said to be equally likely. (d) Favourable cases While tossing a coin, two outcomes (head and tail) are possible. In each throw, if the desired outcome is supposed to be a head, it cannot happen more than once. So is the case with the tail. Here the number of outcomes, which may result in the happening of head (or tail), is one in each case. In such a situation we say that the number cases favourable to head (or tail) is one. (e) Complementary events If the sample space is S = {H, T}, we know at least one of the two events must occur in every trial. But head appears only when tail does not appear and vice versa. In such a situation, the events are said to be complementary to each other. (f) Mutually exclusive events If a coin istossed the head and tail cannotshow up at the same time. In other words, the occurrence of head excludes the occurrence of tail in the same trial. Such events are said to be mutually exclusive. (g) Dependent and independent events In the experiment of tossing a coin, the occurrence of head affects the occurrence of the tail. But, if we toss two coins at the same time, the occurrence of head on one coin does not affect the occurrence of head or tail of the other coin. The two events in the first case are dependent and those in the second case are said to be independent. Statistics and Probability Statistics and Probability 382 Allied The Leading Mathematics-9 Probability 383

Example-1 In a single throw of a die, the sample space is S = {1, 2, 3, 4, 5, 6}. What are the sets representing the events? (a) The event of getting an even number? (b) The event of getting an odd number? (c) The impossible event? Solutions: (a) Here, the event that the number is even is represented by the set {2, 4, 6}. (b) The event that the number is odd is represented by the set {1, 3, 5}. (c) The impossible event is denoted by the empty set φ. Example-2 What is the number of favourable cases for? (a) an event E1 = {1, 3, 5) of getting an odd number in a single throw of a die? (b) an event E of drawing a red card from a pack of 52 cards? (c) an event E of getting a total of 5 in a simultaneous throw of two dice? Solutions: (a) Here, obviously, the event E1 = {1, 3, 5) has three favourable cases. (b) In a full pack of 52 cards there are 26 red cards that are equally likely to be drawn. So, there are 26 favourable cases. (c) In a simultaneous throw of two dice, the event of getting a total of 5 is denoted by the set, E = {(1, 4), (2, 3), (3, 2), (4, 1)}. So, there are 4 favourable cases. Example-3 In an experiment of throwing a single die, the sample space is S = {1, 2, 3, 4, 5, 6}. State whether the following pairs of sets E = {1, 3, 5) and F = {2, 4, 6} are (a) Complementary to each other; (b) Mutually exclusive; (c) Both complementary and mutually exclusive. Solutions: Here, (a) Complementary (b) Mutually exclusive (c) Both complementary and mutually exclusive. Example-4 Two dice are thrown. The events E and F are defined as follows: E = getting an odd number on the first die, F = getting an even number in the second die. State whether the above events are mutually dependent or independent. Solution: Here, the events E and F are independent since the getting an even number in the second die does not depend upon the occurrence of getting an odd number in the first. Statistics and Probability Statistics and Probability 382 Allied The Leading Mathematics-9 Probability 383

PRACTICE 19.1 Keeping Skill Sharp 1. State whether the following is an experiment or not: (a) The earth is round. (b) Acid turns blue litmus paper into red. (c) The number of rotten apples in a box of apples. 2. Which of the following are random experiments? (a) Drawing a red ball from an urn containing 5 red balls and 4 blue balls. (b) Choosing a vowel from the set of the English alphabet. (c) Electrolysis of acidified water. 3. In a single throw of a die, the sample space is S = {1, 2 ,3, 4, 5, 6}. What are the sets representing the events: (a) The event of getting a prime number? (b) The event of getting multiples of 3? (c) The sure event? 4. Suppose a biased or unfair die is rolled. Will all the outcomes be equally likely? 5. What is the number of favourable cases for? (a) an event E1 ={2, 4, 6) of getting an even number in a single throw of a die ? (b) an event E of drawing a black card from a pack of 52 cards? (c) an event E of getting a total of 10 in a simultaneous throw of two dice? 6. In an experiment of throwing a single die, the sample space is S = {1, 2, 3, 4, 5, 6}. State whether the following pairs of sets E ={1, 2, 3) and F = {4, 5, 6} are true or false. (a) Complementary to each other; (b) Mutually exclusive; (c) Both complementary and mutually exclusive. 3. (a) { 2, 3, 5 } (b) { 3, 6 } (c) { x : 1 ≤ x ≤ 6; x ∈ N } 5. (a) 3 (b) 26 (c) 3 Answers Statistics and Probability Statistics and Probability 384 Allied The Leading Mathematics-9 Probability 385

19.2 Empirical and Classical Probabilities At the end of this topic, the students will be able to: ¾ find the probability for the given events. Learning Objectives From what has been described above, we note that a certain experiment may be carried out once, twice, or more as desired. We may expect an event to be successful every time. But this may or may not happen. Suppose an experiment is repeated n1 times and an event (E) occurs m1 times (i.e., we are successful m1times). The number m1 is then called the frequency of the event E in n1 trials (or observations). The ratio s1 n1 is called the relative frequency of the occurrence of the event E for n1 trials. The experiment is then repeated n2 times, and the number of times m2 that the event E occurs is observed. The relative frequency of E in n2 trials is m2 n2 . Further repetitions of this nature; yield a series of relative frequencies: s1 n1 , s2 n2 , s3 n3 ............... In many cases, experience shows that these relative frequencies differ very little from one set of repetitions to another. In other words, the ratio approaches a certain number as the number of such trials is increased very much. This number is known as the experimental or empirical probability of the event (E). It is denoted by P(E). In other words, the empirical or experimental probability of an event (E) is defined by the ratio m:n of the total number (m) of successes to the total number (n) of the experiments, and is denoted by P(E). In symbols, we write P (E) = m n . This probability is sometimes called ‘a posteriori probability’. Note that this probability depends upon the outcomes of the experiments. In a small number of experiments, a particular event may occur more often than other event or events. For instance, if a coin is tossed five times, it may so happen that head may appear one time only, two times, three times four times or all the five times. But, if the experiment is conducted a large number of times, say 1000 times, one can reasonably expect that heads and tails may appear time and again (roughly in equal numbers). An alternative way of defining the probability of an event is as follows: If S is the sample space of a random experiment and E is an event, then the probability P (E) of E is defined by P (E) = m n or n(F) n(S), where m = the number of outcomes favourable to the occurrence of E and n = total number of all possible outcomes or sample points of the sample space of the random experiment. This probability is sometimes called the ‘a priori probability’ or ‘mathematical probability’. In particular, a) If E is an impossible event, m = 0 and so P (E) = 0. b) If E is a sure event, m = n, and so, P (E) = 1. Statistics and Probability Statistics and Probability 384 Allied The Leading Mathematics-9 Probability 385

c) If m is the number of outcomes favourable to the event E and n the total number of possible outcomes, then n – m is the number of outcomes not favourable to E (i.e., the non-occurrence of E denoted by E). Hence, P (E) + P (E) = m n + n – m n = m n + n n – m n =1 Probability Scale Thus, we see that the probability of occurrence of an event is always non-negative and has a maximum value of 1. This fact is often recorded on a scale ranging from 0 to 1. Sure failure Sure success ↓ ↓ 0 1 The value 0 is assigned to the impossible (or sure failure) event and the value 1 to a certain (or sure success) event. The points in between them represent the remaining probabilities. Example-1 A coin is tossed 1000 times. The result is as follows: Result Head Tail Frequency 638 362 Find the experimental probabilities of the occurrence of head (H) and occurrence of tail (T). Solution: Here, total number of occurrence of head (or successes) n(H) = 638 Total number of experiments n(S) = 1000. By definition, Experimental probability of the occurrence of head (H), P (H) = n(H) n(S) = 638 1000 = 0.638. Experimental probability of the occurrence of tail (H), P (T) = n(T) n(S) = 362 1000 = 0.362. Example-2 What is the probability of getting a tail in tossing a coin once? Solution: Here, Sample space, S = {H, T}. Event of getting a tail, E = {T}. Clearly, the number of cases favourable to E, n(E) = 1 Total number of possible cases, n(S) = 2. Now, by definition Probability of getting a tail in tossing a coin once, P(E) = n(E) n(S) = 1 2 . Statistics and Probability Statistics and Probability 386 Allied The Leading Mathematics-9 Probability 387

Example-3 One card is drawn from a pack of 52 cards, each of the card being equally likely to be selected. Find the probability of; (a) the card drawn is red; (b) the card drawn is a king; (c) the card drawn is red and a king. Solutions: Let S denote the sample space. The total number of cards that can be drawn, n(S) = 52. (a) If E denotes the event of drawing a red card, the number of red cards, n(R) = 26. Now, by definition ∴ The probability of drawing red card, P(R) = n(R) n(S) = 26 52 = 1 2. (b) If F denotes the event of drawing a king, then the number of ways a king can be drawn, n(K) = 4. Now, by definition The probability of drawing king card, P(K) = n(K) n(S) = 4 52 = 1 13 . (c) If G denotes the event of drawing a red card which is a king, then the number of red kings, n(R) = 2. Now, by definition The probability of drawing red king card, P(R) = n(R) n(S) = 2 52 = 1 26 . Example-4 A bag contains 3 red balls, 2 black balls and 1 white ball. A ball is drawn out of the bag at random. What is the probability that the ball drawn is: (a) red? (b) black? (c) red or black? Solutions: Here, total number of balls, n(S) = 3 + 2 + 1= 6. (a) If R denotes the event of getting a red ball, then n(R) = 3. Now, we have Probability of getting red ball, P(R) = n(R) n(S) = 3 6 = 1 2. (b) If B denotes the event of getting a black ball, then n(B) = 2. Now, we know that Probability of getting red ball, P(B) = n(B) n(S) = 2 6 = 1 3. (c) If G denoted the event of getting a red or a black ball, then n(G) = 3 + 2 = 5. Probability of getting red ball, P(G) = n(G) n(S) = 5 6. Example-5 Two dice are thrown simultaneously, find; (a) the probability of getting a total of less than 5. (b) the probability of getting a total of 5. (c) the probability of getting a total of greater than 5. Solutions: Here, in a throw of two dice, the total number of possible outcomes is 6 × 6 = 36. So, if S is the sample space, n(S) = 36. (a) Let E be the event of getting a total less than 5. Then, E = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)}. ∴ n(E) = 6. Hence, P(E) = n(E) n(S) = 6 36 = 1 6. Statistics and Probability Statistics and Probability 386 Allied The Leading Mathematics-9 Probability 387

(b) Let F be the event of getting a total of 5. Then, F = {(1, 4), (2, 3), (3, 2), (4, 1)}. ∴ n(F) = 4. Hence, P(F) = n(F) n(S) = 4 36 = 1 9. (c) Let G be the event of getting a total greater than 5. Then, G = {(1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}. ∴ n(G) = 26. Hence, P(G) = n(G) n(S) = 26 36 = 13 18. Example-6 A coin is tossed successively three times. Find the probability of getting exactly one tail or two tails. Solution: Let S be the sample space. Then, S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} Let E be the event of getting exactly one tail or two tails. Then, E = {HHT, HTH, THH, HTT, THT, TTH}. Clearly, n(E) = 6 and n(S) = 8 Hence, P(E) = n(E) n(S) = 6 8 = 3 4. Example-7 The odds in favour of an event are 3:4. Find the probability of the occurrence of the event. Solution: Let E denote the event. Then, the number of cases: Favourable to the occurrence of E = 3k Against the occurrence of E = 4k. So, the total number of possible cases = 3k + 4k = 7k. Hence, P(E) = 3k 7k = 3k 7k = 3 7. Example-8 A card is drawn at random from a well-shuffled pack of 52 card. Find the probability that it is neither a jack nor a queen. Solution: We solve this problem by using the fact that ‘If E denotes an event, P (E) + P (not E) = 1’ In this case, P(neither a jack nor a queen) + P (a jack or a queen) = 1. Since there are four jacks and four queens, the number of favourable outcomes is 4 + 4 = 8. and the total number of possible outcomes is 52, ∴ P(a jack or a queen) = 8 52 = 2 13. Hence, P(neither a jack nor a queen) = 1 – 2 13 = 13 – 2 2 = 11 13. Statistics and Probability Statistics and Probability 388 Allied The Leading Mathematics-9 Probability 389

PRACTICE 19.2 1. A coin is tossed twice. List the sample space. What is the probability that; (a) Both are tails? (b) Getting one head and one tail (c) Getting head at first and tail at second (d) Getting no head (e) At least one of the two is a head? 2. Three coins are thrown. List the sample space. What is the probability of getting; (a) 2 heads and 1 tail? (b) exactly 1 head and 2 tails? (c) at least 1 head and 1 tail? 3. A fair dice is tossed. List the sample space. Find the probability of the event; (a) getting 2 (b) getting 2 or 6 (c) getting an even number (d) getting 1 or 4 or 6 (e) getting more then 6 (f) getting more than 4 (g) a number greater or equal to 2 (h) not getting 6 (i) factors of 6 (j) prime numbers 4. Two dice are thrown simultaneously. List the sample space. Find the probability of; (a) 1 on the first die, (b) showing both die the same number, (c) getting a total of less than 7, (d) getting a total of 7, 5. One card is drawn from a pack of 52 cards. Find the probability of getting; (a) a black card, (b) a spade, (c) a Queen , (d) a card with 8, (e) a face card, (f) a black coloured king, (g) a Jack of heart, (h) no red ace, (i) a queen or a king, (j) a ace or red king. 6. (a) What is the probability that a pregnant women giving birth to a child is on Thursday of a week ? (b) Find the probability of giving birth to a baby by a pregnant woman on Tuesday or Saturday. (c) What is the probability of born two daughters when a married woman gives birth on planning of two children? Find it. . (d) When a couple gives birth to three children, find the probability of giving birth the same sex. 7. A bag contains 5 white balls, 4 black balls and 3 red balls. What is the probability that the ball drawn is; (a) white? (b) black? (c) red? (d) black or red? 8. Tickets numbered form 1 to 30 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number; (a) divisible by 4, (b) a prime number. Statistics and Probability Statistics and Probability 388 Allied The Leading Mathematics-9 Probability 389

9. While rotating the needle on the given spinner as in the figure, find the probability of the needle stopping at 1. (a) 1 (b) odd number 10. (a) A stamping machine turns out 500 parts per hour. Experience has shows that there are approximately 12 defective parts per hour. Find the probability that a part picked at random will be defective. (b) The students of a school planted 1500 saplings of mango. If the probability of survival is 0.8, find the total number of saplings that survived. 11. A die is thrown 500 times. Suppose E denotes the occurrence of one of numbers 1, 2, 3, 4, 5, 6 representing the dots on the face of the die. The result is shown in the following table: Events (E) 1 2 3 4 5 6 Frequency n(E) 75 83 80 92 90 80 Find the probability of getting; (a) an odd number (b) a number less than 3 (c) a number greater than 3. 12. (a) Out of 1000 mangoes the probability of selecting of rotten mango is 0.15. What is the number of good mangoes? Find. (b) Out of 1500 students in a school, the probability of success of girl student is 0.45. Find the number of boys students. 13. (a) If P(A) = 1 3 and P(B) = 2 5, find. (i) P-A_ (ii) P-B_ (iii) P(A) × P (B) (iv) P(A) × P (B) (b) If P(A) = x 2 and P(A) = 3 4, find the value of x. 1. (a) 1 4 (b) 1 2 (c) 1 4 (d) 1 4 (e) 3 4 2. (a) 3 8 (b) 3 8 (c) 3 4 3. (a) 1 6 (b) 1 3 (c) 1 2 (d) 1 2 (e) 0 (f) 1 3 (g) 5 6 (h) 5 6 (i) 2 3 (j) 1 2 4. (a) 1 6 (b) 1 6 (c) 5 12 (d) 1 6 5.(a) 1 2 (b) 1 4 (c) 1 13 (d) 1 13 (e) 3 13 (f) 1 26 (g) 1 52 (h) 25 26 (i) 2 13 (j) 3 26 6. (a) 1 7 (b) 2 7 (c) 3 8 (d) 1 4 7.(a) 5 12 (b) 1 3 (c) 1 4 (d) 7 12 8.(a) 7 30 (b) 1 3 9. (a) 1 4 (b) 1 2 10. (a) 3 125 (b) 1200 11.(a) 49 100 (b) 79 250 (c) 131 250 12. (a) 850 (b) 825 13. (a) 2 3, 3 5, 1 5, 4 15 (d) 1 2 Answers Project Work Roll a dice 50 times and record each outcome in the table. Calculate the probabilities of getting (a) an even numbers, (b) a number less than 3, (c) a number greater than 3. 1 3 4 2 Statistics and Probability Statistics and Probability 390 Allied The Leading Mathematics-9 Probability 391

CONFIDENCE LEVEL TEST - VI Unit VI : Statistics and Probability Class: 9, The Leading Maths Time: 45 mins. FM: 20 Attempt all the questions. 1. The given histogram shows the marks of some students. (a) How many students are getting the high marks?[1] (b) How many students are getting below 40 marks?[1] (c) Make the continuous data fromthe given histogram. [2] (d) What is the difference the students that get high marks and low marks? [1] 2. The marks obtained by 20 students in full marks 75 is given below. Marks 15 25 35 45 55 65 No. of students 2 6 7 3 8 9 (a) In which series, the given data is presented? [1] (b) Construct a cumulative frequency table from the given data. [1] (c) What is the median marks of the class 9 students? Find it. [2] (d) What is the average marks of the students getting more than median score? [2] 3. Observe the weight (in kg) of 7 students below and answer the following questions. 24, 34, 25, 36, 43, 25, 30 (a) What is their average weight ? Find it. [2] (b) Find the modal weight. [1] (c) Find the median weight. [1] (d) Calculate the first and third quartiles of the above data. [2] 4. Three coins are tossed simultaneously at random. (a) Write down its sample space. [1] (b) Find the probability of getting all are heads. [1] (c) Find the probability of getting at least one head. [2] (d) Is the probability of getting all are heads and all are tails equal? Justify. [1] Best of Luck 10 20 30 50 60 80 90 X 40 Marks No. of students 70 30 25 20 15 10 5 0 Statistics and Probability Statistics and Probability 390 Allied The Leading Mathematics-9 Probability 391

Additional Practice – VI 1. The marks obtained by 20 students in full marks 75 is given below. Marks 20 30 40 50 60 70 No. of students 3 4 3 2 4 4 (a) In which series, the given data is presented? (b) Construct a cumulative frequency table from the given data. (c) What is the median marks of the class 9 students? Find it. (d) What is the average marks of the students getting more than median score? Calculate it. 2. The given data shows the marks of 105 students recorded in the form of a table. (a) How many students are getting the most marks? Write in class interval. (b) How many students are getting in the low interval? Write in class interval. (c) Plot the getting marks and number of students on a frequency polygon, representing the continuous data. (d) How much is the average marks of the students. Calculate it.? 3. The time spent on the studying at home by 20 students is given in the table below. Times in hour 1 2 3 4 5 No. of students 8 6 2 3 1 (a) Write the model value of the data. (b) Find the average time spent of the students. (c) Find the median of the given data. (d) If the students who study either 1 hour or 2 hours are removed from the above data, what would be the average study time? Calculate. 4. The heights of trees in a jungle are given below. Height (meter) 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 No. of plants 5 15 20 18 12 7 (a) In which class intervals is the greatest frequency? (b) Construct a cumulative frequency table from the given data. Statistics and Probability Statistics and Probability 392 Allied The Leading Mathematics-9 Probability 393

(c) Construct the histogram of the given data. (d) Prove that the probability of getting the height of tree more than 50 meters is 1 11. 5. Here are given 20 number cards written from 1 to 20. (a) Write the formula for finding probability P(E). (b) Find is the probability of getting the number card which is exactly divisible by 3. (c) What is the probability of getting the number card NOT divisible by 5? Find it. (d) In which conditions can get maximum value of probability? Write with reason. 6. The marks of 60 students out of full marks 75 are as given below. Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 No. of students 5 15 20 18 12 7 (a) Write the model class of the given data. (b) Construct a cumulative frequency table from the given data. (c) Construct the less than ogive from the given data. (d) Find the probability of getting the marks less than 20. 7. In the figure, a cubical die with its six faces is shown. (a) Write down the sample space for a die rolled. (b) Find the probability of getting 3 on the face. (c) Find the probability of NOT getting odd number on the faces of die. (d) In which conditions can get minimum value of probability? Write with example. 8. Three coins are tossed simultaneously at random. (a) Write down its sample space. (b) Find the probability of getting all the heads. (c) Find the probability of getting at least one head. (d) Is the probability of getting all the heads and all the tails equal? Justify. Statistics and Probability Statistics and Probability 392 Allied The Leading Mathematics-9 Probability 393

9. In a pack of cards, there are 52 playing cards as shown in the figure. (a) Write the probability scale of the events. (b) A card is drawn randomly from the well shuffled cards. What is the probability of getting the face card? (c) One card is drawn randomly from the well shuffled cards. What is the probability of NOT getting a spade. (d) In which conditions can we get maximum value of probability? Write with reason. 10. A bag contains 5 red TT balls, 7 white TT balls and 3 black TT balls having the same size. (a) Write the formula for finding the probability of the events. (b) A ball is drawn randomly from the bag. Find the probability getting white balls. (c) A ball is drawn randomly from the bag. What is the probability NOT getting black balls? Calculate it. (d) A marble is drawn randomly from the bag. What is the probability of getting either black or red or white balls? Is it the maximum probability of such events? 1. 50, 65 3. 1, 2.15, 2, 3.83 5. 3 10 , 3 4 7. 1 6 , 1 2 8. 1 8 , 7 8 9. 3 13 , 3 4 10. 7 15 , 4 5, 1 Answers Statistics and Probability Statistics and Probability 394 Allied The Leading Mathematics-9 Probability PB

Trigonometry Trigonometry PB Allied The Leading Mathematics-9 Trigonometry 395 use of trigonometric ratios in problem solving on height and distance Estimated Working Hours : Competency Learning Outcomes At the end of this unit, the students will be able to: construct the concept of trigonometric ratios. find the values of trigonometric ratios of standard angles. solve the behaviour problems including using trigonometric ratios. 7. Trigonometry 7.1 Trigonometric Ratios 7.2 Trigonometric Ratios of Standard Angles Chapters / Lessons TRIGONOMETRY UNIT 8 (Th. + Pr.) HSS Way: Mathematics begins at Home, grows in the Surroundings and takes shape in School. Specification Grid Unit Areas Total working hour Knowledge Understanding Application Higher ability Total number of items Total number of questions Total Marks No. of items Marks No. of items Marks No. of item Marks No. of item Marks VII. Trigonometry 8 1 1 1 1 1 1 1 1 4 1 4 VII

20.1 Trigonometric Ratios At the end of this topic, the students will be able to: ¾ define the trigonometric ratios of an angle of a right-angled triangle and solve the related problems. Learning Objectives Introduction Trigonometry is an important branch of mathematics. It deals with the measurement of sides and angles of a triangle and their use. The word 'trigonometry' is derived from the two words 'trigon' meaning 'triangle' and 'metron' meaning 'measure'. From ancient times, trigonometry was used in the study of movements of stars. Visible stars were divided into 12 groups called signs of zodaic -/ fzL_. The 12 groups are again grouped into 4 bigger groups. Each of them consists of three signs of zodiac forming what is known as 'astrological trigon' or ‘astrological triangle’. The four trigons are: (a) Watery trigon (b) Earthly trigon Cancer, Scorpio, Pisces Taurus, Virgo, Capricorn (c) Airy trigon (d) Fiery trigon Gemini, Libra, Aquarius Aries, Leo, Sagittarius Besides extensive use in astronomy, trigonometry is used in engineering survey, navigation and so on.Ahumble beginning ofthe study oftrigonometry can be donewith reference to an acute angle of a right-angled triangle. Trigonometric Ratios We begin this section with the definition of trigonometric ratios of an acute angle. Consider an acute angle AOB or BOA. We consider a point P on OA (or OB), and then, draw PQ perpendicular on OB (or OA). We denote ∠AOB or ∠BOA by θ (theta, without sign), PQ by p, OQ by b and OP by h. In the right-angled triangle PQO, PQ is opposite to the angle POQ (or QOP) under consideration, OQ is adjacent to the angle POQ (or QOP) and OP is opposite to the right angle PQO (or OQP). O θ A B O Q p h b (P) (p) P (Q) θ A B CHAPTER 20 TRIGONOMETRY Trigonometry Trigonometry 396 Allied The Leading Mathematics-9 Trigonometry 397

The sides PQ, OQ and OP are respectively called the perpendicular, base and hypotenuse with respect to the acute angle ∠POQ (θ or ∠QOP). From the three sides-perpendicular p, base b and hypotenuse h, we can form the following six ratios: p h, b h, p b, b p, h b, h p These six ratios related to the angle θ are respectively called sine, cosine, tangent, cotangent, secant and cosecant of the angle θ. Thus, sine of the angle θ = p h = Perpendicular Hypotenuse cosine of the angle θ = b h = Base Hypotenus tangent of the angle θ = p b = Perpendicular Base cotangent of the angle θ = b p = Base Perpendicular secant of the angle θ = h b = Hypotenuse Base cosecant of the angle θ = h p = Hypotenuse Perpendicular In the beginning, we usually restrict ourselves to acute angles of a right-angled triangle. The ratio thus defined is known as trigonometric ratios for the acute angle θ of a right-angled triangle. In practice, for the acute angle POQ = θ of the right-angled triangle PQO alongside, we write sin θ = Perpendicular Hypotenuse = p h = PQ OP cos θ = Base Hypotenuse = b h = OQ OP tan θ = Perpendicular Base = p b = PQ OQ cot θ = Base perpendicular = b p = OQ PQ sec θ = Hypotenuse Base = h b = OP OQ cosec θ = Hypotenuse Perpendicular = h p = OP PQ Warning: Don’t write (a) Sin θ for sin θ (b) If sin θ = x, θ = x sin and sin = x θ O θ Q P B A p h Perpendicular Hypotenuse b Base Trigonometry Trigonometry 396 Allied The Leading Mathematics-9 Trigonometry 397

An important property of trigonometric ratios is exhibited by the following example: sin θ = 3 5; sin θ = 6 10 = 3 5 cos θ = 4 5; cos θ = 8 10 = 4 5 tan θ = 3 4; tan θ = 6 8 = 3 4 cot θ = 4 3; cot θ = 8 6 = 4 3 sec θ = 5 4; sec θ = 10 8 = 5 4 cosec θ = 5 3; cosec θ = 10 6 = 5 3 Note: (i) For a given angle of reference, each of the six trigonometric ratios (sine, cosine, tangent, cotangent, secant and cosecant) has the same value whatever be the size of the triangle. (ii) The trigonometric ratios (sine, cosine and tangent) of an angle are respectively the reciprocals of cosecant, secant and cotangent of the angle, i.e., sin θ = 1 cosec θ, cos θ = 1 sec θ, tan θ = 1 cot θ . (iii) For the angle C, (a) AB = 4 is the perpendicular (b) BC = 3 is the base (iv) Since hypotenuse is the greatest side in a right-angled triangle, So, sin θ = p h and cos θ = b h are both less than 1. Example-1 In the given triangle ABC, ∠B = 90°. (a) Is sin A = cos C? (b) Is cos A = sin C? (c) Is tan A = cot C? Solutions: (a) Here, sin A = perp. for A hyp. for A = BC AC = base for C hyp. for C = cos C (b) Here, cos A = base for A hyp. for A = AB AC = perp. for C hyp. for C = sin C (c) Here, tan A = perp. for A base for A = BC AB = base for C perp. for C = cot C. O θ Q P 4 5 3 A θ B C 8 10 6 B 3 4 C A B C A Trigonometry Trigonometry 398 Allied The Leading Mathematics-9 Trigonometry 399

Example-2 In the given ∆ABC, if (a) sin A = 3 5, h = 20; find p. (b) cos A = 4 5, b = 8; find h. (c) tan A = 3 4, p = 8; find p. Solutions: Here, (a) sin A = p h = 3 5 and h = 20. So, p 20 = 3 5 ⇒ p = 3 5 × 20 = 12 (b) cos A = b h = 4 5 and b = 8. So, 8 h = 4 5 ⇒ h = 8 × 5 4 = 10 (c) tan A = p b and b = 8. So, p 8 = 3 4, or, p = 3 4 × 8 = 6. Example-3 For the annexed figure, show that: (a) sin A × cosec A = 1 (b) cos A × sec A = 1 (c) tan A × cot A = 1. Solutions: In the given ∆ABC, AC = 8, AB = 4 ∴ BC = AC2 – AB2 [ Pythagoras' theorem, p2 + b2 = h2 ] = 82 – 42 = 64 – 16 = 48 = 16 × 3 = 4 3 (a) Now, sin A = BC AC = 4 3 8 = 3 2 , cosec A = AC BC = 8 4 3 = 2 3 , So, sin A × cosec A = 3 2 × 2 3 = 1. (b) Here, cos A = AB AC = 4 8 = 1 2, sec A = AC AB = 8 4 = 2 1, So, cos A × sec A = 1 2 × 2 1 = 1. (c) Here, tan A = BC AB = 4 3 4 = 3, cot A = AB BC = 4 4 3 = 1 3 , So, tan A × cot A = 3 × 1 3 = 1. 4 B 5 3 C A B 8 4 C A Trigonometry Trigonometry 398 Allied The Leading Mathematics-9 Trigonometry 399

Example-4 From the adjoining figure, find the values of sin θ, cos θ and tan θ. Solution: From the given figure, In right-angled triangle ACD, AD = 17 cm and CD = 8 cm AC = AD2 – CD2 [ Pythagoras' theorem, p2 + b2 = h2 ] = 172 – 82 = 289 – 64 = 225 = 15 cm Again, in right-angled triangle ABC, AC = 15 cm and AB = 9 cm BC = AC2 – AB2 [ Pythagoras’ theorem, p2 + b2 = h2 ] = 152 – 92 = 225 – 81 = 144 = 12 cm Now, we have sin θ = AB AC = 9 cm 15 cm = 3 5, cos θ = BC AC = 12 cm 15 cm = 4 5, tan θ = AB BC = 9 cm 12 cm = 3 4. Example-5 If sin A = 3 5, find the values of cos A and tan A. Solution: Here, sin A = 3 5 = p h = k ∴ p = 3k and h = 5k Now, we have b = h2 – p2 = (5k)2 – (3k)2 = 25k2 – 9k2 = 16k2 = 4k ∴ cos A = b h = 4k 5k = 4 5 and tan A = p b = 3k 4k = 3 4. Example-6 Convert tan θ into sin θ. Solution: We know that sin θ = p h, cos θ = b h and tan θ = p b Now, tan θ = p b = p h b h [ Dividing by h on numerator and denominator] = p h h2 – p2 h [ By Pythagorean relation, b = h2 – p2 ] B C D A 17 cm θ 9 cm 8 cm Trigonometry Trigonometry 400 Allied The Leading Mathematics-9 Trigonometry 401