Math 4 part 1

Lesson 3 Vertex Axis of Minimum/MaximumA. Symmetry value (0, 2) x = 0 or y-axis No. Sign of leading (0, 3) x = 0 or y - Minimum value; y = 2 coefficient axis Minimum value; y = 3 (0, 0) x = 0 or y-axis 1 Positive (0, 5) x = 0 or y-axis Maximum value; y = 0 2 Positive (2, 1) x=2 Maximum value; y = 5 (2, 3) x=2 Minimum value; y = 1 3 Negative (-3, -5) x = -3 Maximum value; y = 3 4 Negative (0, -2) x = 0 or y-axis Minimum value; y = -5 5 Positive (-2, 3) x = -2 Minimum value; y = -2 6 Negative (-1, -2) x = -1 Maximum value, y = 3 7 Positive Minimum value, y = -2 8 Positive 9 Negative 10. PositiveB.No. Direction of Vertex Axis of Minimum/Maximum Opening Symmetry Value (3, 0) x=31 Upward (-2, -1) x=-2 Minimum value, y = 02 Downward (0, 9) x = 0 or y-axis Maximum value, y = -13 Downward (4, 7) x= 4 Maximum value, y = 94 Upward (6, 5) x=6 Minimum value, y = 75 Downward  3 ,− 25  x= 3 Maximum value, y = 56 Upward 2 4  2 Minimum value, y = - 25 (-1, -9)7 Upward  1,− 81 x = -1 48 Upward 2 4  x= 1  3 , 29  Minimum value, y = -99 Downward 2 4  2 Minimum value, y = - 81  − 5 , 49 10 Downward  6 12  x= 3 4 2 Maximum value, y = 29 x=-5 4 6 Maximum value, y = 49 12What have you learned1. b 6. c2. d 7. c3. d 8. d4. a 9. b5. c 10.a 24

Module 1 Statistics What this module is about This module deals with the definition of statistics and terms used in the study ofstatistics. It will also discuss the history and importance of the study of statistics,summation rules, sampling techniques, collection of statistical data and organizingcollected data in a table, constructing frequency distribution tables, and finding themeasures of central tendency for ungrouped data. As you go over the discussion andexercises, you will appreciate more the importance of statistics in daily life. Enjoylearning this module and go over the discussion and examples if you have not yetmastered a concept. What you are expected to learn This module is designed for you to: 1. define statistics, sample, and population. 2. give the history and importance of the study of statistics 3. use the rules of summation to find sums 4. explain the different sampling techniques 5. collect statistical data and organize in a table 6. construct frequency distribution tables 7. find the measures of central tendency using ungrouped data • mean • median • mode

How much do you know1. It is the branch of mathematics concerned with the techniques by which data are collected, organized, analyzed, and interpreted.a. Information Technology c. Trigonometryb. Statistics d. Geometry2. It is a method of selecting the elements of a sample from the population under consideration.a. Sampling c. Organizingb. Drawing d. Collecting3. It is used to present data in a most systematic and organized manner to make its reading and interpretation simple and easy.a. Graph c. Drawingb. Table d. Sampling3. Mrs. Borromeo wants to study the heights and weights of the students in her class. Which of the following samples is most likely to be a good representation of the whole class?a. A sample consisting of all students whose surnames start with E.b. A sample consisting of all athletes in the class.c. A sample consisting of students whose birthdays are from January to June.d. A sample consisting of students whose names were drawn out of a box which contained all the names of the students in the class. 54. Which of the following means ∑ xi ? i=1 a. 1 + 2+ 3 + 4 + 5 b. x + 2x + 3x + 4x + 5x c. x1 + x2 + x3 + x4 + x5 d. none of the above 2

6. The frequency distribution below shows the scores obtained by 300 students in a Mathematics test of 50 items. Score Number of 45-49 Students 40-44 35-39 15 30-34 32 25-29 42 20-24 108 15-19 67 10-14 21 Total 10 5 300 What interval contains the highest frequency? a. 10-14 b. 45-49 c. 30-34 d. 25-297. What class size was used in number 6? a. 5 b. 4 c. 3 d. 25. It is the measure that occurs most in a distribution. a. mean b. median c. mode d. class size6. The following are scores obtained by 10 students in an achievement test.47 45 35 44 48 39 37 29 28 50What is the mean?a. 28 c. 40.2b. 35 d. 457. What is the median in no, 9a. 39 c. 43.5b. 41.5 d. 44 3

What you will do Lesson 1 Definition of Terms Related to Statistics Statistics is a branch mathematics that deals with the collection, classification,description, and interpretation of data obtained by the conduct of surveys andexperiments. Its fundamental purpose is to describe and draw inferences about thenumerical properties of a population. Two important terms that you should understand in studying statistics arepopulation and sample. In statistics, population does not only mean a group of people. Population mayalso mean a defined group or aggregates of objects, animals, materials, measurements,“things”, “events” or “happenings” of any kind. Thus, a sack of rice, a whole pizza pie, ora set of weights and heights are considered population. Since it would be impractical to study the whole population as in the case of asack of rice, then it is necessary to just take a sample of the population. Thus, a handfulof rice is a sample of the population in a sack of rice. Thus, sample is defined as anysubgroup of the population drawn by some appropriate method from the population. Itshould be a representative of the population, that is, the sample will show the propertiesof the population.Try this out Study the following situations. Identify the phrase which represents the sampleand which phrase shows the population. 1. Mrs. Jara wants to know the nutritional status of the first year students in her school so she got 150 first year students to represent the year level. 2. When Sandra bought a sack of rice, she examined a handful from the sack to check if it is the variety she wants. 3. A doctor wants to know what causes the infection in a patient so he requested for the patient’s blood examination. The medical technologist extracted only 10 cubic centimeters of blood from the patient for examination. 4. The chef wants to check if the food being cooked tastes as he wants it to be so he tasted a spoonful of it. 4

5. The school guidance counselor would like to know the course preference of the graduating students in their school so she interviewed 100 of the fourth year students. Lesson 2 History and Importance of Statistics Historical records show that since the beginning of civilization simple forms ofstatistics had already been used. This is manifested by pictorial representations andother symbols used to record numbers of people, animals, and inanimate objects onskins, slabs, sticks of wood, or the walls of caves. Records show that even before 3000B.C., the Babylonians used small clay tablets to record tabulations of agriculturalharvests and of commodities bartered or sold. The Egyptians analyzed the populationand material wealth of their country before they begun building the pyramids in the 31stcentury B.C. Even the biblical books of Numbers and 1 Chronicles show statisticalworks. Numbers contained two separate censuses of the Israelites while 1 Chroniclesdescribed the material wealth of various Jewish tribes. In China, similar numericalrecords existed before 2000 B.C. As early as 594 B.C., the ancient Greeks heldcensuses used as bases for taxation Records also show that the Roman Empire was the first government to gatherextensive data about the population, area, and wealth of the territories that it controlled.In Europe, few comprehensive censuses were made during the Middle Ages. in theearly 16th century, registration of deaths and births begun in England. Then in 1662 thefirst noteworthy statistical study of population was made. In 1691, a similar study ofmortality made in Breslau, Germany was used by the English astronomer EdmondHalley as a basis for the earliest mortality table. In the 19th century, investigatorsrecognized the need to reduce information to numerical values to avoid the ambiguity ofverbal description. At present, statistics is a reliable means of describing accurately the values ofeconomic, political, social, psychological, biological, and physical data. Statistics servesas a tool to correlate and analyze collected data. It is no longer confined to gatheringand tabulating data. Now, it is also a process of interpreting the information that servesas a basis for preparing plans. 5

Lesson 3The Summation Process The study of statistics involves the collection of data or measurement. Thus,there is always a need to add several numbers. The Greek capital letter sigma, Σ isused in the process. The symbol Σ, read as the sum of tells you to add certainnumerical values.Example 1: Consider the scores obtained by 10 students in a 50-items mathematics test.Student No. Score 1 35 2 40 3 29 4 37 5 25 6 33 7 49 8 47 9 28 10 42 For convenience, variables will be used to present the data. Let x = score obtained by each student xi = different values or observations of x xi is read as “x sub i” where i is a subscript which indicates the position of eachvalue in the series. In the given data, there are 10 observations denoted as x1, x2, x3, x4, x5, x6, x7,x8, x9, x10. 10 ∑Hence, xi = x1+ x2+ x3+ x4+ x5+ x6+ x7+ x8+ x9+ x10. i−1 10 ∑The symbol xi is read as “the sum of 10 observations x1 to x10”. i−1 To substitute the data: 10 ∑ xi = 35 + 40+ 29 + 37 + 25 + 33 + 49 + 47 + 28 + 42 i−16

10 ∑ xi = 365 i−1 For large observations, say 50, the summation will be expressed as: 50 ∑ xi = x1+ x2+ x3 + …..+x50. i−1 n ∑In general, xi = x1+ x2+ x3 + …..+xn. i−1 If all the given values of a variable are to be used in finding the sum, the limits ofthe summation are usually omitted, as 10 ∑xi = ∑x i−1Example 2: Given are the ages of the first 4 shoppers at a newly opened conveniencestore in the neighborhood – 12, 24, 30, 45. 1. What will x represent in the information given? 2. What will the subscript i represent? 3. Write an expression for the sum. 4. What are the lower and upper limits of the expression? 5. Write the formula for the summation and find the sum of the given information.Answers: 1. x will represent the ages of the first 4 shoppers in the newly opened convenience store. 2. I will represent the first 4 shoppers in the newly opened convenience store. 4 3. ∑ xi is the expression for the summation. i=1 4. The lower limit is 1 and the upper limit is 4. 4 ∑5. xi = x1+ x2+ x3+ x4 i=1 = 12 + 24 + 30 + 45 = 111 This time, consider 5 observations. If the sum of five observations is written as: 5 ∑ xi = x1+ x2+ x3+ x4+ x5; i=1the sum of the squares of the five observations is represented as: 7

5 ∑ xi2 = x12+ x22+ x32+ x42+ x52; i=1the sum of the products of pairs of five observations is expressed as: 5 ∑aixi = a1x1+ a2x2+ a3x3+ a4x4+ a5x5 i=1Example 3: Consider the first four multiples of 2: 2, 4, 6, 8. Use the corresponding summation formula to find the following:1. the sum of the first four multiples of 22. the sum of the squares of the first four multiples of 23. the sum of the products of pairs of values consisting of the first four counting numbers and the first four multiples of 2.Solutions: 4∑1. xi = x1+ x2+ x3+ x4 i=1 =2+4+6+8 = 20 4∑2. xi2 = x12+ x22+ x32+ x42 i=1 = 22 + 42 + 62 + 82 = 4 + 16 + 36 + 64 =120 4∑3. ai xi = a1x1+ a2x2+ a3x3+ a4x4 i=1 = 1(2) + 2(4) + 3(6) + 4(8) = 2 + 8 + 18 + 32 = 60 6 6Example 4: Find 1. ∑3 2. ∑(−3) i=1 i=1Solutions: 61. ∑ 3 = 3 + 3 + 3 + 3 + 3 + 3 = 6(3) = 18 i=1 8

6 2. ∑(−3) = (-3) + (-3) + (-3) + (-3) + (-3) + (-3) = 6(-3) = -18 i=1 Observe that in example 4, the summation of a constant c is the product of the nconstant and the number of terms n in the summation, that is, ∑ c = nc i=1Try this outExpress each of the following as a sum: 7 1. ∑ xi i=1 5 2. ∑ z j j=1 3 3. ∑ yk k =1 5 4. ∑p j=1 3 ∑5. ak yk k =1Express the following sums in summation notation: 6. x1+ x2+ x3 + …..+x20 7. x12+ x22+ x32+ x42+ x52+ x62+ x72+ x82 8. a1x12+ a2x22+ a3x32+ a4x42 9. 4x12+ 4x22+ 4x32+ 4x42 10. (y1 + z1) +(y2 + z2)+ (y3 + z3)+ (y4 + z4)+ (y5 + z5)Use summation to find the following: 11. the sum of the positive odd integers less than 20 12. the sum of the first ten positive even integers 13. the sum of the squares of the first five positive even integers 14. the sum of the products of the first four counting numbers and the first four multiples of 3. 15. the sum of the products of 5 times the positive odd integers less than 15. 9

Compute: 4 16. ∑5 i=1 10 17. ∑(−2) i=1 8 18. ∑ 4 i=1 Lesson 4 Sampling Techniques The method of drawing samples from a population is of very important. There areseveral ways of doing this. One way is by simple random sampling. This is a procedure where a sample isselected in such a way that every element is as likely to be selected as any otherelement in the population. Another way is by systematic random sampling. This method is a samplingprocedure with a random start. Another method is the stratified random sampling. This is used when thepopulation can be naturally classified into groups or strata. Example: The clinic teacher wants to determine the average height and weight of the first year students. How can she randomly select 50 students consisting of 250 male students and 300 female students to represent the population using (a) simple random technique? (b) systematic random technique? (c) stratified random technique? Answers: The clinic teacher can randomly select the sample using simple random sampling byfollowing these simple steps: 1. Assign the students with numbers. 2. Write the student number with his/her corresponding height in uniform size slips of paper. 3. Roll the pieces of paper uniformly and place them in a box. 10

4. Draw a slip of paper at a time, shaking the box after each draw until 50 samples are taken. The clinic teacher can select the sample using the systematic random samplingusing the steps as follows: 1. Using the same data and with the students assigned with numbers, and arranged chronologically, the clinic teacher with eyes closed, points to a number. If the number pointed to is, let us say, 7, student number 7 becomes a part of the sample (sample number 1). This is a “random start”. 2. From student 7, count 1 to 7 repeatedly until all 50 samples are taken. Numbers which were previously selected will eliminated in the counting. The clinic teacher can select the sample using the stratified random sampling byusing the following procedures: 1. The data should be classified into two groups, male and female. 2. Get a proportional number of samples from each group or strata. The number of samples from the males will be 250 or 5 of 50 which is 23 and from the 550 11 females 27. 3. Place the slips of paper, properly filled up in separated boxes for each group. 4. Draw, one at a time, the required number for each group. Try this out1. Mrs. Lucas is studying the heights and weights of the students in her class. Which of the following samples is most likely to be a good representation of the whole class? Justify your answer. a. A sample consisting of all athletes in the class. b. A sample consisting of all students whose surnames start with D. c. A sample consisting of students whose names were drawn out of a box which contained all the names of the students in the class. d. A sample consisting of students whose birthdays are from January to June.2. For his report in Social Studies, Dennis wishes to wishes to interview a sample of Metro Manila residents to determine their opinion regarding the economic status of the country today.Tell whether he could find a sample that reflects the entire population being studied at 1) a depressed area in Payatas. 2) a shopping mall in Makati. 3) the Starbucks coffee shop.3. A researcher wants to know the average age of teachers in a certain community. Fifty teachers from the elementary and 25 teachers from the secondary levels were 11

interviewed for the purpose. How will the researcher choose a sample size of 20using: a. simple random sampling b. systematic random sampling c. stratified random sampling Lesson 5 Collecting and Organizing Data in a Table The study of statistics begins with the collection of data or measurements. Datacollected should be organized systematically for easier and faster interpretation. Theymaybe presented in any of the following forms: The textual form can be used if the data to be presented if few. The tabular and graphical forms are used when more detailed information aboutthe data is to be presented. A table is used when you want to present a data in a systematic and organizedmanner so that reading and interpretation will be simpler and easier. When a table isused, you must remember the following: 1. The title of the table. 2. Indicate the date of the survey. 3. Arrange the data systematically in columns. The columns must be properly labeled. 4. Identify the source of the data.Example 1: Mahusay National High School Enrolment, SY 2005-2006 Year Level Male Female First 216 267 Second 197 216 Third 187 227 Fourth 176 215 Total 776 925 You will observe that the table above shows clearly the enrolment data inMahusay National High School for the school year 2005-2006. 12

Another type of tabular presentation is the frequency table also known as afrequency distribution. It is an arrangement of the data that shows the frequency ofoccurrence of different values of the variables. A frequency table is constructed by listing the measurements from highest tolowest, then making tally marks to record how often each number occurs. After tallying,count the marks and record them in the proper column.Example 2: The scores of 45 students on a 20-point Science quiz are as follows:17 20 15 18 19 16 11 10 15 1612 12 13 14 11 10 14 13 12 1113 15 14 10 15 16 17 17 18 2020 18 19 19 18 17 16 15 12 1213 14 15 19 20Prepare a frequency table for the set of data.Solution: To prepare a frequency table for the given set of scores, the scores are listed from highest to lowest, tally marks are made and counted. The counted tally marks will then be recorded under the column frequency. Notice that every 5th tally crosses the first four tallies. This is done to make counting of marks easier especially if the number of cases is rather big.Score Tallies Frequency 20 //// 4 19 //// 4 18 //// 4 17 //// 4 16 //// 4 15 //// / 6 14 //// 4 13 //// 4 12 //// 5 11 /// 3 10 /// 3 45TotalTry this out 1. The school budget for Maintenance and Other Operating Expenses of a certain school for Calendar Year 2004 is given below. 13

Expense Item Amount (in Pesos) Power 600 000 Water 95 000 Communication 60 000 Supplies Repair 1 600 000 Others 920 000 100 000a) How much is the total budget of the school for CY 2004?b) Which expense item received the biggest allocation? What percent of the total budget was allocated for it?c) Which expense item received the least allocation? What percent of the total budget was allocated for it?2. The following shows the scores of 15 students in mathematics for the second grading period. Prepare a frequency table given the data below.87 90 89 92 9488 90 91 88 8790 94 92 91 903. The following are heights of male fourth year students in a school. Prepare a frequency table for this set of data.1.36 1.51 1.61 1.61 1.62 1.62 1.62 1.59 1.58 1.611.38 1.49 1.65 1.63 1.58 1.57 1.61 1.62 1.63 1.651.44 1.59 1.57 1.57 1.58 1.60 1.61 1.63 1.64 1.641.55 1.58 1.59 1.65 1.66 1.72 1.56 1.68 1.69 1.63 Lesson 6 Frequency Distribution Tables If the number of measures in consideration is rather big, the presentation of datais further simplified by grouping the measures into class intervals called a frequencydistribution. A frequency distribution is a distribution of the total number of measures orfrequencies over arbitrarily defined categories or classes. The number of measuresfalling under a class is called class frequency.Example 1. 14

The frequency distribution below shows the scores obtained by 300 students inan English test of 50 items.Score Number of45-49 Students40-4435-39 1530-34 3225-29 4220-24 10815-19 6710-14 21Total 10 5 300 In the example above, the symbol 45-49 and the other symbols which follow upto 10-14 are called class intervals. The end numbers are called class limits. Forinstance in the class interval 45-49, 45 is called the lower limit while 49 is called theupper limit. Each class interval has also a lower boundary and a higher boundary. For theclass interval 45-49, the lower boundary is 44.5 while the higher boundary is 49.5.Hence, for the class interval 45-49, 44.5 – 49.5 are called the class boundaries. The size of the class interval, also called class size is the difference between theupper boundary and the lower boundary. Hence, the class size in the given example is5 A class interval has also a midpoint or a class mark. It is obtained by taking halfthe sum of the lower and upper class limit. For instance, the midpoint of the classinterval 45-49 is 45 + 49 or 47. 2 The following are the rules in determining the size of the class interval: 1. The class interval must cover the total range of the observation where the range, R = H – L (H = highest and L = lowest). It is usually between 10 to 20 intervals. 2. Select class intervals with a range of 1, 3, 5, 10, or 20 points since these will meet the requirements of most set of data. 3. Start the class interval at a value which is a multiple of the size of the interval. For example, with a class interval of 3, the intervals should start with the values 3, 6, 9, etc. 15

4. Arrange the class intervals according to the order of magnitude of the observations they include. The class interval containing the largest observations is usually placed at the top.Example 2: The following are scores obtained by a group of 50 students on their English IV examinations. Prepare a frequency distribution for these data using a class interval of 5. 39 93 80 49 41 85 75 59 62 68 34 49 50 46 69 72 73 76 77 54 95 63 66 64 88 90 51 53 56 79 70 70 78 85 86 59 66 72 77 76 71 79 70 65 40 57 82 75 89 82Solution: Since the class interval is already given, and the lowest score is 34 then theclass interval containing the lowest score should be 30-34 since the rule states that theclass interval should start with a number which is divisible by the class size. Afterarranging the class intervals, tally the scores to determine the frequency. Look at theobtained frequency distribution below.Scores Tallies Frequency95-99 / 190-94 // 285-89 //// 5 80-84 3 74-79 /// 9 70-74 //// //// 7 65-69 //// // 5 60-64 //// 3 55-59 4 50-54 /// 4 45-49 //// 3 40-44 //// 2 35-39 /// 1 30-34 // 1 Total / 50 /Try this out1. A sample of fifty shoppers at a newly opened convenience store has been randomly selected. The following data show the shoppers’ ages. Determine 16

the appropriate class interval to use then prepare a frequency distribution for the data. 12 20 17 19 23 32 15 45 60 65 18 22 27 35 37 57 47 38 40 28 13 10 19 24 29 28 38 47 48 57 27 29 33 34 49 76 55 65 37 39 40 14 17 20 32 33 60 65 62 572. The following are the weights of randomly selected 1st year students in kilograms. Prepare a frequency distribution for this set of data. 37 35 40 42 36 57 38 44 60 45 52 64 38 39 40 42 50 56 45 43 38 39 50 41 42 56 57 54 55 60 35 38 40 40 42 53 47 48 39 50 35 37 39 39 50 Lesson 7 Measures of Central Tendency for Ungrouped Data Aside from tables and graphs, another way of describing a set of data is bystating a single numerical value associated with it. This value is where all the othervalues in a distribution tend to cluster. It is called the average or measure of centraltendency. There are three kinds of average: the mean, the median, and the mode.The Mean The mean (also known as the arithmetic mean) is the most commonly usedmeasure of central position. It is the sum of measures divided by the number of _measures in a variable. It is symbolized as x (read as x bar). The mean is used to describe a set of data where the measures cluster orconcentrate at a point. As the measures cluster around each other, a single valueappears to represent distinctively the total measures. It is, however, affected by extrememeasures, that is, very high or very low measures can easily change the value of themean.To find the mean of ungrouped data, use the formula _ = Σx x N 17

where ∑x = the summation of x (sum of the measures) N = number of values of xExample: The grades in Chemistry of 10 students are 87, 84, 85, 85, 86, 90, 79, 82, 78, 76. What is the average grade of the 10 students?Solution: Mean = 87 + 84 + 85 + 85 + 86 + 90 + 79 + 82 + 78 + 76 = 832 = 83.2 10 10The Median The median is the middle entry or term in a set of data arranged in eitherincreasing or decreasing order. The median is a positional measure. Thus the values of the individual measuresin a set of data do not affect it. It is affected by the number of measures and not by thesize of the extreme values. To find the median of a given set of data, take note of the following: 1. Arrange the data in either increasing or decreasing order. 2. Locate the middle value. If the number of cases is odd, the middle value is the median. If the number of cases is even, take the arithmetic mean of the two middle measures.Example 1: The number of books borrowed in the library from Monday to Friday last week were 58, 60, 54, 35, and 97 respectively. Find the median.Solution: Arrange the number of books borrowed in increasing order. 35, 54, 58, 60, 97 The median is 58.Example 2: Cora’s quizzes for the second quarter are 8, 7, 6, 10, 9, 5, 9, 6, 10, and 7. Find the median.Solution: Arrange the scores in increasing order. 5, 6, 6, 7, 7, 8, 9, 9, 10, 10 18

Since the number of measures is even, then the median is the average ofthe two middle scores. Md = 7 + 8 = 7.5 2The Mode The mode is another measure of position. The mode is the measure or valuewhich occurs most frequently in a set of data. It is the value with the greatest frequency.To find the mode for a set of data – 1. select the measure that appears most often in the set;2. if two or more measures appear the same number of times, and the frequency they appear is greater than any other measures, then each of these values is a mode;3. if every measure appears the same number of times, then the set of data has no mode.Example 1: The shoe sizes of 10 randomly selected students in a class are 6, 5, 4, 6,4 1 , 5, 6, 7, 7 and 6. What is the mode? 2Answer: The mode is 6 since it is the shoe size that occurred the most number of times.Example 2: The sizes of 9 classes in a certain school are 50, 52, 55, 50, 51, 54, 55, 53 and 54.Answer: The modes are 54 and 55 since the two measures occurred the same number of times. The distribution is bimodal. Try this outFind the mean, median, and mode (modes) of each of the following sets of data.. 1. 29, 34, 37, 22, 15, 38, 40 2. 5, 6, 7, 7, 9, 9, 9, 10, 14, 16, 20 19

3. 82, 61, 93, 56, 34, 57, 92, 53, 57 4. 26, 32, 12, 18, 11, 12, 15, 18, 21 5. The scores of 20 students in a Physics quiz are as follows: 25 33 35 45 34 26 29 35 38 40 45 38 28 29 25 39 32 37 47 45 Let’s Summarize Statistics is a branch mathematics that deals with the collection, classification, description, and interpretation of data obtained by the conduct of surveys and experiments. Population is a defined group or aggregates of objects, animals, materials, measurements, “things”, “events” or “happenings” of any kind. Sample is defined as any subgroup of the population drawn by some appropriate method from the population. Sampling is the process of selecting the elements of a sample from the population being studied. The methods of sampling include simple random sampling, systematic random sampling, and stratified random sampling. A table is used to present a data in a systematic and organized manner to make its reading and interpretation simple and easy. A frequency distribution is a distribution of the total number of measures orfrequencies over arbitrarily defined categories or classes. The number of measuresfalling under a class is called class frequency. The mean (also known as the arithmetic mean) is the most commonly usedmeasure of central position. It is the sum of measures divided by the number of _measures in a variable. It is symbolized as x (read as x bar). It is used to describe a setof data where the measures cluster or concentrate at a point. The median is the middle entry or term in a set of data arranged in eitherincreasing or decreasing order. It is a positional measure. The values of the individualmeasures in a set of data do not affect the median. It is affected by the number ofmeasures and not by the size of the extreme values. 20

The mode is the value which occurs most frequently in a set of data.What have you learned1. What meaning of this symbol ∑ ? c. product d. quotient a. sum b. difference2. It a procedure where a sample is selected in such a way that every element is likely to be selected as any other element of a population.a. Samplingb. Simple random samplingc. Systematic random samplingd. Stratified random sampling 103. In the expression ∑ xi , what is the upper limit? i=1a. i c. 1b. x d. 104. The enrolment of a certain school for school year 2005-2006 is shown below. Enrolment, SY 2005-2006 Year Level Male Female First 456 497 Second 427 456 Third 487 467 Fourth 356 425How many students enrolled for the school year?a. 953 c. 1845b. 1726 d. 35715. What is x1 + x2 + x3 + . . . .+ x20 using summation notation? 20 20a. ∑ xi ∑b. x20 i=1 i=1 21

20 d. none of the abovec. ∑ x i=16. The frequency distribution below shows the scores obtained by 60 students in a Science. Score Number of 80-89 Students 70-79 60-69 4 50-59 6 40-49 12 30-39 10 20-29 8 10-19 10 Total 7 3 60 What interval contains the lowest frequency? a. 10-19 b. 20-29 c. 30-39 d. 45-497. What class size was used in number 6? a. 5 b. 8 c. 9 d. 108. It is the middle most measure that occurs in a distribution. a. mean b. mode c. median d. class size9. The following are scores obtained by a student in 10 quizzes. 7554897985 22

What is the mode?a. 8 c. 9b. 5 d. 710. What is the median in no. 9? c. 7 d. 8a. 5 d.b. 9 23

Answer keyHow much do you know1. b 6. c2. a 7. a3. d 8. c4. c 9. c5. c 10. bTry this outLesson 1 Sample Population 1 150 first year student First year students 2 Handful of rice One sack of rice 3 10 cubic centimeters of blood Blood 4 Food cooked Spoonful of it 5 100 fourth year students Graduating studentsLesson 31. x1 + x2 + x3 + x4 + x5 + x6 + x72. z1 + z2 + z3 + z4 + z53. y1 + y2 + y34. p1 + p2 + p3 + p4 + p55. a1y1 + a2y2 + a3y3 206. ∑ xi i=1 8∑7. xi2 i=1 42∑8. aixi i=1 429. 4∑ xi i=1 5∑10. (xi + zi ) i=111. 10012. 12013. 5514. 16515. 245

16. 20 17. -20 18. 32Lesson 4 1. c 2. a 3. Using simple random sampling, the following steps should be done: a. Write the names of the teachers in uniform sized slips of paper. b. Roll the slips of paper. c. Put the rolled slips of paper in a box. d. Draw at random 20 slips of paper By systematic random sampling, the following steps should be done: a. Assign to each teacher a number. b. Arrange the numbers in order c. With eyes closed, point to one of the numbers at random. The number pointed to will be the random start. d. Using the number drawn as a random start, count repeatedly until 20 samples are drawn. By stratified random sampling, the following should be done: a. Classify the teachers into two groups, one for elementary and another for secondary. b. Get a proportional number of samples from each group. Elementary :  50 20 = 13  75  Secondary :  25 20 = 7  75  c. Place the slips of paper properly filled up in separate boxes for each group. d. Draw one slip of paper one at a time until 20 samples are drawn.

Lesson 51. a. Php 3, 375, 000 b. supplies – 47.4% c. communication – 1.78%2. Score Frequency 94 2 92 2 91 2 90 4 89 1 88 2 87 2 Total 153. Height Frequency 1.72 1 1.69 1 1.68 1 1.66 1 1.65 3 1.64 2 1.63 4 1.62 4 1.61 5 1.60 1 1.59 3 1.58 4 1.57 3 1.56 1 1.55 1 1.51 1 1.49 1 1.44 1 1.38 1 1.36 1 Total 40

Lesson 6 Ages Frequency 1. 75-79 1 70-74 0 65-69 3 60-64 3 55-59 4 50-54 0 45-49 5 40-44 2 35-39 6 30-34 5 25-29 6 20-24 5 15-19 6 10-14 4 Total 50 2. Weight FrequencyLesson 7 63-65 1 60-62 2 57-59 2 54-56 4 51-53 2 48-50 5 45-47 3 52-44 6 39-41 10 36-38 7 33-35 3 Total 45 Mean Median Mode 1 30.71 34 None 2 10.18 9 3 65 57 9 4 18.33 18 57 5 35.25 35 12,18 Bimodal 45

What have you learned 1. a 2. a 3. d 4. d 5. a 6. d 7. d 8. c 9. b 10. c

Module 1 Triangle Trigonometry What this module is about This module will guide you to determine the kind of equation you will useto solve the missing parts of a right triangle. This will require the use oftrigonometric functions. Here, you will also learn how to solve problems involvingright triangles.What you are expected to learnThis module is designed for you to: 1. determine the equation in solving the missing parts of a right triangle. 2. Apply trigonometric functions to solve right triangle given: a. the length of the hypotenuse and length of one leg b. the length of the hypotenuse and one of the acute angles c. the length of one leg and one of the acute angles d. the length of both sides 3. Solve problems involving right triangle.How much do you knowIn rt. ∆COD angled at O, if ∠D is an acute angle what is its C1. opposite side2. adjacent side do3. hypotenuse OD c

Give the equation that can be used to find the required parts of right triangleDEF.If m ∠D = 70 and f = 25, find: D 70° f = 25 e Ed F4. d5. eUsing another acute ∠F, if m∠F = 60° and d = 20 , find 6. e 7. fSolve right ∆BCA, If ∠B = 80°, and a = 14, B c 80° a =14 C Afind: b8. b9. c10. ∠A 2

What you will do Lesson 1Determine the Equation in Solving the Missing Parts of a Right TriangleConsider the right triangle ACB below. A bc Ca BThe hypotenuse of the triangle is c.The side opposite angle A is a.The side adjacent to angle A is side b.SOH-CAH-TOA is a mnemonic used for remembering the equations:Sin = Opposite HypotenuseCos = Adjacent HypotenuseTan = Opposite AdjacentExamples:1. Identify the opposite and adjacent side as well as the hypotenuse of right triangle STP for any acute angle of the right ∆. s TP pt S 3

a. Using ∠P as the acute angle: p – is the opposite side s – is the adjacent side t – is the hypotenuse of rt. ∆ STP b. Using ∠S as the acute angle: s - is the opposite side p – is the adjacent side t – is the hypotenuse2. Without solving, determine the equation for the missing parts of a rt. ∆. s TP 530 p t = 12SGiven: ∠P = 530 and t = 12.a. Solve for s:Solution: ∠P is the acute angle, t is the length of the hypotenuse, s is the length of the adjacent side of ∠P, we can use CAH. cos P = s the required equation tcos 530 = s 12 s = 12 cos 530b. Solve for p:Solution: ∠P is the acute angle , t is the hypotenuse, and p is the length of the opposite side of ∠P, we can use SOH. 4

sin P = p the required equation tsin 530 = p 12 p = 12 sin 5303. In the figure, if m∠B = 670 and b = 10.6 cma. Solve for a:Solution: ∠B is the acute angle, b is the opposite side and a is theadjacent side of the given acute angle. Use TOA. Btan B = b 67o a tan 670 = 10.6 ac a C b = 10.6 Aa tan 670 = 10.6a = 10.6 the required equation tan 67ob. Solve for c:Solution: ∠B is the acute angle, b is the opposite side and c is the hypotenuse of the given acute angle, we can use SOH. sin B = b the required equation c sin 670 = 10.6 cc sin 670 = 10.6 c = 10.6 sin 67o 5

Try this outUsing the figure below, write the equations that would enable you to solve eachproblem. B14 a A 420 C B1. If A = 15 and c = 37, find a.2. If A = 76 and a = 13, find b.3. If A = 49 13’ and a = 10, find c.4. If a = 21.2 and A = 71 13’, find b.5. If a = 13 and B = 16 , find c.6. If A = 19 07’ and b = 11, find c.7. If c = 16 and a = 7, find b.8. If b = 10 and c = 20, find a.9. If a = 7 and b = 12, find A.10. If a = 8 and c = 12, find B. Lesson 2Solve right triangle given the length of the hypotenuse and length of one leg To solve right triangle means to find the measures of other angles andsides of a triangle. In order to avoid committing errors, maximize the use of thegiven values of the parts of the right triangle.Example:In right triangle BCA angled at C, if c = 23 and b = 17, find ∠ A, ∠B and a. (note: use scientific calculator for values of trigonometric functions) 6

Solution:Sketch the figure: B a c = 23 CA b = 17a. Solve for ∠A : From the given, b is the adjacent side, c is the hypotenuse of a right ∆BCA, so use CAH. Cos A = b c Cos A = 17 23 Cos A = 0.7391 A = 420 20’b. Solve for ∠B: To solve for ∠B, make use of the given parts, b as the opposite sideof ∠B and c the hypotenuse. We can use SOH. Sin B = b c Sin B = 17 23 sin B = 0.7391 B = 47°40’c. Solve for a: 7

Using the Pythagorean theorem:a2 + b2 = c2a2 + (17) 2 = 232a2 + 289 = 529 a2 = 529 – 289 a = 240 a = 15.49Try this outIn the given figure, solve for each right triangle ACB, given the following:1. If b = 17 cm, c = 23 cm B Find : a, ∠A, ∠B2. If c = 16 and a = 7 Find : b, ∠A, ∠B3. If b = 10 and c = 20 ac Find : a, ∠A, ∠B4. If b = 6 and c = 13 C A Find : a, ∠A, ∠B b5. If c = 13 and a = 12 Find : b, ∠A, ∠B Lesson 3Solve Right Triangle Given the Length of the Hypotenuse and the Measure of One Acute AngleIn right triangle BCA angled at C if c = 27 and ∠A = 580, find ∠B, b, a.Solution: B a c = 27 C 58° A 8

a. Solve for∠B: In rt. ∆BCA, ∠B and ∠A are complementary angles. ∠B + ∠A = 900 ∠B = 900 - 580 ∠B = 320 b. Solve for b: Since b is the adjacent side of ∠A and c is the hypotenuse of rt. ∆BCA , wecan use CAH: cos A = b 27 cos 580 = b 27 b = 27 cos 580 b = 27 (0.5299) b = 14.31 c. Solve for a: Since a is the opposite side of ∠A and c is the hypotenuse of rt. ∆BCA then,we can use SOH: sin A = a 27 sin 580 = a 27 a = 27 sin 580 a = 27(0.8480) a = 22.9 9

Try this outSketch the figure and solve each right ∆ ACB angled at C if given the following: 1. If A = 15 and c = 37 Find : B, a, b 2. If B = 64 and c = 19.2 Find : A, a, b 3. If A = 15 and c = 25 Find: B, a, b 4. If A = 45 and c = 7 2 Find : B, a, b 5. If B = 550 55’ and c = 16 Find : A, a, b Lesson 4 Solve Right Triangle Given the Length of One Leg and the Measure of One Acute AngleExample:In rt. ACB angled at C, if ∠A = 630 and a = 11 cm, find ∠B, b, c.Solution: A 63° bc CB a = 11 cma. Solve for ∠B:In rt. ∆ ACB, ∠B and ∠A are complementary angles. ∠B + ∠A = 900 ∠B = 900 – 630 ∠B = 270 10

b. Solve for b: Use TOA, b is the adjacent side of ∠A and a is the opposite side of ∠A. tan A = a b tan 630 = 11 b b tan 630 = 11 b = 11 tan 63o b = 11 1.9626 b = 5.60 cm c. Solve for c: Use SOH. c is the hypotenuse of a rt. ∆ and a is the opposite side of ∠A. Sin A = a c Sin 630 = 11 c c Sin 630 = 11 c = 11 sin 63o c = 11 0.8910 c = 12.35 cmTry this outSketch the figure and solve each right ∆ ACB angled at C given the following: 1. If A = 76 and a = 13 Find : B, b, c 2. If A = 220 22’ and b = 22 Find : B, a, c 11

3. If B = 30 and b = 11 Find : A, a, c4. If B = 18 and a = 18 Find: A, b, c5. If A = 77 and b = 42 Find : B, a, c Lesson 5Solve Right Triangle Given the Length of Both SidesExample: In rt. ∆ ACB angled at C, a = 18.5 cm and b = 14.2 cm, find c, ∠A, ∠B.Solution: A14.2cm= b c C B a = 18.5a. Solve for c: Use Pythagorean theorem: a2 + b2 = c2 (18.5)2+ (14.2)2 = c2 342.25 + 201.64 = c2 543.89 = c2 23.32 = c 12

b. Solve for ∠A: Use TOA, since a and b are opposite & adjacent side of ∠A respectively. Tan A = a b Tan A = 18.5 14.2 Tan A = 1.3028 A = 520 29’ c. Solve for ∠B: Use TOA again because b opposite side and a adjacent side of ∠B. Tan B = b a Tan B = 14.2 18.5 Tan B = 0.7676 B = 370 31’Try this outSketch the figure and solve each right triangle given the following: 1. If a = 15.8, b = 21 Find : ∠A, ∠B, c 2. If a = 7 and b = 12 Find: ∠A, ∠B, c 3. If a = 2 and b = 7 Find : ∠A, ∠B, c 4. If a = 3 and b = √3 Find : ∠A, ∠B, c 5. If a = 250 and b = 250 Find : ∠A, ∠B, c 13

Lesson 6 Problem Solving Solving problems involving right triangles require knowledge of someterms of importance in a particular field. For instance, in surveying the term lineof sight, angle of elevation, and angle of depression are frequently used. So westart with familiarizing in these terms. Line of sight – is an imaginary line that connects the eye of an observer tothe object being observed. If the observer is in a higher elevation than the objectof observation, the acute angle measured from the eye level of the observer tohis line of sight is called the angle of depression. observer angle of depression line of sight object On the other hand, if the situation is reversed, that is, the observer is atthe lower elevation than the object being observed, the acute angle made by theline of sight and the eye level of the observer is called the angle of elevation. object line of sight angle of elevation observer 14

Examples:1. Two hikers are 400 meters from the base of the radio tower. The measurement of the angle of elevation to the top of the tower is 46°. How high is the tower? B 46° A b = 400mSolution:Use the mnemonic TOA. x is the opposite side and b is adjacent side of ∠A.Tan 46° = x 400 x = 400 tan46° x = 400(1.0355) x = 414.2 m2. An observer on a lighthouse 160 ft. above sea level saw two vessels moving directly towards the lighthouse. He observed that the angle of depression are 42° and 35°. Find the distance between the two vessels, assuming that they are coming from the same side of the tower.Illustration: 42° 35° 42° 35° s x 15

Solution: equation 1 Solve for the distance of each boat. equation 2a. For the further boat Tan 35° = 160 s s tan 35° = 160 s = 160 tan 35ob. For the nearer boat tan 42° = 160 s−x (s – x)tan 42° = 160 s – x = 160 tan 42os = x + 160 tan 42oEquate: equation 1 and equation 2160 = x + 160tan 35o tan 42o 160 - 160 = xtan 35o tan 42o228.50 – 177.70 = x 50.81 ft. = x 16

Try this outSolve the following problems. Sketch the figure. 1. If a 150 ft church tower cast a shadow 210 ft. long. Find the measure of the angle of elevation of the sun. 2. From the top of the control tower 250 m tall, a rock is sighted on the ground below. If the rock Is 170 m from the base of the tower, find the angle of depression of the rock from the top of the control tower. 3. From a point on the ground 12 ft. from the base of a flagpole, the angle of elevation of the top of the pole measures 53°. How tall is the flagpole? 4. Ricky’s kite is flying above a field at the end of 65 m of string. If the angle of elevation to the kite measures 70°, how high is the kite above Ricky’s waist? 5. On a hill, inclined at an angle of 19 with the horizontal , stand a mango tree. At a point A 25 meters down the hill from the foot of the tree, the angle of elevation of the top of the mango tree is 45° . Find the height of the mango tree.Let’s summarizeConsider the right triangle ACB below. AbcCa BThe hypotenuse of the triangle is c.The side opposite angle A is a.The side adjacent to angle A is side b. 17

Use SOH-CAH-TOA, a mnemonic for remembering the equations: Sin = Opposite Hypotenuse Cos = Adjacent Hypotenuse Tan = Opposite Adjacent In solving problems, remember this terms:a. Line of sight – is an imaginary line that connects the eye of an observer to theobject being observed. If the observer is in a higher elevation than the object ofobservation, the acute angle measured from the eye level of the observer to hisline of sight is called the angle of depression. observer angle of depressionline of sight objectB. Angle of elevation - the acute angle made by the line of sight and the eyelevel of the observer is called the. object line of sight angle of elevationobserver 18

What have you learnedIn rt.∆XYZ angled at Y, if ∠X is an acute angle what is the 1. opposite side 2. adjacent side 3. hypotenuseGive the equation that could be used to find the missing parts of a rt. ∆ VWM. Ifm ∠V = 48° and m = 17, find: 4. V 5. wUsing another acute ∠M if m∠M = 23° and w = 13, give the equation to be usedto compute: 6. v 7. mSolve rt. ∆ MOP, using the given information; M P o = 35 0 47° P mFind:8. p9. m10. ∠M 19

Answer KeyHow much do you know 1. OC or d 2. OD or c 3. CD or o 4. d =25 tan 70° 5. e = 25 cos 70° 6. e = 20 cos 60° 7. f = 20 tan 60° 8. b = 79.4 9. c = 80.6 10. ∠A = 10°Try this outLesson 11. a = 14 sin 15°2. b = 13 tan 7603. c = 10 sin 49013'4. b = 21.2 tan 71013'5. c = 13 cos 1606. c = 11 cos 19007'7. b = 162 − 728. a = 202 −102 20

9. Tan A = 7 1210.Cos B = 8 12Lesson 21. a = 15.5 ∠A = 42° 20 ∠B = 47° 392. b = 14.4 ∠A = 25° 57’ ∠B = 64° 3’3. a = 17.3 ∠A = 60° ∠B = 30°4. a = 11.5 ∠A = 62° 31’ ∠B = 27° 29’5. b=5 ∠A = 67° 23’ ∠B = 22° 37’Lesson 31. ∠B = 75° a = 9.6 b = 35.72. ∠ A = 26° a = 8.4 b = 17.3 21