MATH 4 part 2

4. -3, 7 5. − 3 , 2 2 Set B 1. 2 , 3 3 2. − 1 , 2 3 3. 3 , − 1 22 4. 2 , 3 32 5. 3 , − 3 44 Set C 1. -8, 5 2. − 1 , 4 2 3. 2, -2 4. 3, 2 5. − 1 ± i 2 Lesson 7 Deriving a Quadratic Function, Given the Table of Values If the table of values representing the quadratic function y = ax2 + bx + c isgiven, then these values should satisfy y = ax2 + bx + c. Examples: Derive a quadratic function, given the table of values 1. x -3 0 2 y -3 0 4 Since (-3, -3), (0, 0), and (2, 4) are points on the graph of y, then theysatisfy y = ax2 + bx + c. That is, 20

Using (-3, -3)y = ax2 + bx + c Eq. 1-3 = a(-3)2 + b(-3) + c ⇔ -3 = 9a – 3b + c Eq. 20 = a(0)2 + b(0) + c ⇔ 0 = c Eq. 34 = a(2)2 + b(2) + c ⇔ 4 = 4a + 2b + cSubstituting Eq. 2 to Eq. 1 and Eq. 3, -3 = 9a – 3b + 0 ⇔ 3a – b = -1 Eq. 4 4 = 4a + 2b + 0 ⇔ 2a + b = 2 Eq. 5Solving Eq. 4 and Eq. 5 by elimination, 3a – b = -1 2a + b = 2 5a = 1 a =1 5Substituting a = 1 to Eq. 5, 5 2  1  + b = 2 5 b= 8 5Hence, the quadratic function is y = 1 x2 + 8 x 552. x0 1 2 y -3 -1 -3 Since (0, -3), (1, -1) and (2, -3) are points on the graph of y, then theysatisfy y = ax2 + bx + c. That is,Using (0, -3)y = ax2 + bx + c ⇔ -3 = c Eq. 1-3 = a(0)2 + b(0) + c ⇔ -1 = a + b + c Eq. 2-1 = a(1)2 + b(1) + c ⇔ -3 = 4a + 2b + c Eq. 3-3 = a(2)2 + b(2) + c 21

Substituting Eq. 1 to Eq. 2 and Eq. 3, -1 = a + b – 3 ⇔ a + b = 2 Eq. 4 -3 = 4a + 2b – 3 ⇔ 4a + 2b = 0 Eq. 5Solving Eq. 4 and Eq. 5 by substitution, a = -b + 2 4(-b + 2) + 2b = 0 -4b + 8 + 2b = 0 -2b = -8 b=4Substituting b = 4 to Eq. 4, a+4=2 a = -2Hence, the quadratic function is y = -2x2 + 4x – 3Try this outDerive the quadratic function, given the zeros of the function.Set A1. x -2 0 2 y 0 2 122. x -2 0 1 y -12 -16 -153. x -5 -2 0 y 0 0 104. x -1 0 2 y -4 -1 -15. x -1 0 1 y 1 03 22

Set B 1. x1 0 6 y -8 36 0 2. x 1 23 y -10 -6 0 3. x -1 0 1 y 4 34 4. x01 2 y 0 3 105. x -2 -1 1 y 15 5 3 Set C 1. x -1 1 2 y 0 49 2. x0 1 2 y -3 -5 -11 3. x -1 0 2 y 4 0 10 4. x -1 0 1 y 0 -1 -4 5. x -1 0 1 y 2 -2 0 23

Lesson 8 Deriving a Quadratic Function, Given the Graph If the graph representing the quadratic function y = ax2 + bx + c is given,identify at least three points on the graph. Use these points to solve the quadraticfunction. Examples: Identify the three points on the graph and derive a quadratic function. 1. Y X0 ● ●● Three of the points on the graph are (4, -2), (5, 0) and (6, -2). They satisfyy = ax2 + bx + c. That is,Using (4, -2)y = ax2 + bx + c Eq. 1-2 = a(4)2 + b(4) + c ⇔ -2 = 16a + 4b + c Eq. 20 = a(5)2 + b(5) + c ⇔ 0 = 25a + 5b + c Eq. 3-2 = a(6)2 + b(6) + c ⇔ -2 = 36a + 6b + cEliminating c in Eq. 1 to Eq. 2 by subtraction, -2 = 16a + 4b + c 0 = 25a + 5b + c -2 = -9a – b b = -9a + 2 Eq. 4Eliminating c in Eq. 2 to Eq. 3 by subtraction, 0 = 25a + 5b + c Eq. 5 -2 = 36a + 6b + c 2 = -11a – b b = -11a – 2 24

Solving Eq. 4 and Eq. 5 simultaneously, -9a + 2 = -11a – 2 2a = - 4 a = -2 Substituting a = -2 in Eq. 4, b = -9(-2) + 2 b = 20 Substituting a = -2 and b = 20 in Eq. 1 -2 = 16(-2) + 4(20) + c c = -50The value of a = -2, b = 20 and c = -50Hence, the quadratic function is y = -2x2 + 20x – 50Try this out Identify the coordinates of the given points in the graph and derive aquadratic function.Set A ●Y 1. ● X●2. Y ● X ●● 25

3. Y ●● ● X4. Y X ●● ●.5. Y ● ● ● XSet B Y 1. ●● X● 26

2. Y X ●3. ● ● X Y ● ●●4. Y● ● X ●5. Y● ● X ● 27

Set C: Given the coordinates, solve for the quadratic functions. Y 1. ●(2, 5/4) X ●● (1,-3/4) (1/2,-1)2. (-6, 4)● Y●(0, 4) X (-1, -1)●3. X ● Y (-2, -4)● ● (0,- 4) ● (-1,-9)4. Y X ● (-2,-1)● ● (0,-1) ● (1,-4) 28

5. Y X (-1,-2) ● (-2,-5)● ● (0,-5) Let’s summarize1. The parabolic structure of the graph of a quadratic function allows it to intersect the x-axis once, twice or none at all.2. If the quadratic expression ax2 + bx + c is factorable, its zeros may be found by factoring it and apply the zero property.3. A non-real number is called imaginary number. The unit imaginary number i is a number whose square root is -1. In symbols, i = −1 → i2 =−14. In solving quadratic equations that are not factorable, the method called completing the square may be used. The principle behind the use of completing the square is to produce a perfect square trinomial so that the square root property can be applied. The following steps are suggested: a. Equate the given quadratic function to zero. b. Transpose the constant term. c. If a ≠1, divide both sides of the equation by a. d. Add to both sides of the equation the square of one-half the coefficient of x. e. Factor the resulting perfect square trinomial. f. Apply the square root property. g. Solve for x.5. The quadratic formula which can be used to solve any quadratic equation is x = −b ± b2 − 4ac 2a6. If the zeros of a quadratic function y are x1 and x2, then the quadratic function is y = (x – x1)(x – x2) 29

7. If the table of values representing the quadratic function y = ax2 + bx + c is given, then those values should satisfy y = ax2 + bx + c.8. If the graph representing the quadratic function y = ax2 + bx + c is given, identify at least three point on the graph, and use these point to form the quadratic function.What have you learned1. What are the zeros of f(x) = 2x2 – 32?2. Solve the equation 2x2 – 5x + 2 = 0.3. Use the quadratic formula in solving x2 – 4x + 3 = 0.4. Determine the roots of 8x2 – 10x + 3 = 0 using the method of completing the square.5. Find a quadratic function whose zeros are -2 and 5.6. Name the quadratic function satisfied by the table below. x -2 -1 0 2 f(x) -4 -5 -4 47. Determine the zeros of the quadratic function whose graph is givenbelow. Y X ●● 8. Determine the quadratic function whose graph is given below. Y (7, 4) ● ● (6, 1) X (5●, 0) 9. What quadratic function has 3 2 and −3 2 as zeros?10. What quadratic function has a vertex at (3,1) and passes through (4,-3)? 30

Answer KeyHow much do you know 1. x = 4 or x = -4 2. x = 2 or x = 3 3. x = −2 − 13 or x = −2 + 13 22 4. x = 1 or x = 1 2 5. f(x) = x2 + x – 6 6. f(x) = x2 – 3x + 2 7. x = -1 or x = 3 8. f(x) = x2 – x – 6 9. f(x) = x2 – 2 10. f(x) = -2x2 + 8x – 7Try this out Lesson 1 Set A 1. -2 and 2. 2. -3 and 1 3. 4 and 8 4. -2 5. No zero Set B 1. -5 and 5 2. -8 and -2 3. No zero 4. -4 and 4 5. 3 Set C 1. -5 and -1 2. 0 3. 5 4. No zero 5. -3.5 and 3.5 31

Lesson 2Set A1. x = -1 or x = 1 62. x = − 6 or x =3. x = -3 or x = 34. x = -3 or x = -15. x = -3 or x = -2Set B1. x = − 7 or x = 7 662. x = -2 or x = -13. x = -3 or x = -44. x = -6 or x = 15. x = 4 or x = -3Set C1. x = 2 or x = 32. x = -7 or x = 23. x = -3 or x = 74. x = -1 or x = 1 25. x = -2 or x = − 2 3Lesson 3Set A1. i 22. 2i3. 6i4. 3i 25. 2i 5Set B1. 9i2. 2i 73. 5 i 3 44. 3 i 3 4 32

5. 1i 31 4Set C1. 2i 132. 3i 53. 1 i 19 74. 1 i 41 45. 9i 2 11Lesson 4Set A1. x = -1 or x = 22. x = −3 + 5 or x = −3 − 53. x = 3 or x = 44. x = 1 or x = 1 25. x = -2 or x = 1 3Set B1. x = 3 or x = 72. x = −2 + 5 or x = −2 − 53. x = 4 + 12 or x = 4 − 124. x = −3 + 14 or x = −3 − 145. x = − 1 + 1 i 5 or x = − 1 − 1 i 5 24 24Set C1. x = -4 or x = 12. x = 1 + 6 or x = 1 − 63. x = − 1 + 5 or x = − 1 − 5 22 224. x = − 1 + 15 or x = − 1 − 15 22 22 33

5. x = −1 + 6 or x = −1 − 6 33Lesson 5Set A1. x = -5 or x = -12. x = −1 + 15 or x = −1 − 15 333. x = 3 + 41 or x = 3 − 41 88 884. x = -1 or x = 3 25. x = − 1 or x = 3 22Set B1. x = -4 or x = 52. x = -6 or x = 23. x = 1 or x = -1 34. x = − 5 or x = 2 25. x = − 2 or x = 1 3Set C1. x = 1 + i 11 or x = 1 − i 11 44 442. x = 1 + 7 or x = 1 − 7 223. x = − 1 + i 11 or x = − 1 − i 11 22 224. x = −1 + 3 2 or x = −1 − 3 2 225. x = −1 + 15 or x = −1 − 15 3 3Lesson 6Set A1. f(x) = x2 – 4x + 3 34

2. f(x) = x2 – x – 123. f(x) = x2 – 10x + 244. f(x) = x2 – 4x – 215. f(x) = 2x2 – x – 6Set B1. f(x) = 3x2 – 11x + 62. f(x) = 3x2 – 5x – 23. f(x) = 4x2 – 4x – 34. f(x) = 6x2 – 13x + 65. f(x) = 16x2 – 9Set C1. f(x) = x2 + 3x – 402. f(x) = 2x2 – 7x – 43. f(x) = x2 - 44. f(x) = x2 – 5x + 65. f(x) = x2 + x - 5 4Lesson 7Set A1. f(x) = x2 + 3x + 22. f(x) = x2 – 163. f(x) = x2 + 7x + 104. f(x) = -x2 + 2x – 15. f(x) = 2x2 + xSet B1. f(x) = x2 – 12x + 362. f(x) = x2 + x – 123. f(x) = x2 + 34. f(x) = 2x2 + x5. f(x) = 3x2 – x + 1Set C1. f(x) = x2 + 2x + 12. f(x) = -2x2 – 33. f(x) = 3x2 – x4. f(x) = -x2 – 2x – 1 35

5. f(x) = 3x2 – x – 2Lesson 8Set A1. f(x) = 2x22. f(x) = x2 – 13. f(x) = x2 + 2x + 34. f(x) = x2 + 2x – 55. f(x) = x2 + 4x + 6Set B1. f(x) = x2 – 2x + 12. f(x) = x2 + 4x + 33. f(x) = x2 – x – 64. f(x) = x2 – 7x + 65. f(x) = 3x2 – 6x + 7Set C1. f(x) = x2 – x – 3 42. f(x) = x2 + 6x – 403. f(x) = 5x2 + 10x – 44. f(x) = -x2 – 2x – 15. f(x) = -3x2 – 6x - 5What have you learned1. x = -4 or x = 42. x = 1 or x = 2 23. x = 1 or x = 34. x = − 1 or x = 3 425. f(x) = x2 – 3x – 106. f(x) = x2 + 2x – 47. -4 and -18. f(x) = x2 – 10x + 259. f(x) = x2 – 1810. f(x) = -4x2 + 24x – 35 36

Module 3 Statistics What this module is about In the previous modules, you learned about the different measures ofcentral tendencies. Now, you will learn about the measures of variabilityspecifically the range and the standard deviation. You will be presented problemswhere you will interpret, draw conclusions and make recommendations. What you are expected to learn This module is designed for you to: 1. give the meaning of measure of variability 2. compute the range and standard deviation of the given ungrouped and grouped data 3. interpret, draw conclusions and make recommendations from the given statistical data. How much do you knowA. Write the letter of the correct answer. 1. It is the difference between the lowest and the highest value in the distribution. a. mean b. mean deviation c. median d. range 2. Which measure of central tendency is generally used in determining the size of the most in demand shoes. a. mean b. median c. mode d. range

3. If the range of a set of scores is 14 and the lowest score is 7, what is the highest score? a. 21 b. 24 c. 14 d. 7 4. The standard deviation of the scores 5, 4, 3, 6 and 2 is __________. a. 2 b. 2.5 c. 3 d. 3.55. The most important measure of variability is _______________ . a. range b. inter quartile range d. mean deviation d. standard deviationB. Find the range and the standard deviation. To assure a uniform product, a company measures each extension wireas it comes off the product line. The lengths in centimeters of the first batch often wires were: 10, 15, 14, 11, 13, 10, 10, 11, 12 and 13.C. On a 20-item quiz, the mean score is 12 and the standard deviation is 4. Find the score that is 1. 5 points above the mean 2. 2 standard deviation above the mean 3. 1.5 standard deviation below the mean 4. Can any of the scores be 20?D. Find the mean and standard deviation of the given data.x 100 - 110 111 - 121 122 - 132 133 - 143 144 - 154 155 – 165f2 1 5 12 9 1 Describe what happen to the mean and standard deviation of a set ofnumbers under these conditions.1. 5 is added to each member in the set2. 1 is subtracted from each member in the set. 2

What you will do Lesson 1 The Range The three measures of central tendencies that you have learned in theprevious module do not give an adequate description of the data. We need toknow how the observations spread out from the average or mean. It is quitepossible to have two sets of observations with the same mean or median thatdiffers in the variability of their measurements about the mean. Consider the following measurements, in liters, for two samples ofapple juice in a tetra packed by companies A and B.Sample A Sample B0.97 1.061.00 1.010.94 0.881.03 0.911.11 1.14Both samples have the same mean, 1.00 liters. It is quite obvious that companyA packed apple juice with a more uniform content than company B. We say thatthe variability or the dispersion of the observations from the mean is less forsample A than for sample B. Therefore, in buying apple juice, we would feel moreconfident that the tetra pack we select will be closer to the advertised mean if webuy from company A. Statistics other than the mean may provide additional information fromthe same data. This statistics are the measures of dispersion. 3

Measures of dispersion or variability refer to the spread of the valuesabout the mean. These are important quantities used by statisticians inevaluation. Smaller dispersion of scores arising from the comparison oftenindicates more consistency and more reliability.The Range The range is the simplest measure of variability. It is the differencebetween the largest and smallest measurement. R=H-Lwhere R = Range, H = Highest measure, L = Lowest Measure The main disadvantage of the range is that it does not consider everymeasure in the data.Examples: 1. The IQs of 5 members of a family are 108, 112, 127, 118 and 113. Find the range. Solution: The range of the IQs is 127 - 108 = 19. 2. The range of each of the set of scores of the three students is as follows:Student A H = 98 L = 92 R = 98 - 92 = 6Student B H = 97 L = 90 R = 97 - 90 = 7Student C H = 97 L = 90 R = 97 - 90 = 7 Observe that two students are “ tie.” This indicates that the range is not areliable measure of dispersion. It is a poor measure of dispersion, particularly ifthe size of the sample or population is large. It considers only the extreme valuesand tells us nothing about the distribution of numbers in between. 4

3. Consider the following two sets of data, both with a range of 12:Set A Set B 3 3 4 7 5 7 6 7 8 8 9 8 10 8 12 9 15 15 In set A the mean and median are both 8, but the numbers vary over theentire interval from 3 to 15. In set B the mean and median are also 8, but most ofthe values are closer to the center of the data. Although the range fails tomeasure the dispersion between the upper and lower observations, it does havesome useful applications. In industry the range for measurements on itemscoming off an assembly line might be specified in advance. As long as allmeasurements fall within the specified range, the process is said to be in control.Disadvantages of the Range: 1. It makes use of very little information: that is, it ignores all but two items only. 2. It is totally dependent on the two extreme values, so it is greatly affected by any changes in these values. 3. It should be used with caution, particularly with data that contain a single extremely large value as this value would have a considerable effect on the range. 4. It cannot identify the differences between two sets of data with the same extreme values, example, the two sets of data are 2, 4, 6, 8, 10, 12, 14, 16, 18 and 2, 2, 2, 2, 2, 2, 2, 2, 18 both have the same range 16. 5

Range of a Frequency Distribution The range of a frequency distribution is simply the difference betweenthe upper class boundary of the top interval and the lower class boundary of thebottom interval.Example: Scores in Second Periodical Test of I – Faith in Mathematics IScores Frequency46 - 50 141 - 45 1036 - 40 1031 - 35 1626 - 30 921 - 25 4Upper Class Boundary = 55.5Lower Class Boundary = 20.5Try this outAnswer the following:A. Compute the range for each set of numbers. 1. ( 12, 13, 14, 15, 16, 17, 18 ) 2. ( 7, 7, 8, 12, 14, 14, 14, 14, 15, 15 ) 3. ( 12, 12, 13, 13, 13, 13, 13, 15, 19, 20, 20 ) 4. ( 12, 13, 17, 22, 22, 23, 25, 26 ) 5. ( 23, 25, 27, 27, 32, 32, 36, 38 ) 6

6. ( 12, 13, 14, 15, 16, 17, 18 )7. ( 7, 7, 8, 12, 14, 14, 14, 14, 15, 15 )8. ( 12, 12, 13, 13, 13, 13, 13, 15, 19, 20, 20 )9. ( 12, 13, 17, 22, 22, 23, 25, 26 )10. ( 23, 25, 27, 27, 32, 32, 36, 38 )B. Compute the range1. Two students have the following grades in six math tests. Pete Ricky 82 88 98 94 86 89 80 87 100 92 94 902. 100–110 111–121 122 - 132 133-143 144-154 155–165 x 2 1 5 12 9 1 fC.1. The reaction times for a random sample of 9 subjects to a stimulant were recorded as 2.5, 3.6, 3.1, 4.3, 2.9, 2.3, 2.6, 4.1 and 3.4 seconds. Calculate range. 2. If the range of the set of scores is 29 and the lowest score is 18, what is the highest score? 3. If the range of the set of scores is 14, and the highest score is 31, what is the lowest score? 7

Lesson 2The Standard Deviation The most important measure of dispersion is the standard deviation. Likethe mean deviation, standard deviation differentiates sets of scores with equalaverages. But the advantage of standard deviation over mean deviation hasseveral applications in inferential statisticsExamples: 1. Compare the standard deviation of the scores of the three students in their Mathematics quizzes.Student A 97, 92, 96, 95, 90Student B 94, 94, 92, 94, 96Students C 95, 94, 93, 96, 92Solution: Student A: Step 1. Compute the mean score. x = 97 + 92 + 96 + 95 + 90 = 94 5Step 2. Find the deviation of each score from the mean. 97 - 94 = 3 92 - 94 = -2 96 - 94 = 2 95 - 94 = 1 90 - 94 = -4 8

Step 3. Square each deviation. ( 3 )2 = 9 ( -2 )2 = 4 ( 2 )2 = 4 ( 1 )2 = 1 ( -4)2 = 16 Step 4. Find the mean of the squared deviations. 9 + 4 + 4 + 1 + 16 = 6.8 5 Step 5. Get the square root of the mean of the squared deviations. This is the standard deviation SD. SD = 6.8 = 2.6Student B: Step 1. Compute the mean score. x = 94 + 94 + 92 + 94 + 96 = 94 5 Step 2. Find the deviation of each score from the mean. 94 - 94 = 0 94 - 94 = 0 92 - 94 = -2 94 - 94 = 0 96 - 94 = 2 9

Step 3. Square each deviation. ( 0)2 = 0 ( 0 )2 = 0 ( 2 )2 = 4 ( 0 )2 = 0 ( 2 )2 = 4 Step 4. Find the mean of the squared deviations. 4 + 4 = 1.6 5 Step 5. Get the square root of the mean of the squared deviations. This is the standard deviation SD. SD = 1.6 = 1.3Student C: Step 1. Compute the mean score. x = 95 + 94 + 93 + 96 + 92 = 94 5 Step 2. Find the deviation of each score from the mean. 95 - 94 = 1 94 - 94 = 0 93 - 94 = -1 96 - 94 = 2 92 - 94 = -2 10

Step 3. Square each deviation. ( 1)2 = 1 ( 0 )2 = 0 ( -1 )2 = 1 ( 2 )2 = 4 ( -2 )2 = 4 Step 4. Find the mean of the squared deviations. 1+0+1+ 4+ 4 = 2 5 Step 5. Get the square root of the mean of the squared deviations. This is the standard deviation SD. SD = 2 = 1.4The steps may be summarized by the formula: Standard Deviation for Ungrouped Data ( )SD = ∑ x − x 2 n where SD = standard deviation x = individual score x = mean n = number of scores The result of the computation of the standard deviation of the threestudents means that student B has more consistent scores. 11

This can also be illustrated by plotting the scores on the number line. x = 94 2.62.690 91 92 93 94 95 96 97Student A’s Scores x = 94 1.3 1.390 91 92 93 94 95 96 97 Student B’s Scores x = 94 1.4 1.490 91 92 93 94 95 96 97 Student B’s Scores Graphically, a standard deviation of 2.6 means most of the scores arewithin 2.6 units from the mean. Standard deviation of 1.3 and 1.4 means thatmost of the scores are within 1.3 and 1.4 units from the mean. Student B is themost consistent among the three students.2. Compute the standard deviation of the following set of scores. 39, 10, 24, 16, 19, 26, 29, 30, 5 x = 5 + 10 + 16 + 19 + 24 + 26 + 29 + 30 + 39 = 22 912

x x- x ( x - x )25 -17 28910 -12 14416 -6 3619 -3 924 2 426 4 1629 7 4930 8 6439 17 289 ( )SD = ∑ x − x 2 9 = 900 9 = 100 SD = 10 The standard deviation is 10. This means that most of the scores arewithin 10 units from the mean or the values are scattered widely about the mean.Standard Deviation for Grouped Data For large quantities, the standard deviation is computed using frequencydistribution with columns for the midpoint value, the product of the frequency andmidpoint value for each interval, the deviation and its square and finally theproduct of the frequency and the squared deviation. To find the standard deviation of a grouped data, use the formula ( )SD = ∑ x − x 2 f n −1In calculating the standard deviation, the steps to follow are: 1. Prepare a frequency distribution with appropriate class intervals and write the corresponding frequency ( f ). 13

2. Get the midpoint ( x ) of each class interval in column 2.3. Multiply frequency ( f ) and midpoint ( x ) of each class interval to get fx.4. Add fx of each interval to get ∑ fx .5. Compute the mean using x = ∑ fx . n6. Calculate the deviation ( x - x ) by subtracting the mean from each midpoint.7. Square the deviation of each interval to get ( x - x )2.8. Multiply frequency ( f ) and ( x - x )2. Find the sum of each product to get( )∑x−x2 . f9. Calculate the standard deviation using the formula ( )SD = ∑ x − x 2 f n −13. The table is the distribution of the number of mistakes 50 students made in factoring 20 quadratic equations. Compute the standard deviation. Number of Frequency Mistakes 18 – 20 2 15 – 17 5 12 – 14 6 10 9 – 11 15 6–8 8 3–5 4 0–2 Total 50 14

Solution:To solve this problem, be sure to follow the different steps.1. Prepare a frequency distribution with appropriate class intervals and write the corresponding frequency ( f ).2. Get the midpoint ( x ) of each class interval in column 2. In the interval 18 – 20, 19 is the midpoint, write this in column 2.3. Multiply frequency ( f ) and midpoint ( x ) of each class interval to get fx. In the interval 18 – 20, the frequency is 2 and the midpoint is 19, fx = 19 ( 2 ) = 38. Write the product in column 3.4. Add fx of each interval to get ∑ fx . ∑ fx = 4385. Compute the mean using x = ∑ fx n x = ∑ fx = 438 = 8.76 n 506. Calculate the deviation ( x - x ) by subtracting the mean from each midpoint.x - x = 19 – 8.76 = 10.24 16 – 8.76 = 7.2413 - 8.76 = 4.24 10 - 8.76 = 1.24 7 - 8.76 = -1.76 4 - 8.76 = -4.76 1 - 8.76 = -7.76 15

7. Square the deviation of each interval to get ( x - x )2.( 10.24 )2 = 104.86 ( 7.24 )2 = 52.42( 4.24 )2 = 17.78 ( 1.24 )2 = 1.54( -1.76 )2 = 3.1 ( -4.76 )2 = 22.66( -7.76 )2 = 60.228. Multiply frequency ( f ) and ( x - x )2. Find the sum of each product to get ( )∑ x − x 2 f .(104.86 ) 2 = 209.72 ( 52.42 ) 5 = 262.1( 17.78 ) 6 = 106.68 ( 1.54 ) 10 = 15.4( 3.1 ) 15 = 46.5 ( 22.66 ) 8 = 181.28( 60.22 ) 4 = 240.88Number of f X Fx (x- x ) ( x - x )2 f ( x - x )2 Mistakes 104.86 209.7218 – 20 2 19 38 10.2415 – 17 5 16 80 7.24 52.42 262.112 – 14 6 13 78 4.24 17.78 106.689 – 11 10 10 100 1.24 1.54 15.46–8 15 7 105 -1.76 3.1 46.53–5 8 4 32 -4.76 22.66 181.280–2 41 5 -7.76 60.22 240.88Total 50 438 1062.56 16

9. Calculate the standard deviation using the formula SD = ∑(x − x)2 f n −1( )SD = ∑ x − x 2 f n −1= (1062.56) 50 −1= 1062.56 49 = 21.68 SD = 4.664. In this example we will use another method of computing the standard deviation of grouped data. Study the two methods and use the one you find easier to use. Determine the standard deviation of the scores in Mathematics IV Test of IV – Emerald.Scores IV – Emerald41 – 45 136 – 40 531 – 35 1026 – 30 1221 – 25 1016 – 20 511 – 15 36 – 10 3 1 1–5 17

Solution: SD = (∑ f )[∑(fX 2 )] − [∑( fX )]2 ∑ f (∑ f −1) where: SD = standard deviation f = frequency x = midpointScores F x Fx x2 fx2 1849 184941 - 45 1 43 43 1444 7220 1089 1089036 - 40 5 38 190 784 9408 529 529031 - 35 10 33 330 324 1620 169 50726 - 30 12 28 336 192 6421 - 25 10 23 230 9 916 - 20 5 18 9011 - 15 3 13 396 - 10 3 8 241-5 1 3 3∑ f = 50 ∑ fx = 1,285 ∑ fx2 = 36, 985 SD = (∑ f )[∑(fX 2 )] − [∑( fX )]2 ∑ f (∑ f −1) = (50)(36,985) − (1,285)2 50(50 −1) = 1,819,250 −1,651,225 50(49) 18

= 168,025 2,450 = 68.58SD = 8.28Try this outA. Compute the standard deviation for each set of numbers.1. ( 12, 13, 14, 15, 16, 17, 18 )2. ( 7, 7, 8, 12, 14, 14, 14, 14, 15, 15 )3. ( 12, 12, 13, 13, 13, 13, 13, 15, 19, 20, 20 )4. ( 12, 13, 17, 22, 22, 23, 25, 26 )5. ( 23, 25, 27, 27, 32, 32, 36, 38 )B. The reaction times for a random sample of 9 subjects to a stimulant were recorded as 2.5, 3.6, 3.1, 4.3, 2.9, 2.3, 2.6, 4.1 and 3.4 seconds. Calculate range and standard deviation.C. The IQ of 100 pupils at a certain high school are given in the following table. IQ Frequency 55 - 64 1 65 - 74 3 75 - 84 7 85 - 94 20 95 - 104 32 105 - 114 25 115 - 124 10 125 - 154 1 135 - 144 1Find the range and the standard deviation. 19

Lesson 3 Characteristics of Data Using the Measures of Variability The measure of central tendency gives us information about thecharacteristics of a set of data but this is not enough because it provides onlyinformation about the average or centers of a distribution. The measure ofvariability provides additional information on the variability or the spread of valuesabout the mean. Measure of variability helps visualize the shape of a data settogether with its extreme values. The standard deviation is considered the best indicator of the degree ofdispersion among the measures of variability because it represents an averagevariability of the distribution. Given the set of data, the smaller the range, thesmaller the standard deviation, the less spread is the distribution. if the population being taken into consideration has a normal distribution,then its histogram is a bell shaped. Ina normal distribution, a width of 6 standarddeviation covers all the distributions. The percentage that indicates the area thatwill be within a certain number of standard deviation units away from the mean isshown in the illustration. 34% 34% 13.5% 13.5%2.5% 2.5%x - 3s x - 2s x - s x x +s x + 2s x - 3s When two groups are comparer, the group having a smaller standarddeviation is less varied. 20

Examples:1. In a certain school, 2000 students took the entrance examination for incoming first year. Suppose the results showed that the mean score is 60.5 and the standard deviation is 9.2.a. How many students had score within the range one standard deviation from the mean?b. How many students had scores within the range one standard deviation from the mean?c. if the school will be very selective and will admit only students who got scores one standard deviation above the mean, how many students will be taken in?Answer:a. The range of score with one standard deviation from the mean 60.5 - 9.2 to 60.5 + 9.2 = 51.3 = 69.7 51 to 70b. The number of students one standard unit away from mean 34% + 34% = 68% or .68 ( 2000 ) = 1360c. The school will only admit ( 13.5% + 2.5% ) = 16% .16( 2000 ) = 3202. The scores in Mathematics Quiz of three students. Student A 97, 92, 96, 95, 90 Student B 94, 94, 92, 94, 96 Students C 95, 94, 93, 96, 92 The computed x of the three students is 94 but they differ in standarddeviation. The standard deviation of Student A is 2.6, Student B is 1.3 and 21

Student C is 1.4. Based on the result we can say that Student B has moreconsistent scores. 3. Consider the scores of two sets of students. Set A 10, 11, 9, 12, 11, 10, 12, 11, 11, 9, 10, 12, 9, 11, 10, 12 Set B 8, 10, 11, 9, 11, 12, 10, 15, 8, 9, 10, 13, 14, 10, 9, 11Set A: x = 10.63 R = 3Set B: x = 10.63 R = 7 Set A tells us that this group of students whose scores are very near eachother have almost the same abilities and therefore more teachable and wouldprogress at the same rate. Set B consists of very slow and very fast learners. They are more difficultto manage because different mental abilities.Try this outAnswer the following. 1. The scores received by Ann and Apple on a ten math quizzes are as follows: Ann: 4, 5, 3, 2, 2, 5, 5, 3, 5, 0 Apple : 5, 4, 4, 3, 3, 1, 4, 0, 5, 5 a. Compute the mean and the standard deviation. b. Which student had the better grade point average? c. Which student was the most consistent. 22

2. The mean IQ score for 2,000 students is 110 with a standard deviation of 12. The scores follow approximately a normal curve a. About how many students have an IQ between 98 and 122? b. About how many students have an IQ below 86? c. About how many students have an IQ below 74 or more than 146?3. In a reading Proficiency Test, the following means and standard deviations were obtained from a sample of 100 students each from the 3rd year and the 4th year.Mean Third Fourth Year YearStandard 80 81.4Deviation 2.5 5.5a. Which group did better in the test?b. Whose scores are more spread? Why?4. Two brands of air conditioning units were compared as to their life span in years and the following data was obtained. Mean Brand Brand A BStandard 9.2Deviation 11.5 3.3 3.1Compare the two brands. Decide which brand is better. 23

Let’s summarize The range is the simplest measure of variability. It is the difference betweenthe largest and smallest measurement. R=H-LWhere, R = Range, H = Highest measure, L = Lowest MeasureStandard Deviation for Ungrouped Data ( )SD = ∑ x − x 2 nWhere, SD = standard deviation x = individual score x = mean n = number of scoresTo find the standard deviation of a grouped data, use the formulaSD = ( )∑ x − x 2 f SD = (∑ f )[∑(fX 2 )] − [∑( fX )]2 n −1 ∑ f (∑ f −1) 34% 34% 13.5% 13.5%2.5% 2.5%x - 3s x - 2s x - s x x +s x + 2s x - 3s 24

What have you learnedAnswer the following completely:1. Find the range for each set of data. a. scores on a quiz: 10, 9, 6, 6, 7, 8, 8, 8, 8, 9 b. Number of points per game: 16, 18, 10, 20, 15, 7, 16, 24 c. Number of VCR’s sold per week: 8, 10, 12, 13, 15, 7, 6, 14, 18, 202. The minimum distance a batter has to hit the ball down the center of the field to get a home run in 8 different stadiums is 410, 420, 406, 400, 440, 421, 402 and 425 ft. Compute the standard deviation.3. Given the set of scores of 2 students received in a series of test Student A: 60, 55, 40, 48, 52, 36, 52, 50 Student B: 62, 48, 50, 46, 38, 48, 43, 39 a. What is the mean score of the student? b. Compute the range. c. Interpret.4. In the entrance examination to an elite school, 9854 took the test. The mean of the applicants was 76.5 with a standard deviation of 8.5. Assume that the results approximated a normal curve. a. How many of the students scored within one standard deviation from the mean? b. If only those applicants who got scores of at least one standard deviation above the mean were accepted and the score of Albert was 87, was he accepted? Why? c. About how many students were accepted in the school? d. If the applicants who got scores 2 standard deviations above the mean were the students given scholarships, about how many students were granted scholarships? 25

Answer KeyHow much do you knowA. 1. d 2. c 3. a 4. d 5. dB. Range = 5 Standard deviation = 1.7C. 1. 17 2. 20 3. 18 4. yesD. 1. The mean will increase by 5 but the standard deviation will remain the same.2. The mean will decrease by 1 but the standard deviation will remain the same.Try this outLesson 1A 1. 6 2. 8 3. 8 4. 14 5. 156. 6 7. 8 8. 8 9. 14 10. 15B. 1. Pete = 18 Ricky = 7 2. Range = 66C. 1. 1.8 2. highest score = 47 3. lowest score = 17Lesson 2.A. 1. Mean = 15 SD = 2 2. Mean = 12.3 SD = 3.113. Mean = 14.82 SD = 2.02 4. Mean = 20 SD = 5.15. Mean = 30 SD = 5B. Mean = 2.9 SD = 0.69C. Range = 90 Mean = 100 fx = 10000 fx2 = 10,180,75 SD = .14 26

Lesson 31. Ann Applea. Mean 3.4 3.4 SD 1.22 1.22b. they have the same point averagec. both of them were consistent with the results of their test, which means that they are equally good.2. a. 1360 b. 50 students c. 20 students3. a. The 4th year group did better in the test.b. The 4th year group because the standard deviation is higher than the 3rd year group.What have you learned1. a. 4 b. 17 c. 142. Mean = 415.5 SD = 11.743. a. Sudent A = 49.13 Student B = 46.75b. range of Student A = 24 range of Student B = 24c. Both Students A and B were consistent with their scores in the test.4. a. 6701 studentsb. Pete was acceptedc. 246 were granted scholarshipsd. 1577 students were accepted 27

Module 3 Circular Functions What this module is about This module is about the properties of the graphs of a circular functions.You will learn how the graphs of circular function look like and how they behavein the coordinate plane. What you are expected to learnThis module is designed for you to:1. describe the properties of the graphs of the functions: • sine • cosine • tangent2. graph the sine, cosine and tangent functions/How much do you know1. What is The period of the sine function y = sin x?a. 2 π b. π c. π d. 3π 2 22. What is the amplitude of a cosine function y = cos x? d. 1a. -2 b. -1 c. 23. What is the value of y = 4 sin 1 x, if x = 5π ? 23

4. What is the value of y = 2 sin x, if x = π ? 4Given the following functions, identify the amplitude of :5. y = 2 cos x6. y = 3 sin x 47. y = -2 sin 3 x 2Given the following functions, determine the period of:8. y = 2 cos 1 x 29. y = 4 sin 2 x 310. y = sin 1 x 511. Which of the following are zeros of y = tan Ө for the interval 0 ≤ Ө ≤ 2π?a. 0, π and 2π c. π and 3π 44b. π and 3π d. π and 5π 22 6612. In which of the following intervals is the cosine function decreasing over the interval [0, 2π]?a. [0, π] c. [π, 2π]b. 0, π  and 3π , 2π d.  π , 3π  2   2  2 2 

What you will do Lesson 1 Graphs of Sine, Cosine and Tangent Circular functions can also be graphed just like the other functions youhave learned before. The difference is that the graphs of circular functions areperiodic. A function is said to be periodic if the dependent variable y takes on thesame values repeatedly as the independent variable x changes. Observe the changes in the values of y = sin θ and y = cos θ for arclengths from -2π to 2π. θ - 3π to -π to - - π to 0 to - 0 to π to π to 3π tosin θ 2 3π 2 π π 2 3π 2 -2π 2 -π 2 2 π 2 2π 1 to 0 0 to 1 -1 to 0 0 to -1 0 to 1 1 to 0 0 to -1 -1 to 0cos θ 0 to 1 -1 to 0 0 to -1 1 to 0 1 to 0 0 to -1 -1 to 0 0 to 1 Using the arc length, θ, as the independent variable and y = sin θ and y =cos θ as the dependent variables, the graphs of the sine and cosine functionscan be drawn. Below is the graph of y = sin θ for -2π ≤ θ ≤ 2π. This was done by plottingthe ordinates on the y-axis and the arc lengths on the x-axis. Observe the properties of this graph. y = sin θ Y maximum point 2 1 amplitude -2π −3π 0 ππ 3π 2π 2 2 -π − π 2 2 -1 period minimum point -2

You can see that the graph is a curve. Call this the sine curve. Observethat the graph contains a cycle. One complete cycle is the interval from -2π to 0and another cycle is the interval from 0 to 2π. This is called the period of thecurve. Hence, the period of y = sin θ is 2π.The amplitude of the graph of y = sin θ is 1. The amplitude is obtained bygetting the average of the maximum value and the minimum value of thefunction. The maximum point is  π ,1 and the minimum point is  3π ,−1 for the 2  2 interval [0, 2π]. The graph crosses the x-axis at (0, 0), (π, 0), and (2π, 0) for theinterval [0, 2π]. Observe also that the sine graph is increasing from 0 to π and 2from 3π to 2π, and decreasing from π to 3π for the interval [0, 2π]. 2 22The Graph of Cosine Function The graph of y = cos θ can be constructed in the same manner as thegraph of y = sin θ, that is, by plotting the abscissa along the y-axis and the arclengths along the x-axis. Observe the properties of the graph of y = cos θ for theinterval -2π ≤ θ ≤ 2π shown below. y = cos θ Y 2 1-2π 3π -π π 0 π π 3π 2π −2 2 −2 2 -1 -2 You will observe that just like the graph of y = sin θ, it is also a curve. Italso has a period of 2π and amplitude 1. For the interval [0, 2π], the minimumpoint is (π, -1), maximum points are (0, 1) and (2π, 1) and the graph crosses the

x-axis at  π , 0 and  3π , 0 . The graph is decreasing from 0 to π and 2  2 increasing from π to 2π over the interval [0, 2π].The Graph of the Tangent Function The graph of y = tan θ can be drawn in the same manner that the graphsof the sine and cosine functions. The value of the tangent of an angle is plottedalong the y-axis and the arc lengths on the x-axis. Observe that the tangent ofthe odd multiples of π (90o) are not defined so that the graph is discontinuous at 2those values. These are denoted by the broken lines (called asymptotes) thatseparate one complete cycle from the others. Thus, the domain of these functionexclude all odd multiples of π while the range is the set of real numbers. 2 Y - 5π - 2π − 3π -π − π 0 π π 3π 2π 5π 2 2 2 2 22 Notice that the period of the graph of the tangent function is π. This is shownby a complete curve in the interval − π , π  . See that the other curves are 2 2 repetitions of the curve for the given interval. The graph of the tangent function isalso said to be an odd function and that the graph is symmetrical with respect tothe origin. You will also see from the graph that it is an increasing function for thedifferent sets of intervals.Try this outA. Refer to the graph of the y = sin θ to answer the following. 1. What is the domain of the sine function?

2. What is its range? 3. Give the intercepts of y = sin θ for the interval [-2π, 0] 4. Determine the interval where the graph of y = sin θ is (a) increasing, (b) decreasing for the interval [-2π, 0].B. Refer to the graph of y = cos θ to answer the following. 1. What is the domain of the cosine function? 2. What is its range? 3. Give the intercepts of y = cos θ for the interval [-2π, 0] 4. Determine the interval where the graph of y = cos θ is (a) increasing, (b) decreasing for the interval [-2π, 0].C. Refer to the graph of y = tan θ 1. What is the domain of the graph of y = tan θ? 2. What is its range? 3. At what values of θ in the graph is tangent not defined? 4. Give the vertical asymptotes of the graph? 5. What are the zeros of y = tan θ? Lesson 2 Properties of Sine and Cosine functions The two properties of Sine and Cosine functions are amplitude and aperiod of a function. This can be determined from a given equations. Thefunction in the form of y = a sin bx and y a cos bx, the amplitude is /a/ and theperiod is 2π . bExamples:Determine the amplitude and the period of the given function: 1. y = 3 sin 2x Solution: