147 Vedanta Excel in Mathematics Teachers' Manual - 9 9) In the given figure, ABCD is a rectangle and ADE is a right angle triangle. Find the size of EC. Solution: i) In rectangle ABCD, AB = CD = 50 3 m ii) In rt. ed 'ABC; tan45q = p b = AB BC or, 1 =50 3m BC ? BC = 50 3m iii) AD = BC = 50 3m [Opposite sides of rectangle ABCD] iv) In rt.ed 'EDA. tan30q = p b = ED AD or, 1 3 = ED 50 3m ? ED = 50 m Hence, EC = ED DC = 50 m 50 3m = 136.6 m Extra Question 1. In the adjoining figure, AB = 6 cm, BC = 8 cm, ABC = 90q, BD A AC and ABD = T, find the value of cosT and tanT. Ans: 4 5 , 3 4 2. Prove that: 1 tan30q 1 tan30q = 1 sin30q 1 sin30q 3. In the given figure, ABCD is a rectangle and EAD is a right angled triangle. Find the length of BE. Ans: 55 cm 4. In the equilateral trianlge ABC, AD A BC and AC = x cm. Show that: sin 60q = 3 2 and tan 30q = 1 3 . 8 cm 6 cm A D C x cm B 50 3cm E A D 30q 5 cm B C
Vedanta Excel in Mathematics Teachers' Manual - 9 148 Allocated teaching periods 10 Competency - To collect, present and analyze the data Learning Outcomes - To construct the frequency distribution table for the collected data - To construct the histogram, line graph and pie-chart for the collected data and solve the related problems - To introduce the less and more than ogive and construct them - To find the mean, median, mode and the quartiles of the ungrouped data. - To collect the real (primary and secondary) data and analyze the data using the appropriate statistical measure. Level-wise learning objectives S.N. LEVELS OBJECTIVES 1. Knowledge (K) - To define histogram - To define pie-chart - To recall the formula of finding the mean, median and quartiles on ungrouped data 2. Understanding (U) - To answer the questions from histogram and line graph - To construct the histogram to represent the given groped data - To find the mean, median of individual series 3. Application (A) - To represent the data through pie-chart - To calculate the mean, median and quartiles of discrete series - To draw the less and more than ogives and identify the median from the graph 4. High Ability (HA) - To collect the real (primary and secondary) data and analyze the data using the appropriate statistical measure and analyze the data Unit 17 Statistics
149 Vedanta Excel in Mathematics Teachers' Manual - 9 Required Teaching Materials/ Resources Colourful chart-paper, colourful markers, chart paper with required formulae, graph-board, graph paper, highlighter etc Pre-knowledge: Frequency distribution table, mean, median etc Teaching Activities 1. Explain about statistics 2. Discuss on data and types of data with real life examples 3. With example, discuss about the frequency tables 4. Explain with examples about the histogram, its importance and procedures of constructing histogram 5. Divide the students and tell to draw histograms from the questions given in exercise and present in class 6. Ask about the pie-chart, its importance and procedures of its construction. 7. Divide the students and tell to show the data in pie-charts from the questions given in exercise and present in class 8. Discuss about the ogives with examples 9. With appropriate examples, discuss on the central tendencies and their calculations 10. With discussion, list out the following formulae (i) Mean (a) For individual data; Mean ( x ) = 6x n (b) For discrete data; Mean ( x ) = 6fx 6f (ii) Median For individual and discrete data; position of median = n + 1 2 th (iii) Mode (a) For individual and discrete data; mode = item having highest frequency (b) For continuous data; mode = Mode (M0 ) = L + f 1 – f0 2f1 –f0 – f2 u c where L = lower limit of model class, fo = frequency of the class preceding to model class, f1 = frequency of model class, f2 = frequency of the class succeeding to model class and c = the width of the class (iv) Quartiles (a) For individual and discrete data; position of Q1 = n + 1 4 th term (b) For individual and discrete data; position of Q3 = 3 n + 1 4 th term
Vedanta Excel in Mathematics Teachers' Manual - 9 150 Solution of selected problems from Vedanta Excel in Mathematics 1) The pie chart given alongside shows the votes secured by three candidates X, Y and Z in an election. If X secured 5760 votes, i) how many votes did Z secure? ii) who secured the least number of votes? How many votes did he secure? Solution: Let the total number of votes secure by three candidates be x. Then, the number of votes secured by x = 5760 or, 120 360 u x = 5760 ? x = 17280 i) No. of votes secured by Z = 140 360 u 17280 = 6720 ii) No. of votes secured by Y = 360 (120 140) 360 u 17280 = 140 360 u 17280 = 4800 Y secured the least number of votes. He secured only 4800 votes. 2) The given pie chart shows the composition of different materials in a type of cloth in percentage. i) Calculate the percentage of each material found in the cloth. ii) Calculate the weight of each material contained by a bundle of 50 kg of cloth. Solution: Materials In Percentage Weight (in kg) Polyster 144 360 u 100 % = 40 % 40 % of 50 kg = 20 kg Cotton 90 360 u 100 % = 25 % 25 % of 50 kg = 12.5 kg Nylon 54 360 u 100 % = 15 % 15 % of 50 kg = 7.5 kg Other 72 360 u 100 % = 20 % 20 % of 50 kg = 10 kg 3. Draw a ‘less than’ ogive from the data given below. Data 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 Frequency 5 8 15 10 6 3 Solution: Z 140° X 120° Y Cotton 90° Nylon 54° Others 72° Polyester 144°
151 Vedanta Excel in Mathematics Teachers' Manual - 9 Less than cumulative frequency table. Marks No. of students (f) Upper limit less than c.f. Coordinates (x, y) 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 5 8 15 10 6 3 10 20 30 40 50 60 5 13 28 38 44 47 (10, 5) (20, 13) (30, 28) (40, 38) (50, 44) (60, 47) less than ogive 10 20 30 0 10 20 30 40 Less than cumulative frequencies Upper limits 40 50 60 70 50 60 (10, 5) (20, 13) (30, 28) (40, 38) (50, 44) (60, 47) 4. The speeds of vehicles recorded in a highway during 15 minutes on a day is given in the following table. The average speed of the vehicles was 54 km per hour. Speed (km/hr) 10 30 50 70 90 No. of vehicles 7 k 10 9 13 (i) Find the value of k. (ii) Find the total number of vehicles counted during the time. Solution: Speed (km/hr) No. of vehicles ( f ) fx 10 30 50 70 90 7 k 10 9 13 70 30k 500 630 1170 Total N = 39 + k 6fx = 2370 + 30k Now, mean ( x ) = 6fx N or, 54 = 2370 + 30k 39 + k or, 2106 + 54k = 2370 + 30k or, 24k = 264 ? k = 11
Vedanta Excel in Mathematics Teachers' Manual - 9 152 i) Required value of k is 11. ii) No. of vehicles counted during the time is 39 k = 39 11 = 50 5. Given that mean in 40 and N = 51, find the missing requencies in the following data. x 10 20 30 40 50 60 f 2 3 - 21 - 5 Solution: Let the missing frequencies be a and b. x No. of vehicles ( f ) fx 10 20 30 40 50 60 2 3 a 21 b 5 20 60 30a 840 50b 300 Total N = 31 a + b 6fx = 1220 30a + 50b Now, N = 51 or, 31 a b = 51 ? a = 20 b .......... (i) Again, mean ( x ) = 6fx N or, 40 = 1220 + 30a 50b 51 or, 2040 = 1220 30a 50b or, 820 = 30a 50b or, 82 = 3a 5b ........... (ii) Substituting the value of a from equation (i) in equation (ii), we get 82 = 3 (20 b) 5b or, 22 = 2b ? b = 11 Putting the value of b in equation (i), we get a = 20 11 = 9 Hecne, the missing frequencies are 9 and 11. 6. Find the first quartile (Q1 ) and third quartile (Q3 ) from the following distribution. Age (in year) 22 27 32 37 42 No. of people 35 42 40 30 24 Solution: Cumulative frequency distribution table, Age (in years) No. of people (f) c.f. 22 27 32 37 42 35 42 40 30 24 35 77 117 147 171 Total N = 171
153 Vedanta Excel in Mathematics Teachers' Manual - 9 Now, Position of (Q1 ) = N + 1 4 th term = 171 + 1 4 th term = 43th term In c.f. column, the c.f. just greater than 43 is 77 and corresponding value is 27. ? The first quartile (Q1 ) = 27 Also, Position of median (Q2 ) = N + 1 2 th term = 171 + 1 2 th term = 86th term In c.f. column, the c.f. just greater than 86 is 117 and its corresponding value is 32. ? Median (Q2 ) = 32 Again, Position of (Q3 ) = = 3 (N + 1) th 4 term = 129th term In c.f. column, the c.f. just greater than 129 is 147 and its corresponding value is 37. ? The third quartile (Q3 ) = 37 Extra Question 1. Construct a histogram from the data given in the table below. Marks 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 No. of student 47853 2. A mobile shop sold the mobile sets of the following brands in a month. Draw a pie chart to show the data. Brand No. of mobile sets Samsung Huawei Oppo Vivo Nokia 130 120 90 40 20 3. Draw a 'less than' ogive from the data given below. Wages (in Rs) 400 - 500 500 - 600 600 - 700 700 - 800 800 - 900 No. of workers 43675 4. The mean of the data given below is 17. Determine the value of m. X 5 10 15 20 25 30 f 2 5 10 m 4 2 [Ans: 7] 5. Find the quartiles from the following distribution. Height (in cm) 90 100 110 120 130 140 No. of students 20 28 24 40 35 18 [Ans: Q1 = 100, Q2 = 120, Q3 = 130]
Vedanta Excel in Mathematics Teachers' Manual - 9 154 Unit 18 Probability Allocated teaching periods 5 Competency - To study the probability in daily life and solve the problems using mathematical structures Learning Outcomes - To introduce probability scale and solve the simple problems related to probability Level-wise learning objectives S.N. LEVELS OBJECTIVES 1. Knowledge (K) - To define probability - To tell the definition of sample space - To recall the probabilities of certain and impossible events - To tell the probability scale 2. Understanding (U) - To find the probability of an event - To solve the simple problems on empirical probability Required Teaching Materials/ Resources Coin, Dice, cards, spinner etc Pre-knowledge: Possibility of raining, having birth of son or daughter etc Teaching Activities 1. Take a coin and ask the following questions (i) How many faces are there? (ii) Which face can be surely placed when the coin is tossed? (iii) Can both faces be shown at once? (iv) What is the possibility of getting head? (v) What is the possibility of getting tail? 2. Similarly, take a die and ask the following questions (i) How many faces are there? (ii) Which face can be surely placed when the coin is tossed? (iii) Is it possible to fix all six faces at a roll?
155 Vedanta Excel in Mathematics Teachers' Manual - 9 (iv) What is the possibility of getting 1? (v) What is the possibility of getting each face separately? (vi) What is the possibility of getting an odd number? (vii) What is the possibility of getting a prime number? 3. Also, discuss on the following questions (i) What is the probability of raining today? (ii) What is the probability of having daughter from a pregnant woman? (iii) What is the probability of getting A+ grade by Ram in Mathematics? 4. Discuss on the following terminologies with proper examples (i) Probability It measures the chances of happening or not happening the event The numerical measurement of the degree of certainty of the occurrence of events Example: When a coin is tossed, it is 50/50 chance that the head or tail occurs. So, the probability of occurrence of head is 1 2 and that of tail is also 1 2. (ii) Random experiment The experiment is an action by which an observation is made. The random experiment is an experiment whose outcome cannot be predicted or determined in advance. Examples: tossing a coin, rolling a die, drawing a card from a well-shuffled pack of 52 playing cards etc. (iii) Sample space The results of random experiments are called outcomes. The set of all possible outcomes in a random experiment is the sample space. Sample space is usually denoted by S. Example: While flipping a coin, the possible outcomes are head (H) or tail (T). ?S = {H, T} (iv) Event Any non-empty subset of a sample space S is called an event. Example: When a die is rolled, the sample space (S) = {1, 2, 3, 4, 5, 6} Here, {1}, {2}, {1, 2, 3}, {2, 4, 6} etc are some events. ‘S’ itself is the sure event and empty subset I is an impossible event. (v) Exhaustive and favourable cases The number of possible outcomes of a random experiment is the exhaustive cases. Example: When a coin is tossed twice, then S = {HH, HT, TH, TT} and exhaustive cases = 4. The number of desirable (expected) outcomes in a random experiment is the favourable cases. Example: When tossing a coin, the favourable cases of getting head = 1 and that of tail = 1 (vi) Equally likely events Two or more events are said to equally likely events, if the chance of occurring any one event is equal to the chance of occurring other events. Example: While tossing a coin, the change of occurring head and tail is equal.
Vedanta Excel in Mathematics Teachers' Manual - 9 156 (vii) Mutually exclusive events Two or more events in a sample space are called mutually exclusive if the occurrence of one event excludes the occurrence of other. Two events A and B of a sample space S are mutually exclusive if AªB =I. Example: While tossing a coin, the occurrence of head excludes the occurrence of tail or HªT = I (viii) Dependent and independent events Two or more events are said to be dependent if the occurrence of one event affects the occurrence of the other events. Example: While drawing a marble in successive trials from a bag containing 3 green and 6 blue marbles without replacement, getting any one coloured ball in the first trial affects to draw another ball in the second trail. Two or more events are said to be independent if the occurrence of one event does not affect the occurrence of the other events. Example: While tossing a coin twice or more, the occurrence of any one event in the first toss does not affect the occurrence of any events in other trials. 5. Explain the probability of an event (E) in an exhaustive case/sample space (S) as P(E) = Favourable number of cases Exhaustive number of cases = n(E) n(S) and P'(E) = 1 – P(E) 6. Show the coin, die and pack of 52 playing cards, and make clear about the facts on coin, dice and playing cards. 7. Discuss about the probabilities of different events including certain and impossible events 8. With examples, make the students discover probability scale 9. Divide the students into groups and give 1/1 coin to each group. Tell them to flip the coin 20/20 times and record the outcomes. Outcomes Head(H) Tail (T) Frequency ... .... 10. Then ask to find the probability of getting head and getting tail. 11. With above experiment, discuss about the empirical probability. Solution of selected questions from Excel in Mathematics 1. In a class of 40 students, 3 boys and 5 girls wear spectacles. If a teacher called one of the students randomly in the office, find the probability that this student is wearing the spectacle. Solution: Here, total number of students, n (S) = 40 Favourable number of cases of wearing glasses, n (E) = 3 + 5 = 8 Now, P (E) = n(E) n(S) = 8 40 = 1 5 Hence, the probability that the student wearing spectacle is 1 5 .
157 Vedanta Excel in Mathematics Teachers' Manual - 9 2. There are 40 students in a class with roll numbers from 1 to 40. The roll number of Bhurashi is 18. If a teacher calls one student with roll number exactly divisible by 3 to do a problem on blackboard, what is the probability that Bhurashi will be selected? Also, find the probability that she will not be selected. Solution: Here, Set of roll numbers which are exactly divisible by 3, (S) = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39} ?n (S) = 13 Favourable number of cases, n (E) = 1 Now, P (E) = n(E) n(S) = 1 13 Thus, the probability that Bhurashi can be selected is 1 13. Hence, the probability that Bhurashi will not be selected, P’ (E) = 1 – 1 13= 12 13 . 3. A card is drawn at random from a deck of 52 cards, what is the probability that the card (i) is an ace (ii) is an ace of spade (iii) is a black ace Solution: Here, total number of cards, n (S) = 52 (i) Favourable number of cases of getting ace, n (E1 ) = 4 ?P (E1 ) = n(E1 ) n(S) = 4 52 = 1 13 Thus, the probability of getting an ace is 1 13. (ii) Favourable number of cases of getting ace of spade, n (E2 ) = 1 ?P (E2 ) = n(E) n(S) = 1 52 Thus, the probability of getting an ace of spade is 1 52 . (iii) Favourable number of cases of getting black ace, n (E3 ) =2 ?P (E3 ) = n(E3 ) n(S) = 2 52 = 1 26 Thus, the probability of getting a black ace is 1 26. 4. A card is drawn at random from a well shuffled pack of 52 cards. Find the probability that the card will be black or face cards. Solution: Here, total number of cards, n (S) = 52 Favourable number of cases of getting black cards = 26 Favourable number of cases of getting faced cards = 12 Favourable number of cases of getting black faced cards = 6 Total number of favourable cases, n(E) = 26 + 12 – 6 = 32 Now, P (E) = n(E) n(S) = 32 52 = 8 13 Thus, the probability that the card will be black or faced card is 8 13 .
Vedanta Excel in Mathematics Teachers' Manual - 9 158 5. One card is drawn at random from the number cards numbered from 10 to 21. Find the probability that the card may be prime or even numbered card. Solution: Here, Sample space (S) = {10, 11, ..., 21} ?n (S) = (21 – 10) + 1 = 12 Favourable number of cases of prime numbers = 4 Favourable number of cases of even numbers = 6 Total number of favourable cases, n (E) = 4 + 6 = 10 Now, P (E) = n(E) n(S) = 10 12 = 5 6 Hence, the probability that the card may be prime or even numbered card is 6 5 . 6. Glass tumblers are packed in cartons, each containing 12 tumblers. 200 cartons were examined for broken glasses and the results are given in the table below: No. of broken glasses 0 1 2 3 4 More than 4 Frequency 164 20 9 4 2 1 If one carton is selected at random, what is the probability that: (i) it has no broken glass? (ii) it has broken glasses less than 3? (iii) it has broken glasses more than 1? (iv) it has broken glasses more than 1 and less than 4? Solution: Here, total number of cartons, n (S) = 200 (i) The number of cartons that have no broken glass = n (E1 ) = 164 P (E1 ) = n(E1 ) n(S) = 164 200 = 41 50 = 0.82 (ii) The number of cartons that have less than 3 broken glasses n (E2 ) = 164+ 20 + 9 = 193 ? P (E2 ) = n(E2 ) n(S) = 193 200 = 0.965 (iii) The number of cartons that have more than 1 broken glasses n (E3 ) = 9+4+2+1 = 16 ? P (E3 ) = n(E3 ) n(S) = 16 200 = 0.08 (iv) The number of cartons that have more than 1 and less than 4 broken glasses, n (E4 ) = 9+4= 13 P (E4 ) = n(E4 ) n(S) = 13 200 = 0.065 7. From a pack of playing cards, two cards are taken, which are not hearts. They are not replaced, and the remaining cards are shuffled. What is the probability that the next card drawn is heart? Solution: Here, total number of cards, n (S) = 52 – 2 = 50 Favourable number of cases of getting the cards of hearts, n (E) = 4
159 Vedanta Excel in Mathematics Teachers' Manual - 9 Now, P (E) = n(E) n(S) = 4 50 = 2 25 8. Three athletes A, B and C are to run a race. B and C have equal chances of winning, but A is twice as likely to win as either. Find the probability of each athlete winning. Solution: Here, let the probability of winning the race by B be x According to question, probability of winning the race by C = x and the probability of winning the race by A = 2x Now, P (A) + P (B) + P (C) = 1 or, 2x + x + x = 1 or, 4x = 1 ?x = 1 4 Hence, the probability of winning the race by A = 2x = 2 × 1 4 = 1 2 and the probability of winning the race by B and C is 1 4 each. 9. A man has 3 pairs of black socks and 2 pair of brown socks. If he dresses hurriedly in the dark, find the probability that (i) the first sock he puts on is brown. (ii) the first sock he puts on is black. (iii) after he put on a black sock, he will then put on another black sock. (iv) that after he has first put on a brown sock, the next sock will also be brown. Solution: Here, total number of socks, n (S) = 2 (3 + 2) = 10 (i) The number of brown socks = n (E1 ) = 4 ? P (E1 ) = n(E1 ) n(S) = 4 10 = 2 5 (ii) The number of black socks = n (E2 ) = 6 ? P (E2 ) = n(E2 ) n(S) = 6 10 = 3 5 (iii) After putting on a black sock, the number of black socks for the next trial = n (E3 ) = 5 and total number of socks, n (S) = 10 – 1 = 9 ? P (E3 ) = n(E3 ) n(S) = 4 50 = 5 9 (iv) After putting on a brown sock, the number of brown socks for the next trial = n (E4 ) = 3 and total number of socks, n (S) = 10 – 1 = 9 ? P (E4 ) = n(E4 ) n(S) = 3 9 = 1 3 Extra Questions 1. Define sample space. What is the probability of a certain event? 2. A die is rolled once. What is the probability that the digit turn off is a prime number. [Ans: 1 2] 3. A bag contains a dozen of identical balls. Among them, 3 are red, 5 are greed and the rest are white. If a ball is randomly drawn, what is the probability of getting the white ball? [Ans: 1 3] 4. A card is drawn at random from a well shuffled pack of 52 cards. Find the probability that the card will be black king or red queen. [Ans: 1 13] 5. Three athletes A, B and C are to run a race. If A is twice as likely to win B and B is thrice as likely to win C. Find the probability of each athlete winning. [Ans: 3 5 , 3 10, 1 10]