GOYAL BROTHERS PRAKASHAN

Transcription

1 Assignments in Mathematics Cass IX (Term 2) 14. STATISTICS IMPORTANT TERMS, DEFINITIONS AND RESULTS The facts or figures, which are numerica or otherwise, coected with a definite purpose are caed data. Data is the pura form of the Latin word datum. Statistics is the area of study deaing with the presentation, anaysis and interpretation of data. The data which are coected by an investigator personay to fufi his objective is caed primary data. The data which are coected by someone ese but used by the investigator for his own purpose is caed secondary data. Putting the data in the form of tabes, in condensed form, is known as the presentation of data. The number of times an observation occurs is caed its frequency. The tabuar arrangement of data, showing the frequency of each observation, is caed a frequency distribution. There are two types of frequency distributions. (a) Excusive form (or continuous form) : A frequency distribution in which upper imit of each cass is excuded and ower imit is incuded, is caed an excusive form. We may consider the casses : 0 10, 10 20, etc. In cass 0 10 we incude 0 and excude 10. In cass 10 20, we incude 10 and excude 20. (b) Incusive form (or discontinuous form) : A frequency distribution in which each upper imit as we as ower imit is incuded, is caed an incusive form. Thus, we have casses of the form 0 10, 11 20, 21 30, etc. In 0 10, both 0 and 10 are incuded. Each group into which the raw data is condensed, is caed a cass interva. Each cass is bounded by two figures, which are caed cass imits. The figure on the eft side of a cass is caed its ower imit and that on the right is caed its upper imit. In the case of excusive casses the upper and ower imits are respectivey known as its true upper imits and true ower imits. But in the case of incusive casses, the true ower and upper imits are obtained by subtracting 0.5 from the ower imit and adding 0.5 to the upper imit. Thus in the case of casses 5 10, 10 15, 15 20,..., True ower imit of the cass = 10 True upper imit of the cass = 15 In the case of cass 1 10, 11 20, 21 30,... True ower imit of the cass = = 10.5 True upper imit of the cass = = 20.5 True upper imits and true ower imits are aso known as boundaries of the casses. The difference between the true upper imit and the true ower imit of a cass is caed the size of the cass. The vaue which ies midway between ower and upper imits of a cass is known as its mid-vaue or cass mark. i.e., Cass mark ower imit of cass + upper imit of cass = 2 The difference of the highest and the owest vaues in the data is caed the range of the data. Bar Diagram or Bar Graph A set of bars (thick ines or narrow rectanges) representing variabe and frequency constitute a bar graph. The foowing points shoud be kept in mind whie drawing bar graphs. (i) The bars shoud be at equa distances from each other. (ii) A the bars in a diagram shoud be of the same width. (iii) Both axes shoud be adequatey abeed. Histogram We use a histogram to represent grouped data by representing cass boundaries aong the horizonta axis and the corresponding frequencies aong the vertica axis. Thus, rectanges are constructed with base as the cass size and their heights representing the frequencies. Frequency Poygon A frequency poygon is obtained by joining the mid-points of the respective tops of the rectanges in a histogram. To compete the poygon join the mid-points of two more casses (caed imaginary casses) one at each end. 1

2 If x 1, x 2, x 3,..., x n are n observations, then Mean ( x ) = x 1 + x 2 + x x n or Σx n n If x 1, x 2, x 3,..., x n are n observations and f 1, f 2, f 3,... f n their corresponding frequencies, then Mean ( x fx 1 1+ fx 2 2+ fx fx n n Σ( fx) ) = = f1 + f2 + f fn Σf x = A + Σd, where A is the assumed mean, n d = x A and n is the number of observations. Median is the midde item in the arrayed data. n + 1 Median = th item, if n is odd. 2 Median n n = Mean of th item th item if n is even. 2 2, Mode is the most frequenty occurring observation. Empirica Formua for mode is Mode = 3 Median 2 Mean SUMMATIVE ASSESSMENT MULTIPLE CHOICE QUESTIONS 1. The mean age of 5 students of a cass is 10 years. If one student is excuded from the cass, the new mean age becomes 9 years. The age of student who is excuded from the cass is : (a) 11 years (b) 12 years (c) 13 years (d) 14 years 2. The ratio between upper imit and ower imit of a cass is 3 : 2. If the cass mark is 15, then ower imit of the cass-interva is : (a) 9 (b) 12 (c) 15 (d) For x = 1, the mode of data x + 1, 0, x + 2, x + 3 and x + 4 is : (a) 0 (b) 1 (c) 2 (d) 3 4. A cricket payer has a mean score of 50 runs in the four innings. How many runs to be scored by the payer in the fifth innings to raise the mean score by five? (a) 25 (b) 50 (c) 75 (d) The mean of the prime factors of 24 is : (a) (b) (c) (d) The mode of the foowing data 44, 40, 40, 41, 45, 43, 45, 40, 41 is : (a) 40 (b) 41 (c) 43 (d) If ower imit and upper imit of a cass are in the ratio 3 : 5 and its cass size is 6, then the cassmark of this cass is : (a) 3 (b) 5 (c) 6 (d) x + 8, x and x 3 are the frequencies of three numbers 5, 8 and 10 respectivey. If their arithmetic mean is 6, then the vaue of x is : (a) 4 (b) 5 (c) 5.5 (d) 6 A. Important Questions [1 Mark] 9. Range of the data 25.7, 16.3, 2.8, 21.7, 24.3, 22.7, 24.9 is : (a) 22 (b) 22.9 (c) 21.7 (d) The median of first six whoe numbers is : (a) 2 (b) 2.5 (c) 3 (d) Cass-marks of (30-40) and (60-70) are : (a) 35 and 60 (b) 30 and 65 (c) 35 and 65 (d) 40 and Two students scored foowing marks (out of 10 marks) in five different subjects : Student I : 2, 3, 4, 5, 6 Student II : 3, 3, 3, 4, 8 Which student scored better? (a) I (b) II (c) both (d) information is not sufficient 13. The mean of perimeters of two squares having sides x units and y units is : (a) (x + y) units (b) ( x+ y) units 2 (c) 2 (x + y) units (d) ( x+ y) units The mean weight of 10 students in a cass is 50 kg. If the sum of the weights of 9 students is 453 kg, the weight of 10th student is : (a) 53 kg (b) 50 kg (c) 47 kg (d) 43 kg 15. For x = 1, the median of data x, x + 1, x 1, x + 3 and x + 2 is : (a) 1 (b) 2 (c) 3 (d) The mean of three numbers is 4. If two of the numbers are 1 and 2, then the third number is : (a) 4 (b) 8 (c) 9 (d) 11 2

3 17. The cass marks of a frequency distribution are given as foows : 15, 20, 25,... The cass corresponding to the cass mark 20 is: (a) (b) (c) (d) In the cass intervas 10-20, 20-30, the number 20 is incuded in : (a) (b) (c) both the intervas (d) none of these 19. If each observation of a data is increased by 5, then their mean : (a) remains the same (b) becomes 5 times the origina mean (c) is decreased by 5 (d) is increased by The cass mark of the cass is : (a) 90 (b) 105 (c) 115 (d) The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 is : (a) 10 (b) 15 (c) 18 (d) In a frequency distribution, the mid vaue of a cass is 10 and the width of the cass is 6. The ower imit of the cass is : (a) 6 (b) 7 (c) 8 (d) The marks obtained by 17 students in a mathematics test (out of 100) are given beow : 91, 82, 100, 100, 96, 65, 82, 76, 79, 90, 46, 64, 72, 68, 66, 48, 49. The range of the data is : (a) 46 (b) 54 (c) 90 (d) The cass-mark of the cass is : (a) 130 (b) 135 (c) 140 (d) The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is : (a) 45 (b) 49.5 (c) 54 (d) For drawing a frequency poygon of a continous frequency distribution, we pot the points whose ordinates are the frequencies of the respective casses and abscissae are respectivey. (a) upper imits of the casses (b) ower imits of the casses (c) cass marks of the casses (d) upper imits of preceeding casses 27. Median of the numbers 4, 4, 5, 7, 6, 7, 7, 12, 3 is: (a) 4 (b) 5 (c) 6 (d) Mode of the data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15 is : (a) 14 (b) 15 (c) 16 (d) The mean temperature of 30 days was taken as 15 C, but ater it was found that temperature of one day was noted as 16 C instead of 13 C. The correct mean is : (a) 14.5 C (b) 14.4 C (c) 14.9 C (d) 13 C 30. The median of first 10 even numbers is : (a) 10 (b) 10.5 (c) 11 (d) Numbers 7, 8, 3x + 2, 5x 4, 40, 41 are given in ascending order. If the median of the data is 15, then the vaue of x is : (a) 3 (b) 3.5 (c) 4 (d) The mean marks obtained by 15 students in a cass is 73. If the marks of each student are doubed, then the new mean marks is : (a) 30 (b) 75 (c) 146 (d) The median of first 49 whoe numbers is : (a) 24 (b) 25 (c) 26 (d) The median of first 49 natura numbers is : (a) 24 (b) 25 (c) 27 (d) If the mean of three consecutive numbers is 3, then the median of these numbers is : (a) 3 (b) 4 (c) 5 (d) The mean age of 5 students of a cass is 10 years. If one student is added to the cass, the new mean age becomes 11 years. The age of the student who is added to the cass is : (a) 12 years (b) 15 years (c) 16 years (d) 14 years 37. If the cass-mark and cass-size of a cass are 6 and 2 respectivey, then the upper imit of the cass is : (a) 2 (b) 5 (c) 6 (d) The mean of first five prime numbers is : (a) 5.6 (b) 6.5 (c) 5.2 (d) The average area of two rectanges having dimensions as (3 units 4 units) and (4 units 5 units) is : (a) 12 sq units (b) 14 sq units (c) 16 sq units (d) 20 sq units 40. The mean of cass-marks of (20 30) and (30 40) is : (a) 20 (b) 30 (c) 40 (d) The mean of 15 numbers is 18 and the mean of 10 numbers is 13. The mean of a the 25 numbers is : (a) 16 (b) 16.5 (c) 12 (d) none of these 42. The sum of first n natura numbers is given by nn ( +1). The mean of first 100 natura numbers 2 is : (a) 101 (b) 50.5 (c) 50 (d) 100 3

4 B. Questions From CBSE Examination Papers 1. If the mean of the data x, x + 1, x + 3, x + 6 is 15, then the vaue of x is : 2 (a) 3 (b) 4 (c) 5 (d) 6 2. The cass mark of the cass is : (a) 100 (b) 125 (c) 130 (d) Cass mark of a particuar cass is 10.5 and cass size is 7, then the cass interva is : (a) (b) (c) (d) The mean of a, b is 8.5 and mean of a, b, c is 7. The vaue of c is : (a) 2 (b) 4 (c) 5 (d) 9 5. Median of the data 5, 9, 8, 6, 3, 5, 7, 12, 15 is : (a) 3 (b) 6 (c) 5 (d) 7 6. If the cass marks in a frequency distribution are SHORT ANSWER TYPE QUESTIONS 1. Is it correct to say that in a histogram, the area of each rectange is proportiona to the cass size of the corresponding cass interva? Justify your answer. 2. A student was asked to find the median of the data : 3, 14, 18, 20, 5. He writes the answer as 18. Is this the correct median? If not, find the correct median of the data. 3. The mean of a data is x. What wi be the new mean if each vaue of the data is increased by a? 4. In a histogram, the areas of the rectanges are proportiona to the frequencies. Can we say that the engths of the rectanges are aso proportiona to the frequencies? 5. Find the median of the foowing numbers. 2, 3, 7, 11, 10, 9,9, 5, The ages (in years) of 8 chidren are given beow: 15, 15, 16, 15, 15, 16, 14, 14. Find the moda age. A. Important Questions 19.5, 26.5, 33.5, 40.5, then the cass corresponding to the cass mark 33.5 is : (a) (b) (c) (d) Cass mark of cass interva is : (a) 60 (b) 70 (c) 65 (d) The mean of 10 numbers is 55. If one number is excuded, their mean becomes 50, the excuded number is : (a) 60 (b) 70 (c) 80 (d) Cass mark of a particuar cass is 6.5 and cass size is 3, then cass interva is : (a) 5 8 (b) (c) (d) The range of the data 25.7, 16.3, 2.8, 21.7, 24.3, 22.7, 24.9 is : (a) 22 (b) 22.9 (c) 21.7 (d) 20.5 [2 Marks] 7. Observe the data : 16, 9, 3, 9, 9, 3, 2. Since, 16 is the highest vaue in the observation, so the mode of the data is 16. Is it correct? 8. Find the mean of first 10 whoe numbers. 9. The mean of 10, 12, 18, 13, x and 17 is 15. Find the vaue of x. 10. What is the median of first 19 natura numbers? 11. The mean of 5 numbers is 27. If one of the numbers is excuded, the mean of remaining numbers is 25. Find the excuded number. 12. Find the median of first ten prime numbers. 13. In a test in mathematics, 12 students scored the foowing marks (out of 100) : 46, 52, 40, 98, 44, 48, 11, 41, 53, 54, 62, 96. Which average wi be a good representative of the above data? Justify your answer. 14. The cass marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44, 1.54 and What wi be the ast cass interva of the distribution? 4

5 1. The weights, in grams of 30 oranges picked at random from a basket are given beow: 70, 110, 119, 81, 158, 144, 125, 120, 155, 90, 102, 62, 98, 156, 145, 150, , 82, 125, 61, 85, 117, 121, 121, 130, 65, 69, 92, 100. Construct the continunous grouped frequency distribution tabe, width of each cass being 20 grams and first cass being (80 not incuded). 2. Find the mean of first 10 natura numbers. 3. Average monthy consumption of petro for first 3 months is 86 itres and for next 9 months is 152 itres. Find the average consumption for the whoe year. 4. Find the mean of first 10 prime numbers. 5. The foowing observations have been arranged in ascending order. If median of the data is 23.5, find the vaue of x. 12, 16, 17, 19, x, x + 3, 27, 37, 38, The mean marks scored by 100 students was 40. Later on, it was discovered that a score of 53 was misread as 83. Find the correct mean marks. 7. Find the mean of 3, 4, 6, 7, 8, 14. If 5 is added to each oservation, what wi be the new mean? 8. Find the mean of the first eight prime numbers. 9. Find the median of data : 26, 56, 32, 33, 60, 17, 34, 29 and 45. If 26 is repaced by 62, find the new median. 10. The mean of first 8 observations is 18 and ast 8 SHORT ANSWER TYPE QUESTIONS B. Questions From CBSE Examination A. Important Questions observations is 20. If the mean of a 15 observations is 19, find the 8th observation. 11. If the mean of the data 6, 8, 10, 3, 7 and m is 7, then find the vaue of m. 12. Give one exampe of a situation in which (i) the mean is an appropriate measure of centra tendency. (ii) the mean is not an appropriate measure of centra tendency but the median is an appropriate measure of centra tendency. 13. The meadian of foowing observations arranged in ascending order is 25. Find x. 11, 13, 15, 19, x + 2, x + 4, 30, 35, 39, Find median of foowing data : 17, 23, 57, 46, 33, 29, 28, 30, 34. If observation 23 is removed from data then find new median. 15. The mean of ten observations is 20. If nine of observations are 16, 20, 18, 17, 22, 16, 15, 20, 17. Find the tenth observation. 16. Find the mean of first five prime numbers 17. Find the median of the foowing data : 15, 35, 18, 27, 19, 23, 29, 20, Find the median of the foowing data : 23, 31, 47, 24, 46, 27, 37, If the mean of 10, 12, 18, 11, p and 19 is 15, find the vaue of p. 20. Find the median of first ten prime numbers. 1. The median of the observations 26, 29, 42, 53, x, x + 2, 70, 75, 82, 93 arranged in ascending order is 65. Find the vaue of x : 2. The foowing data represents height (in cm) of 20 students of a cass : 132, 130, 134, 135, 130, 131, 134, 135, 133, 133, 130, 131, 133, 136, 135, 134, 130, 133, 129, 136. Make a frequency tabe for the above information : [3 Marks] 3. The mean of 70 observations was found to be 150. Whie checking, it was detected that one vaue 210 was wrongy copied as 140, whie cacuating the mean. Find the correct mean. 4. The cass marks of a distribution are 6, 10, 14, 18, 22, 26, 30. Find the cass size and the cass intervas. 5. If the median of the data 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35 and 41 arranged in ascending order is 24, find x. 5

6 6. The points scored by a basket ba team in a series of matches are as foows : 17, 2, 7, 27, 25, 5, 14, 18, 10, 24, 48, 10, 8, 7, 10, 28. Find the median and the mode for the data. 7. Find the mean of the foowing data : x f The mean of 15 numbers is 9. If each observation is mutiped by 4, what wi be the new mean? Give reason for your answer. 9. The mean of 13 observations is 14. If the mean of the first 7 observations is 12, and that of the ast 7 observations is 16, find the seventh observation. 10. For what vaue of x the mode of the foowing data is 15? 15, 16, 17, 13, 17, 16, 15, x + 10, 14, 17, 16, A cass consists of 50 students. Out of which 30 are girs. The mean of marks scored by girs in a test is 73 and that of boys is 71. Find the mean score of the whoe cass. 12. Observe the foowing tabe : x f 3 p If the mean of the data is 25.75, find the vaue of p. 13. If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, find the mean of first three observations. B. Questions From CBSE Examination Papers 1. The foowing observations have been arranged in ascending order where median of the data is 63 : 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95. Find the mean of the data. 2. The foowing tabe gives the ife time 400 neon amps : Life time (in hrs) Number of amps Represent the above information with the hep of a histogram. 3. Find the difference between median and mode of the foowing data : 4, 6, 5, 9, 3, 2, 7, 7, 6, 5, 4, 9, 10, 10, 3, 4, 7, 6, 9, Find the arithmetic mean of the foowing tabe. Marks obtained Number of students If the median of the data in ascending order : 30, 32, 49, 50, 2x, 2x + 2, 73, 78, 85, 96 is 63, find x. Aso, find the mean. 6. Cacuate the arithmetic mean of the foowing data : Marks obtained Number of students The mean of 14 vaues of a data is 23. If one more vaue is introduced in the data, the mean of 15 vaues becomes 25. Find the fifteenth vaue. 8. The mean of 50 observataions of a data was 70. At ater stage, it was noted that a vaue of 85 was wrongy read as 60. Find the vaue of the correct mean. 9. Find the average daiy income of the empoyees of the company. Daiy income (Rs.) Number of empoyess

7 10. Mean of 50 observations was found to be But ater on it was discovered that 96 was misread as 69 at one pace. Find the correct mean. If to each observation a constant vaue k is added, how is the mean affected? 11. Find the mean saary of 50 workers of a factory from the foowing data. Saary (in Rs.) Number of workers The numbers 42, 43, 44, 44, (2x + 3), 45, 45, 46, 47, 49 are written in a ascending order and the meadian is 45. Find x. Hence, find the mode of the above data. 13. Find the mean saary of 60 workers of a factory from the foowing tabe : Saary (in Rs.) No. of workers Tota The pass percentage of students of Cass IX of a schoo for 5 consecutive years is shown beow : Years Pass percentage Draw a bar graph (use graph sheet) to represent the above data. 15. The mean weight per student in a group of 7 students is 55 kg. The individua weights of 6 of them in kg are 52, 54, 55, 53, 56, 54. Find the weight of the seventh student. 16. If the mean of the foowing data is 20.2, find the vaue of p. p f 6 8 p The mean of 25 observations is 36. Out of these observations, the mean of first 13 observations is 32 and that of the ast 13 observations is 40. Find the 13 th observation. 18. Find the mean of the foowing distribution : x f The marks obtained by 40 students of cass IX in an examination are given beow : 18, 8, 12, 6, 8, 16, 12, 5, 23, 2, 16, 23, 2, 10, 20, 12, 9, 7, 6, 5, 3, 5, 13, 21, 13, 15, 20, 24, 1, 7, 21, 16, 13, 18, 23, 7, 3, 18, 17, 16. Present the data in the form of a frequency distribution tabe of the same cass size, one such cass being (where 20 is not incuded.) 20. Find the mean for the weeky pocket money (in rupees) using the foowing data. Pocket money (in Rs.) Frequency Find missing frequency p for the foowing distribution whose mean is 15. x : f : 6 p For what vaue of a the median of the foowing observation arranged in ascending order is 32? 12, 14, 15, 27, a + 2, a + 4, 35, 36, 40, 41 7

8 23. In a mathematics test given to 15 students, the foowing marks out of 100 are recorded : 41, 39, 48, 52, 46, 52, 54, 40, 96, 52, 98, 40, 42, 52, 60. Find the median and mode of this data. 24. The mean of 25 observations is 36. If the mean of the first 13 observation is 32 and that of ast 13 observations is 39, find the 13th observation. 25. In a city, the weeky observation made in a study on the cost of iving index are given in the foowing tabe : Cost of the iving index No. of weeks Tota 52 Draw the frequency poygon for the data given above. 26. The mean of the foowing distribution is 50. x f 17 5p p Find the vaue of p. 27. The foowing are the runs made by 22 payers in a one day cricket series payed between India and Pakistan. 79, 28, 45, 99, 3, 46, 8, 0, 3, 7, 75, 24, 73, 122, 46, 27, 16, 7, 100, 3, 67, 53. Construct a frequency distribution tabe for the above data with equa cass intervas, one of these being 0 20 (20 not incuded). 28. The mean of foowing distribution is 50. Find the vaue of b and hence find the frequency of 30 and 70. x f 17 (5b + 3) 32 (7b 11) The distance (in km) covered by 20 cars in 2 hours are given beow : 125, 120, 130, 134, 100, 90, 96, 100, 110, 120, 112, 116, 104, 96, 100, 98, 120, 122, 124, 92. Represent the data as a continuous frequency distribution tabe with equa cass size, starting from The foowing is the distribution of weights (in kg) of 50 persons : Weights (in kg) Number of persons Draw a histogram for the above data. 31. The foowing tabe gives the distribution of students of two sections according to the marks obtained by them. Section A Section B Marks Frequency Marks Frequency Represent the marks of the students of both the sections on the same graph by two frequency poygons. 32. The heights of 50 students, measured to the nearest centimetres, have been found to be as foows: 135, 162, 173, 151, 176, 165, 162, 145, 171, 172, 157, 161, 172, 158, 163, 175, 148, 173, 163, 159, 160, 162, 172, 175, 176, 168, 167, 170, 172, 173, 165, 151, 149, 169, 173, 138, 156, 148, 159, 166, 176, 151, 139, 146, 164, 173, 141, 142, 150, 159. Represent the data given above by a grouped frequency distribution tabe, taking cass intervas as , etc. 33. The median of the foowing observations arranged in ascending order is 27. Find x. 13, 15, 17, 21, x + 2, x + 4, 32, 41, and Obtain the mean of the foowing data : Variabe (x i ) Frequency (f i ) Construct a histogram for the foowing data : Cass interva Frequency

9 36. The mean mark of 8 students is 65. If marks of 7 students are 60, 70, 55, 50, 60, 65 and 85, then find the marks of 8th student. 37. Find the mean of the foowing frequency distribution showing the marks obtained in a cass test by 20 students. Marks No. Students The foowing data has been arranged in ascending order : 12, 14, 17, 20, 22, x, 26, 28, 32, 36. If the median of the data is 23, find x. In the above data, if 32 is repaced with 23, find the new median. 39. The weight (in kg) of 10 students of a cass are given beow : 41, 39, 48, 52, 35, 28, 30, 35, 46, 32. Find the mean and median for this data. 40. The distribution of weight (in kg) of 100 peope is given beow : weight (kg) Frequency Tota 100 Construct a histogram and frequency poygon for the data. 41. Find the missing frequencies f 1 and f 2 in the foowing frequency distribution, if it is known that the mean of the distribution is x i Tota f i 46 f 1 f Find the vaue of a. 42. The eectricity bis (in Rs.) of 25 houses in a certain ocaity are given beow : 500, 440, 100, 180, 150, 560, 300, 220, 300, 240, 150, 180, 120, 270, 420, 450, 450, 240, 200, 220, 140, 160, 250, 360, 470. Construct a grouped frequency tabe for this data with one of the cass intervas as (200 not incuded). 43. The points scored by a basked ba team in a series of match are foows. 17, 2, 7, 27, 25, 5, 14, 18, 10, 24, 28, 10, 8, 7, 10, 28. Find the median and mode of the data. 44. Find the mean, median and mode of the foowing data : 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, If the mean of the foowing distribution is 6, find the vaue of p. x p + 5 f Given beow are the seats won by different poitica parties in the foowing outcome of state assemby eections. Poitica party A B C D E F Seats won (i) Draw a bar graph to represent the poing resuts. (ii) Which poitica party won the maximum number of seats? 47. Find the mean saary of 20 workers of a factory from the tabe. Saary (in Rs) No. of workers 6, , , ,000 3 Tota The marks obtained by 40 students of cass IX in mathematics are given beow : 81, 55, 68, 79, 85, 43, 29, 68, 54, 74, 47, 35, 72, 64, 95, 44, 55, 77, 64, 35, 79, 52, 45, 54, 70, 83, 62, 64, 72, 92, 84, 76, 63, 43, 54, 38, 73, 68, 52, 54. Prepare a frequency distribution tabe with casssize of 10 marks. 49. Find the median of the foowing data : 19, 25, 59, 48, 35, 37, 30, 32, 51 If 25 is repaced by 52, what wi be the new median 50. Find the mean of the foowing data. Marks Frequency

10 LONG ANSWER TYPE QUESTIONS A. Important Questions 1. Draw a histogram to represent the foowing frequency distribution. Cass-interva Frequency The weights (in grams) of 30 oranges, picked at random from a basket of oranges are given beow : 90, 30, 45, 55, 65, 60, 50, 75, 70, 60, 70, 70, 60, 95, 85, 80, 35, 45, 40, 45, 55, 30, 110, 75, 100, 40, 60, 85, 40, 100 Construct a grouped frequency distrubution tabe with equa cass intervas, one of them being Draw a histogram and frequency poygon (in the same diagram) for the foowing data : Cass Frequency The weights (in kg) of 25 empoyees of an office are given beow : 81, 78, 81, 77, 75, 77, 69, 76, 65, 85, 87, 71, 83, 67, 85, 73, 78, 68, 85, 73, 85, 54, 70, 68, 80 Find the mean, median and mode of the above data. 5. Draw a histogram and a frequency poygon from the foowing data : Cass Frequency The weights (in kg) of 60 persons are given beow. Find the mean weight of a person by (i) direct method (ii) deviation method. Weight (in kg) No. of persons B. Questions From CBSE Examinations Papers 1. Draw a frequency poygon for the foowing data: Cass Frequency [4 Marks] 2. Draw a histogram for the foowing data : Marks No. of students

11 3. Draw a frequency poygon for the foowing data: Cost of iving index No. of weeks Draw a histogram and frequency poygon for the foowing distribution : Marks obtained No. of students Draw a histogram for the foowing data : Age (in years) No. of chidren The foowing tabe presents the number of iterate femaes in a town : Age group No. of femaes Draw a frequency poygon for the above data. 7. Draw a histogram for the foowing data : x f The foowing tabe gives the distribution of students of two sections according to the marks obtained by them. Section A Section B Marks Frequency Marks Frequency Represent the marks of the students of both the sections on the same graph by two frequency poygons. From the two poygons, compare the performance of the two sections. 9. Draw a histogram and frequency poygon on the same graph for the foowing data : Marks No. of students The runs scored by two teams A and B in 7 overs in a cricket match are given. Number of bas Team A Team B Represent the data of both the teams on the same graph by frequency poygon. 11

12 11. Draw a histogram and frequency poygon for the foowing distribution : Marks obtained No. of students The daiy income of 50 doctors is given beow : Daiy income (in Rs) No. of doctors Draw a histogram for the above data. 13. The monthy saary (in thousand rupees) of 50 workers in a factory are given beow : Saary No. of workers Find the mean saary of a worker. 14. Draw a histogram for the foowing data : Weight (in kg) No. of students Foowing tabe shows the frequency distribution for the speeds of cars passing through a particuar point on a highway : Cass interva Frequency Draw a histogram and a frequency poygon to represent the above data. 16. The distribution of weight (in kgs) of 100 peope is given beow. Weight (in kg) Frequency Construct a histogram for the above distribution. 17. The marks obtained by a cass of 80 students are given beow : Marks No. of students Draw a histogram to represent the above data. 18. Make a frequency poygon for given frequency tabe. Cass interva Frequency Draw a histogram for the foowing data : Cass interva Frequency Draw a histogram to represent the foowing distribution. Cass interva Frequency

13 21. Find mean, mode and median for the foowing data : 10, 15, 18, 10, 10, 20, 10, 20, 15, 21, 15 and The daiy wages of 100 workers (in Rs) in a factory are given beow : Daiy wages (in Rs) No of workers Draw a frequency poygon for given data. 23. The foowing tabe presents the number of iitrate femaes in the age group (10 34) in a town. Age group No of femaes Draw a histogram to represent the above data. 24. Construct a frequency poygon for the foowing data : Age Frequency FORMATIVE ASSESSMENT Project The marks obtained by a cass of 80 students (out of 100) are given beow. Construct a histogram to represent the data : Marks Number of students Foowing tabe gives the distribution of the marks obtained by the students of a cass. Marks Number of students Represent the data by frequency poygon. 27. Draw the frequency poygon without constructing the histogram of the foowing observations : Cost of iving index Objective : Coection, tabuation of data obtained by an activity. Representing the information diagrammaticay. Materias Required : Chart paper 50 cm 50 cm, Geometry box, Meter ruer, 10 counters of diameter 2 cm approx, graph sheets. Procedure : 1. On chart paper, draw horizonta and vertica ines to give squares of 5 cm 5 cm. Using sketch pen, write numbers from 1 to 100 as shown in figure Make a tabe as shown in Tabe 1. Number of weeks Tota 52 13

14 3. Pace the chart paper on the foor. 4. Take a the ten counters and drop them simutaneousy on the chart paper from a height of 50 cm (approx) above the centre of chart (approx). 5. Read the numbers written in the squares on the chart over which the counters fa and mark them in the Tay Marks coumn of the tabe. Keep the foowing rues : (i) If a counter fas on the boundary of two squares such that more than haf of it ies in one square and ess than haf in the other, record the number which has the bigger portion of the counter. For exampe, in figure 1 the counter marked I in the diagram shoud be taken as on the number 87, whie the counter marked H shoud be taken as on the number 79. (ii) If a counter fas amost haf on two squares, do not record it and drop the counter again. For exampe, the counter marked J in figure 1. (iii) If a counter fas on the boundary of more than two squares, reject it and drop it again. For exampe, counters marked A, B and C. (iv) If a counter fas outside the chart reject it and drop it again. 6. Repeat the process ten times. 7. Compete the frequency tabe. 8. Draw a histogram to represent the data obtained, in tabe 1 (see figure 2.) 9. Mark the mid-point of the respective tops of the rectanges in the histogram and join them using a ruer. To compete the poygon, join the mid-points of two casses (caed imaginary casses) one at each end. See figure Find the cass mark of each cass in Tabe 1. On a graph sheet, represent cass marks on the x-axis and frequencies on the y-axis. Pot the corresponding points and compete the poygon (see figure 3). Figure 1 14

15 Tabe 1 Cass Tay Marks Frequency Cass Marks Tota 100 Figure 2 Figure 3 (Here we take cass 0 10 to denote greater than 0 and ess than or equa to 10, to denote greater than 10 and ess than or equa to 20 and so on). 15

16 Observations : 1. In figure 2 the rectange in the histogram are of equa width since the cass size is 10 for a the casses. 2. If both a histogram and frequency poygon are to be drawn representing the same data, we use diagram as shown in figure If ony the frequency poygon is to be drawn, we need not draw a histogram and directy draw a frequency poygon as shown in figure 3. Project 2 Objective : Survey of the famiy members of students in a cass. Project Overview : A survey of the famiy members of cass IX students of St. Thomos Pubic Schoo was conducted in the month of Juy The aim of the survey was to find out the types of students with regard to the number of famiy members in their famiies. Such survey has not been conducted so far in this schoo. The schoo is situated in an industria area. Large factories are situated near the schoo. Some government offices are aso situated near the schoo. Most of the students ive in the fats provided to their parents by the factory authorities. This survey is expected to give us a picture of optimum eve of tution fees, transport faciity, items to be sod in the schoo canteen, extra coaching casses and other such things connected to the schoo curricuum. Such surveys may be hepfu in getting an overa view of the type of faciities required at such schoos. Procedure : The data of the survey was coected by directy asking each student of the cass with the hep of the cass teacher Mr. Kabir Saxena. The students were asked to raise their hands according to the number of famiy members in their famiy. Foowing categories were considered as the members of the famiy. (i) Actua reatives iving in their house. (ii) Other members ike paying guests, servents etc. (iii) Brothers, sisters and other reatives who ived in their house during vacation. Data obtained were recorded as beow. No. of members in the famiy No. of famiies Computation : We can represent the above tabe as beow and cacuate the mean. No. of members No. of fx in the famiy famiies (x) (f ) Tota Mean = Σfx Σf = =

17 Diagrams : We can draw bar diagram to represent the above data. Observations : From the bar graph, we observed that (i) 21 famiies have 6 members. (ii) Ony 1 famiy has 9 members. (iii) No famiy has more than 9 members. Aso, average members in a famiy = Do Yoursef : (i) Survey of the pocket money of a cass of students. (ii) Survey of the different modes of transport used by students of a schoo. 17