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Chap 14 : Statistics www.rava.org.in File Revision Date : 20 September 2019 CHAPTER 14 Objective Questions CLASS : 9 th Statistics SUB : Maths For NCERT Solutions Previous Years Chapterwise QB visit www.cbse.online or whatsapp at 8905629969 1. OBJECTIVE QUESTIONS (d) Neither in 40-50 nor in 50-60 Ans : (d) Neither in 40-50 nor in 50-60 1. The mean of six numbers is 30. If one number is 60 is included in neither class interval 40-50 nor in 50-60. excluded, the mean of the remaining numbers is 29. The excluded number is 6. In a morning walk, I took 20 rounds of a park. During this period I came across person A, person (a) 29 (b) 30 B , person C and person D , 11 times, 7 times, 10 times and 5 times respectively. I want to represent this (c) 35 (d) 45 data graphically. Which of the following is the best representation? Ans : (c) 35 (a) Bar graph Sum of 6 numbers = 30 # 6 = 180 (b) Histogram with unequal widths Sum of remaining 5 numbers = 29 # 5 = 145 Hence, Excluded number = 180 - 145 = 35 (c) Histogram with equal widths 2. If each observation of a data is increased by 5, then (d) Frequency polygon their mean (a) is decreased by 5 Ans : (a) Bar graph (b) is increased by 5 (c) becomes 7 times the original mean Bar graph is the simple most and popular graph to (d) remains the same show ungrouped frequency distribution graphically. Ans : (b) is increased by 5 7. The width of each of the five continuous classes in a frequency distribution is 5 and the upper class limit Let xr = x1 + x2 + ..... + xn of the upper class is 60. The lower class limit of the n lowest class is If 5 is added in each observation, then (a) 45 (b) 25 Mean = (x1 + 5) + (x2 + 5) + ..... + (xn + 5) (c) 35 (d) 40 n Ans : (c) 35 = (x1 + ..... + xn) + 5n = xr+ 5 Lower class limit = Upper class limit - width of class n 60 - 5 = 55 3. Tally marks are used to find continuous classes in a frequency distribution is 35-40, 40-45, 45-50, 50-55, 55-60 (a) class intervals (b) range Lower class limit of lowest class is 35. (c) upper limits (d) frequency Ans : (d) frequency 4. The range of the data 15, 20, 6, 5, 30, 35, 92, 35, 90, 8. The class marks of a frequency distribution are 15, 18, 82 is 20, 25, 30, .......... The class corresponding to the class (a) 87 (b) 15 mark 25 is (b) 20.5 - 29.5 (a) 12.5 - 17.5 (c) 18 (d) 26 (c) 18.5 - 21.5 (d) 22.5 - 27.5 (a) 87 Ans : (d) 22.5 - 27.5 Arranging the data in ascending order, we have 5, 6, Class width = 20 - 15 = 5 15, 18, 20, 30, 35, 35, 82, 90, 92 Range = Max. value - Min. value = 92 - 5 = 87 Class mark = 25 Required class is b25 - 5 l - b25 + 5 2 2l 5. In the class intervals 40-45, 50-60 the number 60 is = (25 - 2.5) - (25 + 2.5) included in = 22.5 - 27.5 (a) 50-60 (b) 40-50 (c) Both in 40-50 and 50-60 9. For drawing a frequency polygon of a continuous Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 109

Chap 14 : Statistics www.cbse.online frequency distribution, we plot the points whose 13. If the mean of five observations x , x + 4, x + 8, x + 12 ordinates are the frequencies of the respective classes and abscissa are respectively, the and x + 16 is 15, then the value of x is (a) upper limits of the classes (a) 5 (b) 6 (b) lower limits of the classes (c) 7 (d) 8 (c) class marks of the classes Ans : (c) 7 (d) upper limits of preceding classes Mean = Sum of observations Ans : (c) class marks of the classes Number of observations 15 = x + x + 4 + x + 8 + x + 12 + x + 16 5 10. In a frequency distribution, the mid-value of a class 5x + 40 = 75 5x = 35 is 10 and width of each class is 6. The upper limit of x =7 the class is (a) 13 (b) 7 (c) 8 (d) 12 14. The mean of the marks scored by 40 students was Ans : (a) 13 found to be 35. Later on it was discovered that a score Let the upper limit = u and the lower limit = l of 43 was misread as 34. The correct mean is u-l =6 ...(1) (a) 35.2 (b) 39.4 u+l (c) 39.8 (d) 39.2 2 = 10 u + l = 20 ...(2) Ans : (a) 35.2 Solving (1) and (2), we get Sum of the observation = 35 # 40 = 1400 Once of the observation 43 was misread as 34 u = 13; l = 7 Correct sum = 1400 - 34 + 43 = 1409 11. Let U be the upper class boundary of a class in a Correct mean = 1409 = 35.225 40 frequency distribution and M be the midpoint of the class. Which one of the following is the lower class 15. There are 50 numbers. Each number is subtracted boundary of the class? from 43 and the mean of the numbers so obtained is (M + L) (a) M + 2 (b) L+ M+ L found to be 5. The mean of the given numbers is 2 (c) 2M - U (d) M - 2L (a) 38 (b) 39 Ans : (c) 2M - U (c) 48 (d) 49 Ans : (a) 38 Class mark x1 + ... + x50 50 = Upper class boundary + lower class boundary =x 2 x1 + .... + x50 = 50x M = U + L 2 2M - U = L 43 - x1 + 43 - x2 + ..... + 43 - x50 = Now, 50 New mean 12. The mid-value of a class interval is 25 and the class 43 # 50 # x =5 50 size is 8. The class interval is (a) 37 - 45 (b) 21 - 29 2150 - 50 # x = 50 # 5 (c) 36.5 - 44.5 (d) 36.5 - 46.5 x = 1900 x = 38 50 Ans : (b) 21 - 29 Class mark = 25 16. Mode of the data 15, 14, 71, 15, 91, 2, 51, 19, 41, 51, and Class size = 8 18, 15, 51 is (a) 51 (b) 15 Class mark (c) 16 (d) 17 = Lowerclass limit + Upper classlimit Ans : (a) 51 2 l+4 Arranging the data in ascending order, we have 2, 14, 2 = 25 15, 15, 18, 19, 41, 51, 51, 51, 71, 91 l + u = 50 ...(1) Mode = Highest occurring number upper class limit - lower class limit = class size Mode = 51. u-l =8 ...(2) Solving (1) and (2), we get 17. The median of the numbers 9, 5, 7, 17, 13, 18, 13, 9, u = 29 ,5 , 17, 13, 12, 17 is l = 21 (a) 7 (b) 9 (c) 13 (d) 15 Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 110

Chap 14 : Statistics www.rava.org.in Ans : (c) 13 their mean (a) remains the same Arranging the given numbers in ascending order 5, 5, (b) becomes 7 times the original mean 7, 9, 9, 12, 13, 13, 13, 17, 17, 17, 18. (c) is decreased by 7 (d) is increased by 7 The number of terms = 13 (odd) Ans : (d) is increased by 7 Median = n + 1 th = 13 + 1 th term b 2 b 2 l l x1 + x2 + ..... + xn n = 7th term = 13 Let x = If 7 is added in each observation, then 18. The median of the numbers 45, 34, 65, 48, 93, 54, 22, New mean = x1 + 7 + x2 + 7+ ..... + xn + 7 n 86, 45, 87, is (a) 51 (b) 49.5 = (x1 + ..... + xn) + 7n =x+7 n (c) 54 (d) 56 Ans : (a) 51 Hence the mean is increased by 7. Arranging the given number in ascending order 22, 34, 22. If x is the mean of x1, x2, ....., xn , then for a ! 0, the 45, 45, 48, 54, 65, 86, 87, 93 mean of ax1, ax2, ....., axn , then for a ! 0, the mean x1 x2 xn Number of terms = 10 (even) of ax1, ax2, ...., axn , a , a , ..., a is When number of terms are even, then (a) ba + 1 lx (b) ba + 1 x n th term + n + 1kthterm a a l2 a 2 a 2 Median = k 1 a 2 1 x ba + lx a ln 5th term + 6th term (c) ba + (d) 2n 2 = 1 x a l2 = 48 + 54 = 102 = 51 Ans : (b) ba + 2 2 x = x1 + ..... + xn n 19. If the mean of the observations x , x + 3, x + 5, x + 7 x1 x2 xn and x + 10 is 9, the mean of the last three observations Now, ax1 + ... + axn + a + a + ... + a is 1 2 2n 3 3 (a) 10 (b) 10 a (x1 + x2 + ... + xn) + 1 (x1 + x2 + .... + xn) a (c) 11 1 (d) 11 2 = 2n 3 3 1 1 Ans : (c) 11 3 = a (nx ) + a (nx ) = nx ba + 1 = x ba + 1 2n a 2 a We know, 2n l l Mean = Sum of all the observations 23. The mean of 90 items was found to be 45. Later on Total number. of observation it was discovered that two items were misread as 26 x + x + 3 + x + 5 + x + 7 + x + 10 Mean = 5 and 19 instead of 62 and 09 respectively. The correct mean is 9 = 5x + 25 (a) 49.0 (b) 45.0 5 (c) 45.3 (d) 49.3 x =4 Ans : (c) 45.3 So, mean of last three observations is Sum of the observations = 90 # 45 3x + 22 = 12 + 22 = 34 = 11 1 = 4050. 3 3 3 3 n The observations 62 and 9 are misread as 26 and 19 /20. If x is the mean of x1, x2, x3, ....., xn , then (xi - x ) = Correct sum = 4050 - 26 - 19 + 62 + 9 23 (a) 25 (b) 0 i=1 28 4 = 4076 25 5 (c) (d) Correct mean = 4076 = 45.3 90 Ans : (b) 0 We know that mean of n observations is 24. The mean of 53 observations is 36. Out of these /x 1 n observations, the mean of first 27 observations is 32 n = xi and that of the last 27 observations is 40. The 27th i=1 observation is / /n n (a) 23 (b) 36 nx = xi & (x1 - x ) = 0 (c) 38 (d) 40 i-1 i-1 21. If each observation of a data is increased by 7, then Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 111

Chap 14 : Statistics www.cbse.online Ans : (b) 36 29. The points scored by a kabaddi team in a series of Sum of the observations = 36 # 53 = 1908 matches are as follows: Sum of first 27 observations = 27 # 32 = 864 Sum of last 27 observations = 27 # 40 = 1080 8, 24, 10, 14, 5, 15, 7, 2, 17, 27, 10, 7, 48, 8, 18, 28 We have, x1 + ... + x27 + x27 + x28 + ... + x53 = 864 + 1080 = 1944 Find the median of the points scored by the team. 1908 + x27 = 1944 (a) 12 (b) 14 x27 = 1944 - 1908 = 36 (c) 10 (d) 15 Ans : (a) 12 Arranging the given data in ascending order: 2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48 Number of terms = 16 (even) 25. The marks obtained by 20 students of a class in a test a n th term + a n + 1kth term 2 2 (out of 50) are given below: 40, 44, 45, 46, 50, 42, 41, Median = k 08, 26, 28, 09, 32, 24, 06, 42, 36, 39. 2 The range of the data is = 8th term + 9th term 2 (a) 44 (b) 54 (c) 90 (d) 10 = 10 + 14 = 12 2 Ans : (a) 44 Range = Max. value - Min. value 30. In the class-intervals 30 - 40, 40 - 50 the number 50 = 50 - 6 = 44 is included in (a) 40 - 50 26. The class mark of the class 150-170 is (b) 30 - 40 (c) Both in 30 - 40 and 40 - 50 (a) 130 (b) 135 (d) Neither in 30 - 40 nor in 40 - 50 (c) 140 (d) 160 Ans : (d) Neither in 30 - 40 nor in 40 - 50 Ans : (d) 160 Class mark = upper limit + lower limit 31. Given the class-interval 0-10, 10-20, 20-30, .......... 2 then 20 is considered in class 150 + 170 320 = 2 = 2 = 160 (a) 20-30 (b) 10-20 (c) 10-30 (d) 15-25 27. The mean of eight numbers is 40. If one number is Ans : (a) 20-30 20 is considered in class 20-30. excluded, their mean becomes 30. The excluded number is (a) 30 (b) 130 32. The marks obtained by 12 students of a class in a test (c) 110 (d) 138 are 36, 27, 5, 19, 34, 23, 37, 23, 16, 23, 20, 38. Find Ans : (c) 110 mode. Sum of 8 observations = 40 # 8 = 320 (a) 23 (b) 26 Sum of 7 observations = 30 # 7 = 210 Excluded observation = 320 - 210 = 110 (c) 20 (d) 36 Ans : (a) 23 28. The median of the data arranged in ascending order Arranging the data in ascending order, we have 5, 16, 19, 20, 23, 23, 23, 27, 34, 36, 37, 38 8, 9, 12, 18, (x + 2), (x + 4), 30, 31, 34, 39 is 24. The Mode Marks = Highest occurring marks = 23 value of x is (a) 22 (b) 21 33. In a frequency distribution, the mid value of a class (c) 20 (d) 24 is 10 and the width of the class is 6. The lower limit Ans : (b) 21 of the class is Number of terms = 10 (even) (a) 6 (b) 7 n th n + 1kthterm (c) 8 (d) 12 2 2 Median = a k term + a Ans : (b) 7 2 Let x be the upper limit and y be the lower limit. Since the mid value of the class is 10 = 5th term + 6th term 2 Hence, x + y = 10 2 24 = (x + 2) + (x + 4) 2 x + y = 20 ...(1) 48 = 2x + 6 and x - y = 6 (width of the class = 6) ...(2) x = 21 By solving (1) and (2), we get Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 112

Chap 14 : Statistics www.rava.org.in y =7 the least value of the variate. Hence, lower limit of the class is 7. Ans : range 34. Let xr be the mean of x1, x2, .......... xn and yr the mean 6. .......... is found by adding all the values of the observations and dividing this by the total number of of y1, y2, .......... yn . If xr is the mean of x1, x2, .......... xn , observations. y1, .......... yn then zr is equal to xr+ yr Ans : Mean 2 (a) xr+ yr (b) (c) xr+ yr (d) xr+ yr 7. The .......... of all bars in histogram should be uniform. n 2n Ans : width xr+ yr Ans : (b) 2 x1 + ..... + xn = xr 8. .......... can also be drawn independently without n drawing a histogram. y1 + y2 + ..... + yn = yr Ans : Frequency polygon n zr = xr+ yr 9. If n is an odd number, the median = value of the 2 .......... observation. 35. The mean of 100 observations is 50. If one of the Ans : bn + 1 th 2 observations which was 50 is replaced by 150, the l resulting mean will be (a) 50.5 (b) 51 10. The .......... of a class interval is called its class mark. Ans : mid-point (c) 51.5 (d) 52 Ans : (b) 51 We have xr = 50 3. TRUE/FALSE Sxi = 50 DIRECTION : Read the following statements and write your 100 answer as true or false. Sxi = 5000 1. Arithmetic mean, Geometric mean, Harmonic mean, As, 50 is replaced by 150. Median and Mode are various measures of central tendency. Thus, Ans : True Now, Sxi = 5000 - 50 + 150 = 5100 Resulting mean, = 5100 = 51 100 2. FILL IN THE BLANK 2. The mean of a set of numbers is xr, if each number is increased by k , then mean of new set is xr- k . DIRECTION : Complete the following statements with an Ans : False appropriate word/term to be filled in the blank space(s). New mean = xr+ k 1. The range of the data 15, 20, 6, 5, 30, 35, 93, 34, 91, 17, 83, is .......... 3. The algebraic sum of the deviations of a set of n Ans : 93 - 5 = 88 values from their mean is 0. Ans : True 2. .......... is the value of the middle most observation (s). 4. The space between consecutive bars in bar graph Ans : Median should also be same. 3. Width of the class-interval is called .......... of class Ans : True interval. Ans : size 5. Mean may or may not be the appropriate measure of central tendency. 4. The .......... is the most frequently occurring observation. Ans : True Ans : mode 6. If the arithmetic mean of 7, 5, 13, x and 9 is 10, then 5. The .......... is the difference between the greatest and the value of x is 16. Ans : True 7 + 5 + 13 + x + 9 = 10 5 Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 113

Chap 14 : Statistics www.cbse.online x = 16 (T) Since the number of observation is 10 (even) 7. The data collected by the investigator himself for a Hence, Median = 5th term + 6th term definite plan or purpose is known as primary data. 2 Ans : True 65 = x + x + 2 = 2x + 2 =x+1 2 2 x = 64 8. The data collected by someone and used by any other 2. Match the following : person known as primary data. Ans : False Column-I Column-II It is known as secondary data. (P) For the set of (1) mode + 2 mean numbers 2, 2, 4, 5 4. MATCHING QUESTIONS and 12, the true statement is DIRECTION : Each question contains statements given in two columns which have to be matched. Statements (P, Q, R, S, (Q) 3 median is equal to (2) mid-value of the T) in Column-I have to be matched with statements (1, 2, 3, 4, class 5) in Column-II. (R) In a histogram, each (3) Mean > Mode 1. Match the following : class rectangle is constructed with base as Column-I Column-II (S) A frequency polygon (4) Class-intervals is constructed by (P) Mode of the data 15, 14, (1) 64 plotting frequency 19, 20, 14, 15, 16, 14, 15, of the class interval 18, 14, 19, 20, 15, 17, 15 and the is Ans : P–3, Q–1, R–4, S–2 (Q) The range of the data 25, (2) 105 (P) Mean = 2 + 2 + 4+ 5 + 12 =5 18, 20, 22, 16, 6, 17, 15, 5 12, 30, 32, 10, 19, 8, 11, 20 is Mode = 2 (Mean>Mode) (Q) Mode = 3 median - 2 mean (R) The class mark of the (3) 17.5 - 22.5 class 90 - 120 is 3 medium = mode + 2 mean (S) The class marks of a (4) 15 3. Match the following : frequency distribution are given as follows: 15, Column-I Column-II 20, 25, .......... The class corresponding to the class (P) The class marks of the class (1) 13 mark 20 is interval 145-150 is (T) The following (5) 26 (Q) In a frequency distribution, (2) 110 the mid-value of a class is observations are arranged 10 and width of each class is 6. The upper limit of the in ascending order: 26, class is 29, 42, 53, x , x + 2, 70, 75, 82, 93. If the median is 65, then the value of x (R) The mean of eight numbers (3) 22.5–27.5 is 40. If on number is is excluded, their mean becomes 30. The excluded Ans : P–4, Q–5, R–2, S–3, T–1 number is (P) Most occurring observation is 15. (Q) Highest data value is 32 and the lowest is 6 Hence, Range = highest value -lowest value (S) The class marks of a (4) 147.5 frequency distribution are = 32 - 6 = 26 15, 20, 25, 30, .......... The class corresponding to the (R) Class-mark = 90 + 120 = 210 = 105 class mark 25 is 2 2 (S) The class corresponding to the class mark 20 is given as Ans : P–4, Q–1, R–2, S–3 15 + 20 = 35 = 17.5 (P) Class - Marks = 145 + 150 = 295 2 2 2 2 and 20 + 25 = 45 = 22.5 = 147.5 2 2 Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 114

Chap 14 : Statistics www.rava.org.in (Q) Let the upper limit = x 5. Match the following: Hence, Lower limit = y List-I List-II and x-y =6 (P) Data which is collected for (1) Secondary x+y = 10 the first time by the statis- data 2 tical investigator or with the x + y = 20 help of his workers is called On solving both the equations, we get x = 13 and y=7 (Q) These are the data already (2) Variable collected by a person or a (R) Sum of the 8 observation = 40 # 8 = 320 society and these may be in published form. These data Sum of the 7 observation . = 30 # 7 = 210 should be carefully used. Excluded observation is 320 - 210 = 110 (R) When the data is compiled (3) P r i m a r y (S) Class - width = 20 - 15 = 5 in the same form and order Data Class mark = 25 in which it is collected, it is Hence, Required class = b25 - 5 l - b25 + 5 l known as 2 2 (S) A quantity which can vary (4) Raw Data = 22.5 - 27.5 from one individual to another is called 4. Match the following : List-I List-II P QR S 4 (P) Mean of first 10 odd prime (1) 27.5 (a) 3 12 2 numbers is 4 (b) 3 14 2 (Q) Mean of first 10 multiples of 5 is (2) 15.8 (c) 1 32 (R) mean of first 9 doublets of natu- (3) 11 (d) 1 34 ral numbers is Ans : (b) P - 3, Q - 1, R - 4, S - 2 (S) Mean of first 10 even numbers is (4) 55 P QR S 5. ASSERTION AND REASON (a) 1 234 (b) 1 243 DIRECTION : In each of the following questions, a statement (c) 2 143 of Assertion is given followed by a corresponding statement (d) 2 134 of Reason just below it. Of the statements, mark the correct answer as Ans : (c) P - 2, Q - 1, R - 4, S - 3 (a) Both assertion and reason are true and reason is (P) First 10 odd prime numbers are 3, 5, 7, 11, 13, 17, the correct explanation of assertion. 19, 23, 29, 31 (b) Both assertion and reason are true but reason is not the correct explanation of assertion. Mean = Sum of observations = 158 = 15.8 Number of observations 10 (c) Assertion is true but reason is false. (d) Assertion is false but reason is true. (Q) First 10 multiples of 5 are 5, 10, 15, 20, 25, 30, 1. Assertion : If the mean of five observations x , x + 2 , x + 4, x + 6, x + 8 is 11, then mean of last three 35, 40, 45, 50 observations is 8. Reason : Mean of n observations Mean= Sum of observation = 275 = 27.5 Sum of observations Number of observations 10 Number of observations (R) First 9 doublets of natural numbers are 11, 22, 33, Ans : (d) Assertion is false but reason is true. 44, 455, 66, 77, 88, 99 Mean= Sum of observation = 495 = 55 Mean = Sum of observation Numbers of observations 9 Numbers of observations (S) First 10 even numbers are 2, 4, 6, 8, 10, 12, 14, 11 = x+x+2+x+4+x+6+x+8 5 16, 18, 20 Mean= Sum of observation = 110 = 11 55 = 5x + 20 Numbers of observations 10 5x = 35 & x = 7 Mean of last three observations = 11 + 13 + 15 = 39 = 13 3 3 Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 115

Chap 14 : Statistics www.cbse.online 2. Assertion : The range of the first 6 multiples of 6 is 9. 7. Assertion : Median of the given data 34, 31, 42, 43, 46, Reason : Range = Maximum value - Minimum value Ans : (d) Assertion is false but reason is true. 25, 39, 45, 32, is 39. First 6 multiples of 6 = 6, 12, 18, 24, 30, 36 Reason : When the number of observations (n) is even, Range = 36 - 6 = 30 n an2 + 1kth the median is the mean of the a 2 th and k observations. 3. Assertion : The median of 83, 37, 70, 29, 45, 63, 41, Ans : (b) Both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion. 70, 34, 54, is 49.5. 8. Assertion : The difference between the maximum and minimum values of a variable is called its range. Reason : The median of n odd number of observations Reason : The number of times a variate (observation) n + 1 th occurs in a given data is called range. is b2l term. Ans : (c)Assertion is correct but Reason is incorrect. Ans : (b) Both assertion and reason are true but The number of times a variate (observation) occurs in reason is not the correct explanation of assertion. a given data is called frequency of that variate. 29, 34, 37, 41, 45, 54, 63, 70, 70, 83 Number of terms = 10 (even) For even number of observation we have Median = ^ n hth term + ^ n + 1hth term 9. Assertion : If the median of the given data 26, 29, 42, 2 2 53, x , x + 2, 70, 75, 82, 93, is 65 then the value of x 2 is 64. 5th term + 6th term = 2 Reason : When the number of observations (n) is odd n + 1 + the median is the value of the b 2 th observation. = 45 2 54 = 99 = 49.5 2 l Ans : (b) Both Assertion and Reason are correct, but 4. Assertion : The median of the following observation Reason is not the correct explanation of Assertion. 0, 1, 2, 3, x , x + 2, 8, 9, 11, 12 arranged in ascending Assertion : Given, number of observation is 10 (even) order is 63, then the value of x is 62. Reason : Median of n even observations is Hence Median = x + (x + 2) 2 ^ n hth term + ^ n + 1hth term 65 = 2x + 2 =x+1 2 2 2 2 Ans : (a) Both assertion and reason are true and 65 = x + 1 reason is the correct explanation of assertion. x = 64 Number of terms = 10(even) 10. Assertion : Mean may or may not be the appropriate Median = ^ n hth term + ^ n + 1hthterm measure of central tendency. 2 2 2 Reason : If the number of observations are even then 5thterm + 6thterm median is b n + 1 th term. 2 2 l + x+ Ans : (c) Assertion is correct but Reason is incorrect. 2 = x 2 = 63 Assertion is correct and Reason is incorrect. 2x + 2 = 126 WWW.CBSE.ONLINE x = 62 NO NEED TO PURCHASE ANY BOOKS 5. Assertion : The following is the data of wages per day: For session 2019-2020 free pdf will be available at 8, 4, 7, 5, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8, then the mode of www.cbse.online for the data is 8. 1. Previous 15 Years Exams Chapter-wise Question Reason : Mode = Highest observation - lowest observation. Bank 2. Previous Ten Years Exam Paper (Paper-wise). Ans : (c) Assertion is true but reason is false. 3. 20 Model Paper (All Solved). 4. NCERT Solutions 8 is the most frequent value. All material will be solved and free pdf. 6. Assertion : Mode of the given data 110, 120, 130, 120, Disclaimer : www.cbse.online is not affiliated to Central 110, 140, 130, 120, 140, 120, is 120. Board of Secondary Education, New Delhi in any manner. Reason : The observation that occurs most frequently, www.cbse.online is a private organization which provide i.e., the observation with maximum frequency is called free study material pdfs to students. At www.cbse.online mode. CBSE stands for Canny Books For School Education. Ans : (a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. Since the value 120 occurs maximum number of times. Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 116