Marks obtained 0-10 10-20 20-30 30-40 40-50 No. of students 8 10 22 40 20 Hence determine:  (i) the median  (ii) the pass marks if 85% of the students pass.  (iii) the marks which 45% of the...

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 3 8 12 14 10 6 5 2 Also state the median class Solution: Arranging the data in cumulative frequency table. Marks Frequency Cumulative...

Age in years 12 13 14 15 16 17 18 No. of students 2 3 5 6 4 3 2 Solution: Age in years xi No. of students fi Cumulative frequency fixi 12 2 2 24 13 3 5 39 14 5 10 70 15 6 16 90 16 4 20 64 17 3 23 51...

Solution: Here n = 9 which is odd. Median = ((n+1)/2)th term Median = ((9+1)/2)th term Median = (10/2)th term Median = 5 th term Median = 5 Mode is the number which appears most often in a set of...

Solution: Arranging data in ascending order 1,3,5,6,8,9,10,12,13,15,17,18,20,21,21,23 Here n = 16 which is even (i) So median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (16/2 th term +...

Solution: Arranging numbers in ascending order 17,26,29,32,33,34,45,56,60 Here n = 9 which is odd. So Median =( (n+1)/2) th term Median = ((9+1)/2)th term Median = (10/2)th term Median = 5th term...

Solution: Arranging numbers in ascending order 3,4,5,x,8,9,11 Here n = 7 which is odd Given median = 6 So Median =( (n+1)/2) th term 6 = ((7+1)/2)th term 6 = ((8/2)th term 6 = 4th term 6 = x Hence...

Diameter in mm 32-36 37-41 42-46 47-51 52-56 57-61 62-66 No. of screws 15 17 p 25 q 20 30 Solution: Given mean = 51.2 mm The given distribution is not continuous. Adjustment factor = (37-36)/2 = ½ =...

Expenditure in Rs 140-160 160-180 180-200 200-220 220-240 No. of families 5 25 f1 f2 5 Solution: Given mean = 188 Class Frequency fi Class mark xi fixi 140-160 5 150 750 160-180 25 170 4250 180-200...

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Students 2 4 5 16 20 10 6 8 4 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Marks Students fi Class mark xi fixi...

Variate (xi) 13 15 17 19 20+p 23 Frequency (fi) 8 2 3 4 5p 6 Solution: Variate (xi) Frequency (fi) fi xi 13 8 104 15 2 30 17 3 51 19 4 76 20+p 5p 5p2+100p 23 6 138 Total Ʃfi = 23+5p Ʃfi xi =...

Variate (xi) 10 15 20 25 35 Frequency (fi) 3 10 p 7 5 Solution: Variate (xi) Frequency (fi) fx 10 3 30 15 10 150 20 p 20p 25 7 175 35 5 175 Total Ʃfi = 25+p Ʃfi xi = 530+20p Mean = Ʃfx/Ʃf 20.6 =...

Height (in cm) 65 66 67 68 69 70 71 72 73 No. of children 1 4 5 7 11 10 6 4 2 Calculate the mean height for this distribution correct to one place of decimal.  Solution: Height x No. of children f...

No. of matches 41 42 43 44 45 46 No. of boxes 5 8 13 12 7 5 Calculate the mean number of matches per box. Solution: No. o f matches x No. of boxes f fx 41 5 205 42 8 336 43 13 559 44 12 528 45 7 315...

Solution: Total number of students = 50 No. of boys = 40 No. of girls = 50-40 = 10 Average weight of 50 students = 44 kg So sum of weight = 44×50 = 2200 kg Average weight of girls = 40 kg So sum of...

Solution: Average height of 30 students = 150 cm So sum of height = 150×30 = 4500 Difference between correct value and wrong value = 165-135 = 30 So actual sum = 4500+30 = 4530 So actual mean =...

Solution: Given the mean of 20 numbers = 18 Sum of numbers = 18×20 = 360 If 3 is added to each of first 10 numbers, then new sum = (3×10)+360 = 30+360 = 390 New mean = 390/20 = 19.5 Hence the mean...

Solution: Marks scored in English = 36 Marks scored in Civics = 44 Marks scored in Mathematics = 75 Marks scored in Science = x No. of subjects = 4 Average marks = sum of marks / No. of subjects =...

Height in cm 121-130 131-140 141-150 151-160 161-170 171-180 No. of pupils 12 16 30 20 14 8 Draw the ogive for the above data and from it determine the median (use graph paper). Solution: We write...

Height 140-145 145-150 150-155 155-160 160-165 165-170 170-175 175-180 No. of students 12 20 30 38 24 16 12 8 Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2...

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No. of students 5 9 16 22 26 18 11 6 4 3 Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for ogive...

Weight 40-45 45-50 50-55 55-60 60-65 65-70 70-75 75-80 Frequency 5 17 22 45 51 31 20 9 Use your ogive to estimate the following:  (i) The percentage of students weighing 55 kg or more.  (ii) The...

Monthly income No. of employees 6000-7000 20 7000-8000 45 8000-9000 65 9000-10000 95 10000-11000 60 11000-12000 30 12000-13000 5 Draw an ogive of the given distribution on a graph sheet taking 2 cm...

marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No. of students 5 11 10 20 28 37 40 29 14 6 Draw an ogive for the given distribution taking 2 cm = 10 marks on one axis and 2 cm =...

Wages in Rs 400-450 450-500 500-550 550-600 600-650 650-700 700-750 No. of workers 2 6 12 18 24 13 5 Use a graph paper to draw an ogive for the above distribution. ( a scale of 2 cm = Rs 50 on x-...

Scores 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No. of shooters 9 13 20 26 30 22 15 10 8 7 Use your graph to estimate the following:  (i) The median.  (ii) The interquartile...

Weight in kg 50-60 60-70 70-80 80-90 90-100 100-110 110-120 No. of workers 4 7 11 14 6 5 3 Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis , and 2 cm = 5...

Age( yrs) 5-15 15-25 25-35 35-45 45-55 55-65 65-75 No. of casualities due to accidents 6 10 15 13 24 8 7 (i) Construct the ‘less than’ cumulative frequency curve for the above data, using 2 cm = 10...

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 No. of students 3 8 12 14 10 6 5 2 Solution: We write the given data in cumulative frequency table. Marks No of students f Cumulative frequency...

Height ( in cm ) 150-155 155-160 160-165 165-170 170-175 175-180 180-185 No. of workers 6 12 18 20 13 8 6 (i) Determine the cumulative frequencies. (ii) Draw the cumulative frequency curve on a...

Marks obtained 24-29 29-34 34-39 39-44 44-49 49-54 54-59 No. of students 1 2 5 6 4 3 2 Solution: We write the given data in cumulative frequency table. Marks obtained No of students Cumulative...

Class intervals 1-10 11-20 21-30 31-40 41-50 51-60 Frequency 3 5 8 7 6 2 Solution: The given distribution is not continuous. Adjustment factor = (11-10)/2 = ½ = 0.5 We subtract 0.5 from lower limit...

Height ( in cm ) 150-160 160-170 170-180 180-190 190-200 No. of students 8 3 4 10 2 Solution: We write the given data in cumulative frequency table. Height in cm No of students Cumulative frequency...

Mid value 12 18 24 30 36 42 48 Frequency 20 12 8 24 16 8 12 Also state the modal class. Solution: Mid value Frequency 12 20 18 12 24 8 30 24 36 16 42 8 48 12 Here mid value and frequency is given....

Pocket expenses (in Rs) Number of students (Frequency ) 0-5 10 5-10 14 10-15 28 15-20 42 20-25 50 25-30 30 30-35 14 35-40 12 Draw a histogram representing the above distribution and estimate the...

Classes 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 2 8 10 5 4 3 Solution: Construct histogram using given data. Classes 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 2 8 10 5 4 3 Represent classes...

Marks 50-60 60-70 70-80 80-90 90-100 No. of students 4 8 14 19 5 Draw a histogram for the above data using a graph paper and locate the mode. (2011) Solution: Construct histogram using given data....

Class 0-5 5-10 10-15 15-20 20-25 25-30 Frequency 2 5 18 14 8 5 Solution: Construct histogram using given data. Class 0-5 5-10 10-15 15-20 20-25 25-30 Frequency 2 5 18 14 8 5 Represent class on...

Weekly wages (in Rs) 50-55 55-60 60-65 65-70 70-75 75-80 80-85 85-90 No. of workers 5 20 10 10 9 6 12 8 Calculate: (i) The mean. (ii) the modal class (iii) the number of workers getting weekly wages...

Class interval 50-55 55-60 60-65 65-70 70-75 75-80 80-85 85-90 Frequency 5 20 10 10 9 6 12 8 Solution: (i) Class mark (xi) = (upper limit + lower limit)/2 Let assumed mean (A) = 67.5 Class size (h)...

Marks obtained 5 6 7 8 9 10 No. of students 3 9 6 4 2 1 Solution: We write the marks in cumulative frequency table. Marks x Frequency (f) fx Cumulative frequency 5 3 15 3 6 9 54 12 7 6 42 18 8 4 32...

Marks 0 1 2 3 4 5 No. of students 1 3 6 10 5 5 Calculate the mean, median and mode of the above distribution. Solution: We write the data in cumulative frequency table. Marks x Frequency (f)...

x 10 11 12 13 14 15 f 1 4 7 5 9 3 Solution: We write the data in cumulative frequency table. x Frequency (f) Cumulative frequency 10 1 1 11 4 5 12 7 12 13 5 17 14 9 26 15 3 29 Here number of...

Solution: (i)We arrange given marks in ascending order 7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19 16 appears maximum number of times. Hence his modal mark is 16. (ii)Here number of observations, n =...

Solution: Given data in ascending order: 13, 35, 43, 46, x, x +4, 55, 61,71, 80 Given median = 48 Number of observations, n = 10 which is even. median = ½ ( n/2 th term + ((n/2)+1)th term) 48 = ½...

Solution: We arrange given data in ascending order 1, 2, 3, 3, 3, 4, 5, 5, 6, 7 Mean = Ʃxi/n = (1+2+3+3+3+4+5+5+6+7)/10 = 39/10 = 3.9 Here number of observations, n = 10 which is even. So median = ½...

Solution: We arrange given data in ascending order 6, 6, 7, 8, 10, 10, 10, 11, 13 Mean = Ʃxi/n = (6+6+7+8+10+10+10+11+13)/9 = 81/9 = 9 Hence the mean is 9. Here number of observations, n = 9 which...

Solution: We arrange given data in ascending order 1, 1, 2, 2, 3, 3, 3, 6 Mean = Ʃxi/n = (1+1+2+2+3+3+3+6)/8 = 21/8 = 2.625 Hence the mean is 2.625. Here number of observations, n = 8 which is even....

Solution: Mode is the number which appears most often in a set of numbers. (i)Given set is 3, 2, 0, 1, 2, 3, 5, 3. In this set, 3 occurs maximum number of times. Hence the mode is 3. (ii) Given set...

Variate 25 31 34 40 45 48 50 60 Frequency 3 8 10 15 10 9 6 2 Solution: We write the variates in cumulative frequency table. Variate Frequency (f) Cumulative frequency 25 3 3 31 8 11 34 10 21 40 15...

Variate 15 18 20 22 25 27 30 Frequency 4 6 8 9 7 8 6 Solution: We write the variates in cumulative frequency table. Variate Frequency (f) Cumulative frequency 15 4 4 18 6 10 20 8 18 22 9 27 25 7 34...

Solution: Arranging the observations in ascending order 21, 23, 25, 26, 27, 28, 29, 29, 30, 31, 31, 35, 35, 38, 41, 42, 45, 47, 53 Here n = 19 which is odd. (i)Median = ((n+1)/2)th term = (19+1)/2 =...

Number 5 10 15 20 25 30 35 Frequency 1 2 5 6 3 2 1 Solution: We write the numbers in cumulative frequency table. Marks (x) No. of students (f) Cumulative frequency fx 5 1 1 5 10 2 3 20 15 5 8 75 20...

Marks 20 70 50 60 75 90 40 No. of students 8 12 18 6 9 5 12 Calculate the median marks. Solution: We write the marks in ascending order in cumulative frequency table. Marks No. of students (f)...

Marks 35 45 50 64 70 72 No. of students 3 5 8 10 5 5 Solution: We write the distribution in cumulative frequency table. Marks No. of students (f) Cumulative frequency 35 3 3 45 5 8 50 8 16 64 10 26...

Wages per day in Rs. 38 45 48 55 62 65 No. of workers 14 8 7 10 6 2 Solution: We write the distribution in cumulative frequency table. Wages per day in Rs. No. of workers (f) Cumulative frequency 38...

Solution: Observation are as follows : 11, 12, 14, (x-2), (x+4), (x+9), 32, 38, 47 n = 9 Here n is odd. So median = ((n+1)/2)th term = (9+1)/2 )th term = 5th term = x+4 Given median = 24 x+4 = 24 x...

Solution: Arranging the numbers in ascending order : 0, 1, 1, 2, 2, 3, 3, 3, 4, 5 Here, n = 10 which is even Median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (10/2 th term + ((10/2)+1)th term) = ½...

Solution: (a) Arranging the numbers in ascending order : 0, 1, 2, 2, 3, 4, 5, 5, 5, 7, 8, 9 Here, n = 12 which is even Median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (12/2 th term +...

Solution: Arranging the data in the ascending order 1,2,2,3,4,5,5,6,7,7,8 Here number of terms, n = 11 Here n is odd. So median = [(n+1)/2 ]th observation = (11+1)/2 = 12/2 = 6th observation Here...

Solution: Class mark, xi = (upper class limit + lower class limit)/2 Assumed mean, A = 45 Class interval frequency (fi) Class mark (xi) di = xi – A fidi 20-30 3 25 -20 -60 30-40 5 35 -10 -50 40-50...

Life time in days No. of tubes Less than 50 8 Less than 100 23 Less than 150 55 Less than 200 81 Less than 250 93 Less than 300 100 Find the mean life time of electricity tubes. Solution: Class mark...

Class intervals 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 7 P 12 q 8 5 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class intervals Frequency fi Class mark xi fixi 0-20...

Class intervals 0-20 20-40 40-60 60-80 80-100 Frequency 17 P 32 q 19 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class intervals Frequency fi Class mark xi fixi 0-20 17 10...

Daily pocket allowance in Rs. 11-13 13-15 15-17 17-19 19-21 21-23 23-25 No. of children 3 6 9 13 f 5 4 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Daily pocket allowance in...

Class intervals 0-8 8-16 16-24 24-32 32-40 40-48 Frequency 5 3 10 P 4 2 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class intervals Frequency fi Class mark xi fixi 0-8 5 4...

No. of days 0-6 6-10 10-14 14-20 20-28 28-38 38-40 No. of students 11 10 7 4 4 3 1 Solution: Class mark, xi = (upper class limit + lower class limit)/2 No. of days Frequency fi Class mark xi fixi...

Marks 11-20 21-30 31-40 41-50 51-60 61-70 71-80 No. of students 2 6 10 12 9 7 4 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Assumed mean, A = 45.5 Marks No. of students (fi)...

Wages in Rs. 45-50 50-55 55-60 60-65 65-70 70-75 75-80 No. of workers 5 8 30 25 14 12 6 Calculate their mean by short cut method. Solution: Class mark, xi = (upper class limit + lower class limit)/2...

Marks 0-10 10-20 20-30 30-40 40-50 50-60 No. of students 10 9 25 30 16 10 Solution: Class mark (xi) = (upper limit + lower limit)/2 Let assumed mean (A) = 25 Class size (h) = 10 Class Interval No....

Class interval 0-10 10-20 20-30 30-40 40-50 Frequency 10 6 8 12 5 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class interval Frequency fi Class mark xi fixi 0-10 10 5 50...

Variate (xi) 5 7 9 11 _ 15 20 Frequency (fi) 4 4 4 7 3 2 1 Solution: Let the missing variate be x. Variate (xi) Frequency (fi) fixi 5 4 20 7 4 28 9 4 36 11 7 77 x 3 3x 15 2 30 20 1 20 Total Ʃfi =25...

Variate 5 6 7 8 9 10 11 12 Frequency 20 17 f 10 8 6 7 6 Solution: Variate (x) Frequency (f) fx 5 20 100 6 17 102 7 f 7f 8 10 80 9 8 72 10 6 60 11 7 77 12 6 72 Total Ʃf = 74+f Ʃfx = 563+7f Given mean...

Number 5 10 15 20 25 30 35 Frequency 1 2 5 6 3 2 1 Solution: Number (x) Frequency (f) fx 5 1 5×1 = 5 10 2 10×2 = 20 15 5 15×5 = 75 20 6 20×6 = 120 25 3 25×3 = 75 30 2 30×2 = 60 35 1 35×1 = 35 Total...