Daily wages in Rs 0-10 10-20 20-30 30-40 40-50 50-60 No. of employees 1 8 10 5 4 2 Estimate the modal daily wages for this distribution by a graphical method. Solution: Daily wages in Rs. No. of...
Solution: Here n = 10, which is even Median = 25 So Median = ½ ( n/2 th term + ((n/2)+1)th term) 25 = ½ (( 10/2 )th term + (10/2)+1)th term) 25 = ½ (( 5 )th term + (6)th term) 25 = ½ (x+2 + x+4) 25...
Class 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 5 8 p 12 7 8 Solution: Class Frequency fi Class mark xi fixi 0-20 5 10 50 20-40 8 30 240 40-60 p 50 50p 60-80 12 70 840 80-100 7 90 630 100-120...
Age in years 25-29 30-34 35-39 40-44 45-49 50-54 55-59 No. of persons 4 14 22 16 6 5 3 Solution: The given distribution is not continuous. Adjustment factor = (30-29)/2 = ½ = 0.5 We subtract 0.5...
Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No. of students 3 7 12 17 23 14 9 6 5 4 Draw an ogive on a graph sheet and from it determine the : (i) median (ii) lower quartile ...
Weight (gm) 50-60 60-70 70-80 80-90 90-100 100-110 110-120 120-130 Frequency 8 10 12 16 18 14 12 10 (i) Calculate the cumulative frequencies. (ii) Draw the cumulative frequency curve and from it...
Wt. in kg 40-44 45-49 50-54 55-59 60-64 65-69 No. of students 2 8 12 10 6 4 Hence estimate the modal weight. Solution: The given distribution is not continuous. Adjustment factor = (45-44)/2 = ½ =...
Score 0 1 2 3 4 5 No. of shots 0 3 6 4 7 5 (i) What was his modal score ? (ii) What was his median score ? (iii) What was his total score ? (iv) What was his mean ? Solution: We write the marks in...
Solution: Given data is 0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8 Number of observations, n = 16 Mean = Ʃxi/n = (0+0+2+2+3+3+3+4+5+5+5+5+6+6+7+8)/16 = 64/16 = 4 Hence the mean is 4. Here n = 16...
Class intervals 0-50 50-100 100-150 150-200 200-250 250-300 frequency 4 8 16 13 6 3 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class interval Frequency fi Class mark xi...
Class interval 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 8 5 12 35 24 16 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class interval Frequency fi Class mark xi fixi 0-10 8...
Marks 5 6 7 8 9 No. of students 6 a 16 13 b If the mean of the distribution is 7.2, find a and b. Solution: Marks (x) No. of students (f) fx 5 6 30 6 a 6a 7 16 112 8 13 104 9 b 9b Total Ʃf = 35+a+b...
Category A B C D E F G Wages (in Rs) per day 50 60 70 80 90 100 110 No. of workers 2 4 8 12 10 6 8 Category Wages in Rs. (x) No. of workers f fx A 50 2 100 B 60 4 240 C 70 8 560 D 80 12 960 E 90 10...
Numbers 60 61 62 63 64 65 66 Cumulative frequency 8 18 33 40 49 55 60 Solution: Numbers (x) Cumulative frequency Frequency (f) fx 60 8 8 60×8 = 480 61 18 18-8 = 10 61×10 = 610 62 33 33-18 = 15 62×15...
No. of matches 35 36 37 38 39 40 41 No. of boxes 6 10 18 25 21 12 8 (i) Calculate, correct to one decimal place, the mean number of matches per box. (ii) Determine how many extra matches would have...
Solution: Mean of 10 numbers = 13 Sum of numbers = 13×10 = 130 Mean of remaining 15 numbers = 18 Sum of numbers = 15×18 = 270 Sum of all numbers = 130+270 = 400 Mean = sum of numbers/25 = 400/25 =...
Solution: (a)Given mean age = 13 Number of students = 33 Sum of ages = mean ×number of students = 13×33 = 429 After a girl leaves, the mean of 32 students becomes = 207/16 Now sum of ages =...
Solution: (a) Given observations are 5.7, 6.6, 7.2, 9.3, 6.2. Number of observations = 5 Mean = sum of observations / number of observations Mean = (5.7+6.6+7.2+9.3+6.2)/5 = 35/5 = 7 Hence the mean...
Solution: In a circle with center O and radius 3cm and p is at a distance of 5cm. That is OT = 3 cm, OP = 5 cm OT is the radius of the circle OT ⊥ PT Now in right ∆ OTP, by Pythagoras axiom, OP2 =...
Solution: Given height of the cone, h = 24 cm Radius, r = 7 cm We know that l2 = h2+r2 l2 = 242+72 l2 = 576+49 l2 = 625 l = √625 l = 25 Curved surface area = rl = (22/7)×7×25 = 22×25 = 550 cm2 So...