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- 2. Calculate the arithmetic of the following data 3, 4, 5, 9, 3, 4, 7 Answer is 5 Arithmetic mean is defined as the sum of all the observations in the distribution divided by the number of observations
- 3. 1. Calculate arithmetic mean of weights of 25 children Weight in kg(x) 15 16 17 18 19 20 total No. of children(f) 3 4 4 6 5 3 25 fx 45 64 68 108 95 60 440
- 4. Summation fx divided by summation f 440/25 Answer is 17.6
- 5. 3. Calculate the arithmetic mean of marks of 50 students Marks 10-20 20-30 30-40 40-50 50-60 60-70 Total No. of students(f) 5 3 7 15 10 10 Mid value(x) 15 25 35 45 55 65 fx 75 75 245 675 550 650
- 7. 4. Calculate the arithmetic mean of the monthly incomes of 50 families Monthly income (in Rs.) 10000-15000 15000-20000 20000-25000 25000-30000 30000-35000 No. of families 7 8 20 10 5
- 8. 5. Calculate arithmetic mean of following observations • 20, 30, 40, 30, 30, 40, 50, 40, 30, 70 • The Answer is 38
- 9. 6. Calculate the arithmetic mean of following observations, the answer is 35.2 X 30 32 34 36 38 40 f 6 8 8 12 10 6
- 10. Median • Median is defined as the value of the middle observation when the observations are arranged in the order of their magnitude • Median denoted by m
- 11. 7.The marks of 9 students are 7, 9,8,5, 7,7,6,8,9 • solution 5,6, 7,7,7,8, 8,9,9 N=9 M= =5 The value of 5th observation is 7 Median marks is 7
- 12. 8. The weights of 8 children are given as (kgs) 12, 11.5,13, 13.5, 10.5,14,12,15. find the median • Ascending order 10.5, 11.5, 12,12,13,13.5, 14,15 • N=8, N is even • M=N/2 • 8/2=4 • The average of 4th and 5th observation values is our median 12+13/2= 12.5 M=12.5
- 13. 9. Calculate the median weight of a group of children Weight in kgs( X) 30 31 32 33 34 35 36 37 Total No. of children (f) 8 12 15 25 20 12 5 2 99 Cumulative frequency 8 20 35 60 80 92 97 99
- 14. • N=99, 99+1/2 • 50 • Is between 36th to 60th observation • 33 • Median is 33
- 15. 10. Calculate median weight for the following group of persons Weight in kgs (x) 50-55 55-60 60-65 65-70 70-75 75-80 Total No. of persons (f) 8 10 25 35 15 7 100 Cumulative frequency (cf) 8 18 43 78 93 100
- 16. • N= 100, m=100/2=50 (65+7/35(5) 65+0.2(5) 65+1=66 Median weight is 66 kgs
- 17. 11. Calculate the median Age in years 20-25 25-30 30-35 35-40 40-45 45-50 50-55 55-60 No. of teachers 3 7 8 9 10 11 8 2
- 18. Mode- Z • It is the value of the series with maximum occurrence • It is the value of the variable with the highest repetition in the given series of data
- 19. 12 find out mode • 7, 8, 9, 7, 6, 7, 6, 6, 9, 8,7, 5,7,4 Ascending order- 4,5,6,6,6,7,7,7,7,7,8,8,9,9 From the above series of the variable provided the maximum occurrence is 7 Therefore Mode is 7
- 20. 13. Calculate the modal size of shoes Size of shoes 5 6 7 8 9 10 Number of pairs 48 52 56 50 47 48
- 21. • here the maximum frequency is 56 against the size 7. therefore modal size is 7.
- 22. Mode in a continuous distribution 𝑙1+ 𝑓1−𝑓0 (𝑓1−𝑓0)+(𝑓1−𝑓2) (𝑙2−𝑙1)
- 23. 14: Calculate the modal life Here the maximum frequency is 100, corresponding to the class interval 1200-1300 Therefore the modal class is 1200- 1300 1200+100-80/(100-80)+(100-60)*(1300- 1200) =1200+ 100/3 1233.33 Modal life of bulbs is 1233.33 hours Life in hours 1000- 1100 1100- 1200 1200- 1300 1300- 1400 1400- 1500 1500- 1600 No. of bulbs 40 80 100 60 60 50
- 24. 15. Lives of two models of refrigerators in a survey was found to be Life (No. of yrs) 0-2 2-4 4-6 6-8 8-10 10-12 Model A 5 16 13 7 5 4 Model B 2 7 12 19 9 1 • ANSWER - 5.12, 6.16 What is the average life of each model of these refrigerators?
- 25. Measures of dispersion • Dispersion mean differences or deviation spread over the certain values from the central value • Measure means method of ascertaining the values • Measures of dispersion means the various possible methods of measuring the dispersion or the differences of the different values in the series of data
- 26. Range • Range is the simplest method of calculating the dispersion or deviation. It is defined as the difference between the largest and the smallest values of the data • Range = Largest value – Smallest value (R=L-S) • Coefficient of range = largest value – smallest value/ largest value + smallest value
- 27. 16. Calculate the range for the following data giving the daily sales of a shop for a week sales in Rs. 160,130, 125, 127, 143, 150, 155 • L=160, S= 125 • RANGE= LARGEST VALUE – SMALLEST VALUE • 160 -125 =35 • COEFFICIENT OF RANGE= LARGEST VALUE – SMALLEST VALUE/ LARGEST VALUE + SMALLEST VALUE • 160-125/ 160 +125= 0.12 • COEFFICIENT OF RANGE IS 0.12
- 28. 17. The following are the marks obtained by the students. Calculate the range and the coefficient of the range No. of students Marks of student 1 20 2 25 3 80 4 30 5 90 6 45
- 29. • L= 90, S= 20 • RANGE= L-S, 90-20= 70 • COEFFICIENT OF RANGE= L-S/L+S • 90-20/90+20= 0.64 • Coefficient of range is 0.64
- 30. 18. Find the range and coefficient of range from the following data Weight No. of persons 95-105 20 105-115 25 115-125 80 125-135 30 135-145 90 145-155 45
- 33. 19. Calculate the standard deviation of the marks of 8 students Students no. 1 2 3 4 5 6 7 8 Marks (x) 4 5 7 7 8 8 8 9
- 34. Solution: Student no. Marks (x) 1 4 -3 (4-7) 9 2 5 -2 (5-7) 4 3 7 0 (7-7) 0 4 7 0 (7-7) 0 5 8 1 (8-7) 1 6 8 1 (8-7) 1 7 8 1 (8-7) 1 8 9 2 (9-7) 4 TOTAL 8 56 20
- 35. • N=8 • 56/8=7 • = • 1.58 • = 1.58
- 36. 20. Calculate the standard deviation for the following data Weights in kgs 3 4 5 6 7 No. of babies 10 25 30 25 10
- 37. Weights in kgs(x) No. of babies f fx f 3 10 30 -2 4 40 4 25 100 -1 1 25 5 30 150 0 0 0 6 25 150 1 1 25 7 10 70 2 4 40 100 500 130
- 38. • = =500/100= 5 • = • 1.14 kgs
- 39. 21. Calculate the standard deviation for the following data giving the salary of 1000 employees of a firm Salary in 1000 Rs. 10 12 14 16 18 20 No. of employees 18 25 20 15 12 10
- 40. Salary in 1000 Rs. (x) No. of employees f fx 10 18 180 1800 12 25 300 3600 14 20 280 3920 16 15 240 3840 18 12 216 3888 20 10 200 4000 TOTAL 1416 21048
- 41. • = • 1416/100= 14.16 • = 14.16
- 42. Correlation 22. Find the coefficient of correlation between advertising expenditure (1000 Rs.) and actual sales (in 1000 Rs.) given below Advertising expenditure 3 7 4 2 1 4 1 2 Sales 11 16 9 4 7 6 3 8
- 43. Advertising expenditure x Sales y ( ( 3 11 0 3 0 9 0 7 16 4 8 16 64 32 4 9 1 1 1 1 1 2 4 -1 -4 1 16 4 1 7 -2 -1 4 1 2 4 6 1 -2 1 4 -2 1 3 -2 -5 4 25 10 2 8 -1 0 1 0 0 24 64 0 0 28 120 47
- 44. • = 3, = 8 • = = 1.87 • = = 3.87 • r = = 47/8(1.87)(3.87) • 0.81 • The coefficient of correlation is 0.81 • There is very strong correlation between Advertising expenditure and sales • The value of r lies between -1 and +1 • If 0<r<1, the correlation is positive • If r=1, the correlation is positive perfect • If -1<r<0, the correlation is negative • If r= -1, the correlation is negative perfect • r is 0.3, 0.4 weak correlation, • 0.5, 0.6,0.7 strong correlation, • 0.8 very strong correlation • 0.9 extremely strong correlation
- 45. 23. Calculate coefficient of correlation from the following data X 12 9 8 10 11 13 7 Y 14 8 6 9 11 12 3 r is 0.95