STANDARD DEVIATION (2018) (STATISTICS)
- 1. STANDARD DEVIATION SUMAN MATHEWS GRADE 11 MATH COLLEGE MATH STATISTICS 𝜎
- 2. EXPERIENCE: PREVIEW: WE DISCUSS THE DIRECT METHOD, SHORT CUT METHOD AND STEP DEVIATION METHOD TO CALCULATE STANDARD DEVIATION. ALL THE 3 TYPES OF PROBLEMS ARE DISCUSSED WITH FORMULAE . SO TAKE YOUR PAPER AND PEN AND WORK THROUGH THE VIDEO WITH ME. BONUS: FORMULAE OF MEAN USING STEP DEVIATION METHOD ALSO GIVEN
- 3. THE DEGREE TO WHICH NUMERICAL DATA IS SPREAD OVER AN AVERAGE VALUE IS CALLED DISPERSION OF DATA. STANDARD DEVIATION IS THE MOST IMP0RTANT MEASURE OF DISPERSION AND IS WIDELY USED IN MANY STATISTICAL FORMULAE.
- 4. 𝜎2 = 𝐼=1 𝑛 (𝑥𝑖 − 𝑥)2 𝑛 𝜎2 = 𝑥𝑖 2 𝑛 − 𝑥2 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = 𝜎 = 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
- 5. Question 1 ( standard deviation) Find the standard deviation of the following data 6,8, 10, 12, 14, 16, 18,20, 22, 24 x 𝒙 𝟐 6 36 8 64 10 100 12 144 14 196 16 256 18 324 20 400 22 484 24 576 150 2580 𝜎2 = 2580 10 − (15)2 = 258-225=33 𝜎 = 33 = 5.74
- 6. QUESTION 2 Find the mean and standard deviation for the following data 𝑥𝑖 𝑓𝑖 𝑓𝑖 𝑥𝑖 𝑓𝑖 𝑥𝑖 2 6 2 12 72 10 4 40 400 14 7 98 1372 18 12 216 3888 24 28 192 4608 28 4 112 3136 30 3 90 2700 40 760 16176 𝜎 = 𝑓𝑖 𝑥𝑖 2 𝑓𝑖 − ( 𝑓𝑖 𝑥𝑖 𝑓𝑖 )2 = 16176 40 − ( 760 40 )2 For a frequency distribution, 𝜎 = 𝑓𝑖(𝑥𝑖 − 𝑥)2 𝑓𝑖 Direct method
- 7. = 404.4 − 192 = 43.4 = 6.59𝜎 𝑥 = 𝑓𝑖 𝑥𝑖 𝑓𝑖 = 19
- 8. Question 3 Find the standard deviation for the following Class interval 25-35 35-45 45-55 55-65 65-75 frequency 64 132 153 140 51 We take the midpoint of the class interval as the x values We use the short cut method. Let A be the assumed mean 𝑑𝑖 = 𝑥𝑖 − 𝐴 Statistics- grade 11 and college level
- 9. 𝜎 = 𝑓𝑖 𝑑𝑖 2 𝑓𝑖 − ( 𝑓𝑖 𝑑𝑖 𝑓𝑖 )2 C I f 𝒙𝒊 𝒅𝒊 = 𝒙𝒊 − 𝟓𝟎 𝒇𝒊 𝒅𝒊 𝒇𝒊 𝒅𝒊 𝟐 25-35 64 30 -20 -1280 25600 35-45 132 40 -10 -1320 13200 45-55 153 50 0 0 0 55-65 140 60 10 1400 14000 55-65 51 70 20 1020 20400 540 -180 73200 = 73200 540 − (− 180 540 )2 = 135.55 − 0.111 = 135.44 = 11.64 TAKE A= 50
- 10. QUESTION 4 Calculate the mean, variance and standard deviation of the following data using step deviation method C I 30-40 40-50 50-60 60-70 70-80 80-90 90-100 f 3 7 12 15 8 3 2 𝑝𝑢𝑡 𝑡𝑖 = 𝑥𝑖 − 𝐴 𝑐 A is the assumed mean, c is the width of the class interval
- 11. C I f 𝒙𝒊 𝒕𝒊 = 𝒙𝒊 − 𝟔𝟓 𝟏𝟎 𝒇𝒊 𝒕𝒊 𝒇𝒊 𝒕𝒊 𝟐 30-40 3 35 -3 -9 27 40-50 7 45 -2 -14 28 50-60 12 55 -1 -12 12 60-70 15 65 0 0 0 70-80 8 75 1 8 8 80-90 3 85 2 6 12 90-100 2 95 3 6 18 50 -15 105 𝜎 = 𝑐 𝑓𝑖 𝑡𝑖 2 𝑓𝑖 − ( 𝑓𝑖 𝑡𝑖 𝑓𝑖 )2 = 10 105 50 − ( −15 50 )2 = 14.18 Variance = 201 Put A = 65
- 12. 𝑚𝑒𝑎𝑛 = 𝐴 + 𝑐 𝑓𝑖 𝑡𝑖 𝑓𝑖 = 65 + 10 −15 50 = 62
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