Miscellaneous

Chapter 13 Class 11 Statistics
Serial order wise

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### Transcript

Misc 5 The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) If wrong item is omitted. Given that number of observations (n) = 20 Incorrect mean ( 𝑥﷯) = 10, Incorrect standard deviation (σ) = 2 We know that 𝑥﷯ = 1﷮𝑛﷯ 10 = 1﷮20﷯ 10 × 20 = 200 = = 200 ∴ Incorrect sum of observations = 200 Finding correct sum of observations , incorrect observation 8 is removed So, Correct sum of observations = Incorrect sum – 8 = 200 – 8 = 192 Hence, Correct mean = Correct sum of observations﷮Number of observations ﷯ = 192﷮20 − 𝟏﷯ = 192﷮𝟏𝟗﷯ = 10.1 Now, Incorrect Standard deviation (σ) = 1﷮N﷯ ﷮N ×(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) – 𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi﷯﷯2﷯﷯ 2 = 1﷮20﷯ ﷮20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) – 𝑰𝒏𝒄𝒐𝒓𝒓𝒆𝒄𝒕 ﷮﷮xi﷯﷯2﷯﷯ 2 × 20 = ﷮20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) – 𝟐𝟎𝟎﷯2﷯﷯ 40 = ﷮20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) – 200﷯2﷯﷯ Squaring both sides (40)2 = ﷮20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 )– 200﷯2﷯﷯﷯﷮2﷯ 1600 = 20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) – 200﷯2﷯ 1600 = 20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) –﷯ 40000 1600 + 40000 = 20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) ﷯ 41600 = 20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) ﷯ 20(𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) ﷯ = 41600 𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ﷯ = 1﷮20﷯ × 41600 𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ﷯ = 2080 Since 8 is removed So, (𝐶𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) ﷯ = 2080 – (8)2 = 2080 – 64 = 2016 Now, Correct Standard deviation (σ) = 1﷮N﷯ ﷮N ×(𝐶𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 )– 𝐶𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi﷯﷯2﷯﷯ = 1﷮𝟏𝟗﷯ ﷮𝟏𝟗 2016﷯ − 192﷯2﷯ = 1﷮19﷯ ﷮38304−36864﷯ = 1﷮19﷯ ﷮1440﷯ = 1﷮19﷯ ﷮144 ×10﷯ = 1﷮19﷯ ﷮144﷯ × ﷮10﷯ = 1﷮19﷯ ﷮ 12﷮2﷯﷯ × ﷮10﷯ = 1﷮19﷯ × 12 × ﷮𝟏𝟎﷯ = 1﷮19﷯ × 12 × 3.162 = 1.99 Hence, correct Standard deviation = 1.99 Misc 5 The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (ii) If it is replaced by 12. From part (i), Incorrect sum of observations = 200 Finding correct sum of observations , incorrect observation 8 is removed & 12 is added So, Correct sum of observations = Incorrect sum – 8 + 12 = 200 – 8 + 12 = 204 Hence, Correct mean = Correct sum of observations﷮Number of observations ﷯ = 204﷮20﷯ = 10.2 Finding correct Standard deviation From last part 𝐼𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ﷯ = 2080 Since 8 is removed & 12 is added So, (𝐶𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 ) ﷯ = 2080 – (8)2 + (12)2 = 2080 – 64 + 144 = 2160 Now, Correct Standard deviation (σ) = 1﷮N﷯ ﷮N ×(𝐶𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi2 )– 𝐶𝑜𝑟𝑟𝑒𝑐𝑡 ﷮﷮xi﷯﷯2﷯﷯ = 1﷮20﷯ ﷮20 2160﷯ − 204﷯2﷯ = 1﷮20﷯ ﷮43200 −41616﷯ = 1﷮20﷯ ﷮1584﷯ = 1﷮20﷯ × 39.799 = 1.98 Hence, Correct Standard deviation = 1.98

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.