Misc 3 - Mean, standard deviation of six observations are 8, 4 - Indirect questions - Multiplication of observation

Misc 3 - Chapter 15 Class 11 Statistics - Part 2
Misc 3 - Chapter 15 Class 11 Statistics - Part 3 Misc 3 - Chapter 15 Class 11 Statistics - Part 4

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Misc 3 The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations. Let the observations be 𝑥﷮1﷯, 𝑥﷮2﷯, 𝑥﷮3﷯, ..., 𝑥﷮6﷯ and 𝑥﷯ be their mean. Given that Mean = 𝑥﷯ = 8, Standard deviation = 4 If each observation is multiplied by 3 , we get new observations, Let the new observations be 𝑦﷮1﷯, 𝑦﷮2﷯, 𝑦﷮3﷯, ..., 𝑦﷮6﷯ where 𝑦﷮𝑖﷯ = 3( 𝑥﷮𝑖﷯) Calculating new mean New mean = 1﷮𝑛﷯ ﷮﷮ 𝑦﷮𝑖﷯﷯ 𝑦﷯ = 1﷮6﷯ ﷮﷮ 3𝑥﷮𝑖﷯﷯ 𝑦﷯ = 3 × 1﷮6﷯ ﷮﷮ 𝑥﷮𝑖﷯﷯ 𝑦﷯ = 3 𝑥﷯ 𝑦﷯ = 3 × 8 𝑦﷯ = 24 So, New Mean = 24 Calculating new standard deviation First we find variance of the new observations i.e. New Variance = 1﷮n﷯ ﷮﷮( 𝑦﷮𝑖﷯﷯− 𝑦﷯)﷮2﷯ Given Old Standard deviation = 4 So, Old Variance = 42 = 16 Now, Old Variance = 1﷮𝑛﷯ ﷮﷮( 𝑥﷮𝑖﷯﷯− 𝑥﷯)﷮2﷯ 16 = 1﷮6﷯ ﷮﷮( 𝑥﷮𝑖﷯﷯− 𝑥﷯)﷮2﷯ 16 × 6 = ﷮﷮( 𝑥﷮𝑖﷯﷯− 𝑥﷯)﷮2﷯ 96 = ﷮﷮( 𝑥﷮𝑖﷯﷯− 𝑥﷯)﷮2﷯ ﷮﷮( 𝑥﷮𝑖﷯﷯− 𝑥﷯)﷮2﷯ = 96 ﷮﷮( 1﷮3﷯ 𝑦﷮𝑖﷯﷯− 1﷮3﷯ 𝑦﷯)﷮2﷯ = 96 ﷮﷮( 1﷮3﷯ (𝑦﷮𝑖﷯﷯− 𝑦﷯))﷮2﷯ = 96 1﷮3﷯﷯﷮2﷯ ﷮﷮ ( 𝑦﷮𝑖﷯﷯− 𝑦﷯)﷮2﷯ = 96 1﷮9﷯﷯ ﷮﷮( 𝑦﷮𝑖﷯﷯− 𝑦﷯)﷮2﷯ = 96 ﷮﷮ ( 𝑦﷮𝑖﷯﷯− 𝑦﷯)﷮2﷯ = 96 × 9 ﷮﷮ ( 𝑦﷮𝑖﷯﷯− 𝑦﷯)﷮2﷯ = 864 So, New Variance = 1﷮𝑛﷯ ﷮﷮( 𝑦﷮𝑖﷯﷯− 𝑦﷯)﷮2﷯ = 1﷮6﷯ × 864 = 144 Hence, New standard deviation = ﷮𝑁𝑒𝑤 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒﷯ = ﷮144﷯ = 12

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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.