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Ex 15.1, 3 - Find mean deviation about median - Class 11 - Mean deviation about median - Ungrouped

Ex 15.1, 3 - Chapter 15 Class 11 Statistics - Part 2


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Ex 15.1, 3 Find the mean deviation about the median for the data. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17 Arranging data in ascending order, 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18 Here, n = number of observations = 12 (even) Since n is even Median = 𝑛īˇŽ2īˇ¯īˇ¯īˇŽđ‘Ąâ„Žīˇ¯đ‘œđ‘đ‘ đ‘’đ‘Ÿđ‘Ŗ𝑎𝑡𝑖𝑜𝑛 + 𝑛īˇŽ2īˇ¯ + 1īˇ¯īˇŽđ‘Ąâ„Žīˇ¯đ‘œđ‘đ‘ đ‘’đ‘Ÿđ‘Ŗ𝑎𝑡𝑖𝑜𝑛īˇŽ2īˇ¯ M = 12īˇŽ2īˇ¯īˇ¯īˇŽđ‘Ąâ„Žīˇ¯đ‘œđ‘đ‘ đ‘’đ‘Ÿđ‘Ŗ𝑎𝑡𝑖𝑜𝑛 + 12īˇŽ2īˇ¯ + 1īˇ¯īˇŽđ‘Ąâ„Žīˇ¯đ‘œđ‘đ‘ đ‘’đ‘Ÿđ‘Ŗ𝑎𝑡𝑖𝑜𝑛īˇŽ2īˇ¯ M = 6īˇŽđ‘Ąâ„Žīˇ¯đ‘œđ‘đ‘ đ‘’đ‘Ÿđ‘Ŗ𝑎𝑡𝑖𝑜𝑛 + 7īˇŽđ‘Ąâ„Žīˇ¯đ‘œđ‘đ‘ đ‘’đ‘Ÿđ‘Ŗ𝑎𝑡𝑖𝑜𝑛īˇŽ2īˇ¯ M = 13 + 14īˇŽ2īˇ¯ = 27īˇŽ2īˇ¯ = 13.5 Thus, for 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18 Median = 13.5 Mean deviation about median = īˇŽīˇŽ đ‘Ĩ𝑖 −Mīˇ¯īˇ¯īˇŽ12īˇ¯ = 10 − 13.5īˇ¯ + 11 − 13.5īˇ¯ + 11 − 13.5īˇ¯ + 12 − 13.5īˇ¯ + 13 − 13.5īˇ¯+ 13 − 13.5īˇ¯īˇŽ+ 14 − 13.5īˇ¯ + 16 − 13.5īˇ¯ + 16 − 13.5īˇ¯ + 17 − 13.5īˇ¯ + 17 − 13.5īˇ¯+ 18 − 13.5īˇ¯īˇ¯īˇ¯īˇŽ12īˇ¯ = −3.5īˇ¯ + −2.5īˇ¯ + −2.5īˇ¯ + −1.5īˇ¯ + −0.5īˇ¯ + −0.5īˇ¯īˇŽ+ 0.5īˇ¯ + 2.5īˇ¯ + 2.5īˇ¯+ 3.5īˇ¯ + 3.5īˇ¯ + 4.5īˇ¯ īˇ¯īˇ¯īˇŽ12īˇ¯ = 3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5īˇŽ12īˇ¯ = 28īˇŽ12īˇ¯ = 2.33

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.