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

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  1. Chapter 15 Class 11 Statistics
  2. Serial order wise
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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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