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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 = đ‘›īˇŽ2īˇ¯īˇ¯īˇŽđ‘Ąâ„Žīˇ¯đ‘œđ‘đ‘ đ‘’đ‘Ÿđ‘Ŗđ‘Žđ‘Ąđ‘–đ‘œđ‘› + đ‘›īˇŽ2 + 1īˇ¯īˇŽđ‘Ąâ„Žīˇ¯đ‘œđ‘đ‘ đ‘’đ‘Ÿđ‘Ŗđ‘Žđ‘Ąđ‘–đ‘œđ‘›īˇŽ2 M = 12īˇŽ2īˇ¯īˇ¯īˇŽđ‘Ąâ„Žīˇ¯đ‘œđ‘đ‘ đ‘’đ‘Ÿđ‘Ŗđ‘Žđ‘Ąđ‘–đ‘œđ‘› + 12īˇŽ2 + 1īˇ¯īˇŽđ‘Ąâ„Žīˇ¯đ‘œđ‘đ‘ đ‘’đ‘Ÿđ‘Ŗđ‘Žđ‘Ąđ‘–đ‘œđ‘›īˇŽ2 M = 6īˇŽđ‘Ąâ„Žīˇ¯đ‘œđ‘đ‘ đ‘’đ‘Ÿđ‘Ŗđ‘Žđ‘Ąđ‘–đ‘œđ‘› + 7īˇŽđ‘Ąâ„Žīˇ¯đ‘œđ‘đ‘ đ‘’đ‘Ÿđ‘Ŗđ‘Žđ‘Ąđ‘–đ‘œđ‘›īˇŽ2 M = 13 + 14īˇŽ2 = 27īˇŽ2 = 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¯īˇŽ12 = 10 − 13.5 + 11 − 13.5 + 11 − 13.5 + 12 − 13.5 + 13 − 13.5+ 13 − 13.5īˇ¯īˇŽ+ 14 − 13.5 + 16 − 13.5 + 16 − 13.5 + 17 − 13.5 + 17 − 13.5+ 18 − 13.5īˇ¯īˇ¯īˇ¯īˇŽ12 = −3.5 + −2.5 + −2.5 + −1.5 + −0.5 + −0.5īˇ¯īˇŽ+ 0.5 + 2.5 + 2.5+ 3.5 + 3.5 + 4.5 īˇ¯īˇ¯īˇŽ12 = 3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5īˇŽ12 = 28īˇŽ12 = 2.33

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