Example 3 - Number of wickets taken by bowlers in one-day

Example 3 - Chapter 14 Class 10 Statistics - Part 2

Example 3 - Chapter 14 Class 10 Statistics - Part 3
Example 3 - Chapter 14 Class 10 Statistics - Part 4

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Example 3 (Method 1 – Direct Method) The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify? Here Class Size is not same, So, we solve by Direct Method Mean(π‘₯ Μ…) = (βˆ‘β–’π‘“π‘–π‘₯𝑖)/(βˆ‘β–’π‘“π‘–) π‘₯ Μ… = 6880/45 π‘₯ Μ… = 152.89 Thus, Mean signifies that on average, 45 bowlers take 152.89 wickets Example 3 (Method 2 – Step Deviation Method) The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify? Here Class Size is not same… So, in Step Deviation Method Mean(π‘₯ Μ…) = a + h Γ— (βˆ‘β–’π’‡π’Šπ’–π’Š)/(βˆ‘β–’π’‡π’Š) We use a value of h which can divide 𝑑𝑖 = π‘₯𝑖 βˆ’ a Mean(π‘₯ Μ…) = a + h Γ— (βˆ‘β–’π’‡π’Šπ’–π’Š)/(βˆ‘β–’π’‡π’Š) Where a = Assumed Mean Let h = Class interval Also, βˆ‘β–’π’‡π’Š = 45 βˆ‘β–’π’‡π’Šπ’–π’Š = βˆ’212 Putting values in formula Mean(𝒙 Μ…) = a + h Γ— (βˆ‘β–’π’‡π’Šπ’–π’Š)/(βˆ‘β–’π’‡π’Š) π‘₯ Μ… = 200 + 10 Γ— (βˆ’212)/45 π‘₯ Μ… = 200 – 47.11 𝒙 Μ… = 152.89 Thus, Mean signifies that on average, 45 bowlers take 152.89 wickets

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.