Example 3 (Method 1 β 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 ) = (ββπππ₯π)/(ββππ)
π₯ Μ = 6880/45
π₯ Μ = 152.89
Thus, Mean signifies that on average, 45 bowlers take 152.89 wickets
Example 3 (Method 2 β 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 sameβ¦
So, in Step Deviation Method
Mean(π₯ Μ ) = a + h Γ (ββππππ)/(ββππ)
We use a value of h which can divide
ππ = π₯π β a
Mean(π₯ Μ ) = a + h Γ (ββππππ)/(ββππ)
Where
a = Assumed Mean
Let h = Class interval
Also,
ββππ = 45
ββππππ = β212
Putting values in formula
Mean(π Μ ) = a + h Γ (ββππππ)/(ββππ)
π₯ Μ = 200 + 10 Γ (β212)/45
π₯ Μ = 200 β 47.11
π Μ = 152.89
Thus, Mean signifies that on average, 45 bowlers take 152.89 wickets
Made by
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 and Computer Science at Teachoo
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