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Question
For a frequency distribution, the mean is 200, the coefficient of variation is 8% and Karl Pearsonian’s coefficient of skewness is 0.3. Find the mode and median of the distribution.
Solution
Mean = `bar"x"` = 200, Coefficient of variation,
C.V. = 8%, Skp = 0.3
C.V. = `sigma/bar"x" xx 100`, where σ = standard deviation
∴ 8 = `sigma/200 xx 100`
∴ `sigma = (8 xx 200)/100` = 16
Now, Skp = `("Mean"-"Mode")/"S.D."`
∴ 0.3 = `(200 - "Mode")/16`
∴ 0.3 × 16 = 200 – Mode
∴ Mode = 200 – 4.8 = 195.2
Since, Mean – Mode = 3(mean – Median)
∴ 200 – 195.2 = 3(200 – Median)
∴ 4.8 = 600 – 3 Median
∴ 3 Median = 600 – 4.8 = 595.2
∴ Median = `(595.2)/3` = 198.4
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