If for a group of 20 items, ∑x = 1452, ∑x² = 144280 and mode = 63.7, then Pearsonian coefficient of skewness is equal to:
0.202
Given: n = 20, \(\sum x = 1452\), \(\sum x^2 = 144280\), Mode = 63.7.
Step 1 — Mean: \(\bar{x} = \dfrac{\sum x}{n} = \dfrac{1452}{20} = 72.6\).
Step 2 — Standard deviation: \(\sigma = \sqrt{\dfrac{\sum x^2}{n} - \bar{x}^2} = \sqrt{\dfrac{144280}{20} - (72.6)^2} = \sqrt{7214 - 5270.76} = \sqrt{1943.24} \approx 44.08\).
Step 3 — Karl Pearson's coefficient of skewness (since mode is given): \(S_k = \dfrac{\text{Mean} - \text{Mode}}{\sigma} = \dfrac{72.6 - 63.7}{44.08} = \dfrac{8.9}{44.08} \approx 0.202\).
Hence, the Pearsonian coefficient of skewness is 0.202.
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Important Questions from Measures of skewness
Which one of the following statements is true?
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