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Exercises · Q13
Q.

Find the standard deviation of the following continuous frequency distribution by the step-deviation method:

Class0–2020–4040–6060–8080–100
ff461064
Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Take class mid-points xx, assumed mean A=50A = 50, class width h=20h = 20, d=x−5020d = \dfrac{x - 50}{20}.

Classxxffddfdfdfd2fd^2
0–20104−2-2−8-816
20–40306−1-1−6-66
40–605010000
60–80706166
80–1009042816
Total30044

Here N=30N = 30, ∑fd=0\sum fd = 0, ∑fd2=44\sum fd^2 = 44.

Standard deviation:

σ=h∑fd2N−(∑fdN)2=204430−0=201.4667≈20×1.2111=24.22.\sigma = h\sqrt{\frac{\sum fd^2}{N} - \left(\frac{\sum fd}{N}\right)^2} = 20\sqrt{\frac{44}{30} - 0} = 20\sqrt{1.4667} \approx 20 \times 1.2111 = 24.22. …

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