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NCERT Exemplar · Q10
Q.

For the frequency distribution:

xx234567
ff491614116

Find the standard distribution.

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For this frequency distribution the mean is xˉ=4.617\bar{x}=4.617 and the standard deviation is σ≈1.38\sigma\approx 1.38.

Formula

σ=∑fixi2N−(∑fixiN)2,N=∑fi.\sigma=\sqrt{\frac{\sum f_i x_i^{2}}{N}-\left(\frac{\sum f_i x_i}{N}\right)^{2}},\qquad N=\sum f_i.

Step 1 — Build the fxfx and fx2fx^{2} columns

xxfffxfxfx2fx^{2}
24816
392781
41664256
51470350
61166396
7642294
Total602771393

Step 2 — Mean

xˉ=∑fixiN=27760=4.6167.\bar{x}=\frac{\sum f_i x_i}{N}=\frac{277}{60}=4.6167.

Step 3 — Variance

σ2=139360−(27760)2=23.2167−21.3136=1.9031.\sigma^{2}=\frac{1393}{60}-\left(\frac{277}{60}\right)^{2}=23.2167-21.3136=1.9031. …

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