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NCERT Class XI Mathematics - Statistics - Solutions

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Question : 20 of 34
Marks: +1, -0
Classes 0 - 1010 - 2020 - 30 30 - 40 40 - 50
Frequencies 58 15 16 6
Solution:  
Classes Mid values
xix_i		 fif_i uiu_i = xi−2510\frac{x_i-25}{10} fiuif_i u_i fiui2f_i u_i^2
0 - 10 5 5 - 2 - 10 20
10 - 20 15 8 - 1 - 8 8
20 - 30 25 15 0 0 0
30 - 40 35 16 1 16 16
40 - 50 45 6 2 12 24
50 10 68
Let assumed mean (A) = 25
Mean (x−)\left(\overset{-}{x}\right) = A + ΣfiuiN\frac{\Sigma f_i u_i}{N} × h = 25 + 1050\frac{10}{50} × 10 = 25 + 2 = 27
Variance (σ2)(\sigma^2) = h2N2[NΣfiui2−(Σfiui)2]\frac{h^2}{N^2} [N\Sigma f_i u_i^2 - (\Sigma f_i u_i)^2] = (10)2(50)2\frac{(10)^2}{(50)^2} [50×68−(10)2][50 \times 68 - (10)^2]
= 1002500\frac{100}{2500} [3400 - 100] = 125\frac{1}{25} × 3300 = 132
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