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NCERT Class XI Mathematics - Statistics - Solutions

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Question : 30 of 34
Marks: +1, -0
The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
Solution:  
Let x1,x2,x3,x4,x5,x6x_1,x_2,x_3,x_4,x_5,x_6 be the six observations, then
x1+x2+x3+x4+x5+x66\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6} = 8
⇒ x1+x2+x3+x4+x5+x6x_1+x_2+x_3+x_4+x_5+x_6 = 48
Now if each observation is multiplied by 3 then
New mean = 3x1+3x2+3x3+3x4+3x5+3x66\frac{3x_1+3x_2+3x_3+3x_4+3x_5+3x_6}{6}
= 3(x1+x2+x3+x4+x5+x6)6\frac{3(x_1+x_2+x_3+x_4+x_5+x_6)}{6} = 12\frac{1}{2} × 48 = 24
Also ,
16(x12+x22+x32+x42+x52+x62)−(8)2\frac{1}{6}(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2)-(8)^2
= 16
⇒ x12+x22+x32+x42+x52+x62x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2 = 480
If each observation is multiplied by 3, then
New variance
=
16(9x12+9x22+9x32+9x42+9x52+9x62)−(24)2\frac{1}{6}(9x_1^2+9x_2^2+9x_3^2+9x_4^2+9x_5^2+9x_6^2)-(24)^2
= 96\frac{9}{6} × 480 - 576 = 720 - 576 = 144
∴ New S.D. = 144\sqrt{144} = 12
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