NCERT Solutions
Class 11 Maths
Sequences and Series

Ex.Misc.Q.3
The mean and standard deviation of six observations is 8 and 4, respectively.
If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
Let the observations be x1, x2, x3, x4, x5, and x6.
It is given that mean is 8 and standard deviation is 4.
Mean, = (x1 + x2 + x3 + x4 + x5 + x6) ÷ 6 = 8 ……………… (1)
If each observation is multiplied by 3 and the resulting observations are yi, then
yi = 3xi
i.e., xi = yi ÷ 3 for i = 1 to 6
So, New Mean, = (y1 + y2 + y3 + y4 + y5 + y6) ÷ 6 = 8
= 3(x1 + x2 + x3 + x4 + x5 + x6) ÷ 6
= 3 × 8
= 24
Standard deviation σ = √ [(1 ÷ n) × Σi=16 ( – x)2]
=> 42 = Σi=16 (xi – )2 ÷ 6
=> Σi=16 (xi – )2 = 96 …………. (2)
From equation (1) and (2), it can be observed that,
= 3
=> =
÷ 3
Substituting the values of xi and x in equation (2), we obtain
=> Σi=16 (( ÷ 3) – (y ÷ 3))2 = 96
=> Σi=16 (yi – y )2 = 864
Therefore, variance of new observations = 864 ÷ 6 = 144
Hence, the standard deviation of new observations is √144 = 12