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Question
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 the observations be x1, x2, x3, x4, x5 and x6
It is given that mean is 8 and the standard deviation is 4.
⇒ Mean = `overline x = (x_1 + x_2 + x_3 + x_4 + x_5 + x_6)/6 = 8` ....(i)
If each observation is multiplied by 3 and the resulting observation are yi, then
yi = 3xi, for i = 1 to 6
∴ New mean `overline y = (y_1 + y_2 + y_3 + y_4 + y_5 + y_6)/6`
= `(3(x_1 + x_2 + x_3 + x_4 + x_5 + x_6))/6`
= 3 × 8 .....[Using (i)]
= 24
Standard deviation σ = `sqrt(1/n sum_(i = 1)^6 (x_ i - overline x)^2)`
∴ `(4)^2 = 1/6 sum_(i = 1)^6(x_ i - overline x)^2`
`sum_(i = 1)^6(x_ i - overline x)^2 = 96` ...(ii)
From (i) and (ii) it can be observed that
`overline y = 3overline x`
`overline x = 1/3 overliney`
Substituting the values of xi and `overline x` in (ii) we obtain
`sum_(i = 1)^6(1/3 y_i - 1/3 overline y)^2 = 96`
⇒ `sum_(i = 1)^6 (y - overline y)^2 = 864`
Therefore, variance of new observation = `(1/6 xx 864) = 144`
Hence, the standard deviation of new observation is `sqrt144 = 12`
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