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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.
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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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