1. Tardigrade
  2. Question
  3. Mathematics
  4. Let x1, x2, ldots, xn be n observations. Let wi=lxi+k for i=1,2, ldots, n, where l and k are constants. If the mean of xi 's is 48 and their standard deviation is 12, the mean of wi 's is 55 and standard deviation of wi 's is 15 . The values of l and k should be

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Q. Let $x_1, x_2, \ldots, x_n$ be $n$ observations. Let $w_i=l_{x_i}+k$ for $i=1,2, \ldots, n$, where $l$ and $k$ are constants. If the mean of $x_i$ 's is 48 and their standard deviation is 12, the mean of $w_i$ 's is 55 and standard deviation of $w_i$ 's is 15 . The values of $l$ and $k$ should be

Statistics

Solution:

Given, $w_i=l x_i+k$
$M\left(x_j\right)=\bar{x}=48, \sigma\left(x_i\right)=12$
$M(\bar{w})=55$ and $\sigma(w)=15$
$M\left(w_i\right) =I M\left(x_i\right)+M(k) $
$55 =I \times 48+k ....$(i)
and $\sigma\left(w_i\right)=\operatorname{lo}\left(x_i\right)+\sigma(k)$
$\Rightarrow 15 =I(12)+0$
$\Rightarrow I =\frac{15}{12} $
$=1.25$
From Eq. (i),
$ 55 =1.25 \times 48+k$
$\Rightarrow k =55-60$
$\therefore k =-5$