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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 $x_i^{\prime} \operatorname{sis} 48$ and their standard deviation is 12 , the mean of $w_i$ 's is 55 and standard deviation of $w_i^{\prime}$ 's is 15 , then the value of $l$ and $k$ should be
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The correct answer is:
$l=1.25, k=-5 & \text$
$l=1.25, k=-5 & \text$
Since, $w_i=l x_l+k, \bar{x}_i=48, \sigma x_i=12, w_i=55$ and $\sigma w_i=15$
So, $\bar{w}_i=l \bar{x}_i+k$
$\Rightarrow 55=48 l+k$
Now, $\sigma \omega_i=l \sigma x_i \Rightarrow l=\frac{15}{2}=1.25$
After putting the value of 1 in equation (i), we get $k=55-1.25 \times 48=-5$
So, $\bar{w}_i=l \bar{x}_i+k$
$\Rightarrow 55=48 l+k$
Now, $\sigma \omega_i=l \sigma x_i \Rightarrow l=\frac{15}{2}=1.25$
After putting the value of 1 in equation (i), we get $k=55-1.25 \times 48=-5$
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