Question

Let $x_1,x_2,\dots ,x_n$ be 𝑛 observations. Let 𝑤𝑖=𝑙𝑥𝑖 + 𝑘 for 𝑖 = 1, 2, …𝑛 , where 𝑙 and 𝑘 are constants. If the mean of $w_i^{\prime }s$is 48 and their standard deviation is 12, the mean of 𝑤𝑖′𝑠 is 55 and standard deviation of $w_i^{\prime }s$is 15, the values of 𝑙 and 𝑘 should be

• A. l=1.25, k=-5
• B. l=-1.25, k=5
• C. l=2.5, k=-5
• D. l=2.5, k=5

Answer 1

The correct option is A l=1.25, k=-5

Given: $w_i=l{x}_i+k$

For n terms

$\sum w_i=\sum (i{x}_i+k)$

$\sum w_i=(l\sum x_i)+nk...\left(i\right)$

Given: mean of $x_i=48$

$\therefore \sum w_i=48nl+nk....\left(ii\right)$

Also, given mean of $w_i=55$

$\therefore \sum w_i=55n....\left(iii\right)$

from equation (ii) and (iii), we get

$\Rightarrow 55n=48nl+nk$

$\Rightarrow 55=48l+k....\left(iv\right)$

Now,$w_i=l{x}_i+k$

Standard deviation of $w_i=l\times \left(standarddeviationof{x}_i\right)$

$(\therefore$ Standard deviation do not depend on constant 𝑘)

$\Rightarrow 15=l\times 12$

$\Rightarrow l=\frac{15}{12}=1.25$

Now, from equation(iv),we get

$k=55-1.25\times 48$

$\Rightarrow k=-5$

Hence, option (𝐴) is correct