Let X=x N:1 x 19 and for some a,b R,Y=ax+b:x X.

Question

Let $X={x\in N:1\le x\le 19}$ and for some $a,b\in R,Y={ax+b:x\in X}$. If the mean and variance of the elements of Y are 30 and 750 , respectively, then the sum of all possible values of $b$ is

Accepted Answer

$Y={ax+b:x\in X}.X={x\in N:1\le x\le 19}$

mean for $Y$ is 30.

$	herefore $ $(\vec{Y}=30)$

$\vec{Y}=\frac{\sum _{i=1}^{19}{Y}_{i}}{19}$

$\Rightarrow \frac{\sum _{i=1}^{19}a{x}_{i}+b}{19}=30$

$\Rightarrow$ $\sum _{i=1}^{19}a{x}_{i}+\sum _{i=1}^{19}b=19	imes 30$

$\Rightarrow$ $a(1+2+\cdots +19)+19b=19	imes 30$

$\Rightarrow$ $a	imes \frac{19	imes 20}{2}+19b=570$

$\Rightarrow$ $10a+b=30\Rightarrow a=\frac{30-b}{10}$ .......(1)

Variance for $Y$ is 750

${\sigma }^{2}=\frac{\sum _{i=1}^{19}{Y}_{2}^{i}}{19}-(\vec{Y}{)}^{2}=750$

$\Rightarrow$ $\frac{\sum _{i=1}^{19}{(a{x}_{i}+b)}^{2}}{19}-(30{)}^{2}=750$

$\Rightarrow$ $\sum _{i=1}^{19}{a}^{2}{x}_{2}^{i}+{b}^{2}+2a{x}_{i}=(750+900)19$

$\Rightarrow$ ${a}^{2}\sum _{i=1}^{19}({x}_{2}^{i}+{b}^{2}+2a\sum _{i=1}^{19}{x}_{i})=1650	imes 19$

$\Rightarrow$ ${a}^{2}({1}^{2}+{2}^{2}+\cdots +{19}^{2})+\sum _{i=1}^{19}{b}^{2}+2ab(1+2+\cdots +19)=1650	imes 19$

$\Rightarrow$ ${a}^{2}\cdot \frac{19	imes 20	imes 39}{6}+19{b}^{2}+\frac{2ab	imes 19	imes 20}{2}=1650	imes 19$

$\Rightarrow$ $130{a}^{2}+{b}^{2}+20ab=1650$ ......(2)

Substitute value of $a$ from (1) in equation (2)

$130{(\frac{30-b}{10})}^{2}+{b}^{2}+20b(\frac{30-b}{10})=1650$

$\Rightarrow$ $\frac{13}{10}(900+{b}^{2}-60b)+{b}^{2}+60b-2{b}^{2}=1650$

$\Rightarrow$ $1170+1.3{b}^{2}-78b+10{b}^{2}+600b-20{b}^{2}=16500$

$\Rightarrow$ $3{b}^{2}-180b-4800=0$

$\Rightarrow$ ${b}^{2}-60b-1600=0$

Let roots of quadratic equation is ${b}_{1},{b}_{2}$

So sum of possible values of $b=$ sum of roots

${b}_{1}+{b}_{2}=\frac{-b}{{b}^{2}}=60$

Solutions · 1