Let x1,x2,…,x10 be 10 observations such that ∑i=110 xi−5=10 and ∑i=110 xi−52=40. If mean and variance of observations x1−3,x2−3,…,x10−3 are λ and μ respectively, then (λ, μ)=

# Let ${x}_{1},{x}_{2},\dots ,{x}_{10}$ be $10$ observations such that $\sum _{i=1}^{10} \left({x}_{i}-5\right)=10$ and $\sum _{i=1}^{10} {\left({x}_{i}-5\right)}^{2}=40.$ If mean and variance of observations $\left({x}_{1}-3\right),\left({x}_{2}-3\right),\dots ,\left({x}_{10}-3\right)$ are $\lambda$ and $\mu$ respectively, then

1. A

2. B

3. C

4. D

Register to Get Free Mock Test and Study Material

+91

Verify OTP Code (required)

### Solution:

We have,

$\sum _{i=1}^{10} \left({x}_{i}-5\right)=10$ and $\sum _{i=1}^{10} {\left({x}_{i}-5\right)}^{2}=40$

$⇒\sum _{i=1}^{10} \left[\left({x}_{i}-3\right)-2\right]=10$ and $\sum _{i=1}^{10} {\left\{\left({x}_{i}-3\right)-2\right\}}^{2}=40$

$⇒\sum _{i=1}^{10} \left({x}_{i}-3\right)-20=10$ and $\sum _{i=1}^{10} \left\{{\left({x}_{i}-3\right)}^{2}-4\left({x}_{i}-3\right)+4\right\}=40$

$⇒\sum _{i=1}^{10} \left({x}_{i}-3\right)=30$ and $\sum _{i=1}^{10} {\left({x}_{i}-3\right)}^{2}-4=\sum _{i=1}^{10} \left({x}_{i}-3\right)+40=40$

$⇒\frac{1}{10}\sum _{i=1}^{10} \left({x}_{i}-3\right)=3$ and $\frac{1}{10}\sum _{i=1}^{10} {\left({x}_{i}-3\right)}^{2}-4×3=0$

and $\frac{1}{10}\sum _{i=1}^{10} {\left({x}_{i}-3\right)}^{2}=12$

$⇒\lambda =3$ and $\frac{1}{10}\sum _{i=1}^{10} {\left({x}_{i}-3\right)}^{2}-{\left\{\frac{1}{10}\sum _{i=1}^{10} \left({x}_{i}-3\right)\right\}}^{2}=12-9$

$⇒\lambda =3$ and $\mu =3.$

Hence. $\left(\lambda ,\mu \right)=\left(3,3\right)$

ALITER Let $U$ and $V$ be two variables taking values ${u}_{1},{u}_{2},\dots ,{u}_{10}$ and ${v}_{1},{v}_{2},\dots ,{v}_{10}$ respectively such that

${u}_{i}={x}_{i}-5$ and ${v}_{i}={x}_{i}-3={u}_{i}+2,i=1,2,\dots ,10$.

It is given that

$\sum _{i=1}^{10} \left({x}_{i}-5\right)=10$ and $\sum _{i=1}^{10} {\left({x}_{i}-5\right)}^{2}=40$

and $\sum _{i=1}^{10} {u}_{i}^{2}=40$

and $\frac{1}{10}\sum _{i=1}^{10} {u}_{i}^{2}=4$

and $\frac{1}{10}\sum _{i=1}^{10} {u}_{i}^{2}-{\left(\frac{1}{10}\sum {u}_{i}\right)}^{2}=4-{1}^{2}$

and

Now, ${v}_{i}={u}_{i}+2,i=1,2,\dots ,10$

and $\mathrm{Var}\left(V\right)=\mathrm{Var}\left(U\right)$

and $\mu =3⇒\left(\lambda ,\mu \right)=\left(3,3\right)$  Register to Get Free Mock Test and Study Material

+91

Verify OTP Code (required)