13.Statistics
hard

माना $X=\{x \in N : 1 \leq x \leq 17\}$ और $Y=\{a x+b: x \in X$ और $a, b \in R , a>0\}$ यदि $Y$ के अवयव का माध्य और प्रसरण क्रमश $17$ और $216$ है तो $a+b$ बराबर है

A

$-7$

B

$7$

C

$9$

D

$-27$

(JEE MAIN-2020)

Solution

$\sigma^{2}=$ variance

$\mu=$ mean

$\sigma^{2}=\frac{\sum_{i=1}^{n}\left(x_{i}-\mu\right)^{2}}{n}$

$\mu=17$

$\Rightarrow \frac{\sum_{x=1}^{17}(a x+b)}{17}=17$

$\Rightarrow \quad 9 a+b=17$

$\sigma^{2}=216$

$\Rightarrow \quad \frac{\sum_{x=1}^{17}(a x+b-17)^{2}}{17}=216$

$\Rightarrow \frac{\sum_{x=1}^{17} a^{2}(x-9)^{2}}{17}=216$

$\Rightarrow \quad a^{2} 81-18 \times 9 a^{2}+a^{2} 3 \times(35)=216$

$\Rightarrow \quad$ From $(1), b=-10$

So, $a+b=-7$

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.