JEE Main 2023 (Online) 10th April Evening ShiftQuestion 1
Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution
$x_i$ : $0,1,2,3,4,5$
$f_i$ : $k+2,\ 2k,\ k^{2}-1,\ k^{2}-1,\ k^{2}+1,\ k-3$
where $\sum f_i=62$. If $[x]$ denotes the greatest integer $\le x$, then $\left[\mu^{2}+\sigma^{2}\right]$ is equal to
Statistics
Solution
$\sum f_i=(k+2)+2k+2(k^2-1)+(k^2+1)+(k-3)=3k^2+4k-2$ (the two $k^2-1$ and one $k^2+1$ give $3k^2$). Setting $=62$: $3k^2+4k-64=0\Rightarrow k=4$ (reject $-\frac{16}{3}$). Frequencies: $6,8,15,15,17,1$, $N=62$. $\sum fx=8+30+45+68+5=156$, $\mu=\frac{78}{31}$; $\sum fx^2=8+60+135+272+25=500$. $\mu^2+\sigma^2=\frac{\sum fx^2}{N}=\frac{500}{62}\approx8.06$, so $[\cdot]=8$. Option C.