- Home
- Standard 11
- Mathematics
13.Statistics
hard
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 $\leq x$, then $\left[\mu^2+\sigma^2\right]$ is equal $.........$.
A
$8$
B
$7$
C
$6$
D
$9$
(JEE MAIN-2023)
Solution
$\sum f _{ i }=62$
$3 k ^2+16 k -12 k -64=0$
$k =\text { or }-\frac{16}{3}(\text { rejected) }$
$\mu=\frac{\sum f _{ i } x _{ i }}{\sum f _{ i }}$
$\mu=\frac{8+2(15)+3(15)+4(17)+5}{62}=\frac{156}{62}$
$\sigma^2=\sum f _{ i } x _{ i }^2-\left(\sum f _{ i } x _{ i }\right)^2$
$=\frac{8 \times 1^2+15 \times 13+17 \times 16+25}{62}-\left(\frac{156}{62}\right)^2$
$\sigma^2=\frac{500}{62}-\left(\frac{156}{62}\right)^2$
$\sigma^2+\mu^2=\frac{500}{62}$
${\left[\sigma^2+\mu^2\right]=8}$
Standard 11
Mathematics
Similar Questions
medium
