| Class: | $0-6$ | $6-12$ | $12-18$ | $18-24$ | $24-30$ |
| Frequency : | $a$ | $b$ | $12$ | $9$ | $5$ |
If mean $=\frac{309}{22}$ and median $=14$, than value $(a-b)^{2}$ is equal to $.....$
| Class | Frequency | $X_i$ | $F_i\,X_i$ |
| $0-6$ | $a$ | $3$ | $3a$ |
| $6-12$ | $b$ | $9$ | $9b$ |
| $12-18$ | $12$ | $15$ | $180$ |
| $18-24$ | $9$ | $21$ | $189$ |
| $24-30$ | $5$ | $27$ | $135$ |
| $N=(26+a+b)$ | $(504+3a+9b)$ |
Mean $=\frac{3 a+9 b+180+189+135}{a+b+26}=\frac{309}{22}$
$\Rightarrow 66 a+198 b+11088=309 a+309 b+8034$
$\Rightarrow 243 a+111 b=3054$
$\Rightarrow 81 a+37 b=1018 ....(1)$
Now, Median $=12+\frac{\frac{a+b+26}{2}-(a+b)}{2} \times 6=14$
$\Rightarrow \frac{13}{2}-\left(\frac{a+b}{4}\right)=2$
$\Rightarrow \frac{a+b}{4}=\frac{9}{2}$
$\Rightarrow a+b=18 \rightarrow(2)$
From equation $(1)\, and\,(2)$
$a=8, b=10$
$\therefore(a-b)^{2}=(8-10)^{2}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\sin ^{-1}\left(\sum_{i=1}^{\infty} x^{i+1}-x \sum_{i=1}^{\infty}\left(\frac{x}{2}\right)^i\right)=\frac{\pi}{2}-\cos ^{-1}\left(\sum_{i=1}^{\infty}\left(-\frac{x}{2}\right)^i-\sum_{i=1}^{\infty}(-x)^i\right)$
lying in the interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$ is. . . . .
(Here, the inverse trigonometric functions $\sin ^{-1} x$ and $\cos ^{-1} x$ assume values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $[0, \pi]$, respectively.)