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STD 11 - 13. statistics — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 13. statistics4 Marks
MCQ
Consider the following frequency distribution :

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 $.....$

  • A
    $5$
  • B
    $6$
  • ✓
    $7$
  • D
    $11$
✓

Answer

Correct option: C.
$7$
c
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 $(1)\, and\,(2)$

$a=8, b=10$

$\therefore(a-b)^{2}=(8-10)^{2}$

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