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If the standard deviation of the numbers 2,3,a and 11 is 3.5, then which of the following is true?

$3a^{2}-26a+55=0$

$3a^{2}-32a+84=0$

$3a^{2}-34a+91=0$

$3a^{2}-23a+44=0$

Option B

We know that , if x₁ , x₂..........x_n are n observations , then their standard deviation is given by

$\sqrt{\frac{1}{n}\sum x_i^2}-(\frac{\sum x_{i}}{2})^{2}$

We have , $(3.5)^{2}=\frac{(2^{2}+3^{2}+11^{2})}{4}-(\frac{2+3+a+11}{4})$

$\Rightarrow \frac{49}{4}=\frac{(4+9+a^{2}+121)}{4}-(\frac{16+a}{4})^{2}$

$\Rightarrow \frac{49}{4}=\frac{(134+a^{2})}{4}-(\frac{256+a^{2}+32a}{16})$

$\Rightarrow \frac{49}{4}=\frac{4a^{2}+536-256-a^{2}-32a}{16}$

$\Rightarrow 49\times4=3a^{2}-32a+280$

$\Rightarrow 3a^{2}-32a+84=0$

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