The variance of the numbers& 2, 3, 11 and x is (49)/(4) , then the values of x & are

Correct answer: A. 6or(14)/(3) From the given data, we make the following table x x^ 2 2 4 3 9 11 121 x x^ 2 ∑ x = 16 + x ∑ x^ 2 = 134 + x^ 2 But we know that, variance = ∑ x^ 2 n - ((∑ x)/(n)) ^ 2 ⇒ 134 + x^ 2 4 - ((16 + x)/(4)) ^ 2 = (49)/(4) (given) ⇒ 134 + x^ 2 4 - (256 + x^ 2 + 32x) 16 = (49)/(4) ⇒ 3x^ 2 - 32x + 280 16 = (49)/(4) ⇒ 3x^ 2 - 32x + 280 = 196 ⇒ 3x^ 2 - 32x + 84 = 0 ⇒ (x - 6)(3x - 14) = 0 ⇒ x = 6, x = (14)/(3) Therefore, the values of x are 6 or (14)/(3)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Statistics

NTA Abhyas 2020 · Mathematics · Statistics

The variance of the numbers& 2, 3, 11 and x is (49)/(4) , then the...

The variance of the numbers&

2, 3, 11

and

x

\dfrac{49}{4}

, then the values of

x

& are

Options

A $6or\dfrac{14}{3}$
B $6or\dfrac{14}{5}$
C $6or\dfrac{16}{3}$
D $4or\dfrac{13}{5}$

Correct answer

A. $6or\dfrac{14}{3}$

Step-by-step solution

From the given data, we make the following table

$x$	$x^{2}$
$2$	$4$
$3$	$9$
$11$	$121$
$x$	$x^{2}$
$∑ x = 16 + x$	$∑ x^{2} = 134 + x^{2}$

But we know that, variance

= \dfrac{∑ x^{2}}{n} - {\left(\dfrac{∑ x}{n}\right)}^{2}

⇒ \dfrac{134 + x^{2}}{4} - {\left(\dfrac{16 + x}{4}\right)}^{2} = \dfrac{49}{4}

(given)

⇒ \dfrac{134 + x^{2}}{4} - \dfrac{\left(256 + x^{2} + 32x\right)}{16} = \dfrac{49}{4}

⇒ \dfrac{3x^{2} - 32x + 280}{16} = \dfrac{49}{4}

⇒ 3x^{2} - 32x + 280 = 196 ⇒ 3x^{2} - 32x + 84 = 0

⇒ \left(x - 6\right)\left(3x - 14\right) = 0 ⇒ x = 6, x = \dfrac{14}{3}

Therefore, the values of

x

are

6

\dfrac{14}{3}

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