Question

If the standard deviations of the numbers - 1, 0, 1, K is $\sqrt5$, where $k>0$, then k is equal to

• A. $\frac{4 \sqrt{5}}{3}$
• B. $\frac{2\sqrt{10}}{3}$
• C. $\sqrt6$
• D. $2\sqrt6$

Answer 1

Answer: (D)

$2\sqrt6$

Answer 2

Standard deviation = $\sqrt{\frac{1}{n} \sum (x_i - \bar{x})^2}$
We are given that the standard deviation of the numbers -1, 0, 1, and k is $\sqrt5$.
So we can set up the equation as follows:
$\sqrt{5} = \sqrt{\frac{1}{4} \left[ (-1 - \bar{x})^2 + (0 - \bar{x})^2 + (1 - \bar{x})^2 + (k - \bar{x})^2 \right]}$
Simplifying the equation:
$\sqrt{5} = \sqrt{\frac{1}{4} \left[ (1 + \bar{x}^2) + (\bar{x}^2) + (1 + \bar{x}^2) + (k^2 - 2k\bar{x} + \bar{x}^2) \right]} \sqrt{5}$

= $\sqrt{\frac{1}{4} \cdot (3\bar{x}^2 + k^2 - 2k\bar{x} + 2)}$
Squaring both sides of the equation:
$5 = \frac{1}{4} \cdot (3\bar{x}^2 + k^2 - 2k\bar{x} + 2) 20$

= $3\bar{x}^2 + k^2 - 2k\bar{x} + 2$
Rearranging the terms: $3\bar{x}^2 - 2k\bar{x} + k^2 = 18$
Now, we need to find the value of k that satisfies this equation.
Given that $k > 0$, we can consider the discriminant of the quadratic equation:
Discriminant$(D) = (-2k)^2 - 4(3)(k^2 - 18)$

Simplifying:
$D = 4k^2 - 4(3k^2 - 54)\ D$
= $4k^2 - 12k^2 + 216 = D$
= $-8k^2 + 216$
For the quadratic equation to have real solutions, the discriminant D must be greater than or equal to $-8k^2 + 216 \geq 0$
Dividing both sides by -8 (and flipping the inequality sign):
$k^2 - 27 \geq 0 \quad \text{and} \quad (k - \sqrt{27})(k + \sqrt{27}) \geq 0$
Since $k > 0$, we need to consider the range $k > \sqrt{27}$
The only option that satisfies $k > \sqrt{27}$ is option (4) $2\sqrt{6}$.
Therefore, k is equal to $2\sqrt{6}.$