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

The Correct Option is D

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 \(2\sqrt{6}.\)

Answer 2

We are given the numbers -1, 0, 1, K and their standard deviation is \(\sqrt{5}\), with \(k>0\).

First, calculate the mean (\(\mu\)) of the numbers:

\(\mu = \frac{-1 + 0 + 1 + k}{4} = \frac{k}{4}\)

The variance (\(\sigma^2\)) is the average of the squared differences from the mean:

\(\sigma^2 = \frac{(-1 - \frac{k}{4})^2 + (0 - \frac{k}{4})^2 + (1 - \frac{k}{4})^2 + (k - \frac{k}{4})^2}{4}\)

The standard deviation (\(\sigma\)) is the square root of the variance, so \(\sigma = \sqrt{\sigma^2}\). We are given that \(\sigma = \sqrt{5}\), therefore \(\sigma^2 = 5\).

Substitute \(\sigma^2 = 5\) into the variance equation:

\(5 = \frac{(-1 - \frac{k}{4})^2 + (-\frac{k}{4})^2 + (1 - \frac{k}{4})^2 + (\frac{3k}{4})^2}{4}\)

Multiply both sides by 4:

\(20 = (1 + \frac{k}{2} + \frac{k^2}{16}) + \frac{k^2}{16} + (1 - \frac{k}{2} + \frac{k^2}{16}) + \frac{9k^2}{16}\)

Simplify:

\(20 = 2 + \frac{12k^2}{16}\)

\(18 = \frac{3k^2}{4}\)

\(k^2 = \frac{18 * 4}{3}\)

\(k^2 = 6 * 4\)

\(k^2 = 24\)

\(k = \sqrt{24} = \sqrt{4 * 6} = 2\sqrt{6}\)

Since k > 0, we take the positive root. Therefore, k = \(2\sqrt{6}\).

Answer:

\(2\sqrt6\)