In a series of 2n observations, half of them are equal to `a^(2)` and the remaining half are equal to `-a^(2)`. If the standard deviation of the observation is 2, then `|a|` is equal to

Options

A. `(1)/(n)`
B. `sqrt2`
C. 2
D. `(sqrt2)/(n)`

Answer

To solve the problem, we need to find the value of \(|a|\) given that in a series of \(2n\) observations, half are equal to \(a^2\) and the other half are equal to \(-a^2\), and the standard deviation of the observations is 2.

### Step 1: Understand the Observations
We have \(2n\) observations:
- \(n\) observations are equal to \(a^2\)
- \(n\) observations are equal to \(-a^2\)

### Step 2: Calculate the Mean
The mean (\(\bar{x}\)) of the observations can be calculated as follows:

\[ \bar{x} = \frac{1}{2n} \left( \sum_{i=1}^{n} a^2 + \sum_{j=n+1}^{2n} (-a^2) \right) \]

Since there are \(n\) terms of \(a^2\) and \(n\) terms of \(-a^2\):

\[ \bar{x} = \frac{1}{2n} \left( n a^2 + n (-a^2) \right) = \frac{1}{2n} (n a^2 - n a^2) = \frac{0}{2n} = 0 \]

### Step 3: Calculate the Variance
The variance (\(\sigma^2\)) is given by:

\[ \sigma^2 = \frac{1}{2n} \sum_{i=1}^{2n} (x_i - \bar{x})^2 \]

Since \(\bar{x} = 0\), this simplifies to:

\[ \sigma^2 = \frac{1}{2n} \sum_{i=1}^{2n} x_i^2 \]

Calculating \(x_i^2\):
- For the \(n\) observations equal to \(a^2\), we have \(n \cdot (a^2)^2 = n a^4\).
- For the \(n\) observations equal to \(-a^2\), we have \(n \cdot (-a^2)^2 = n a^4\).

Thus,

\[ \sum_{i=1}^{2n} x_i^2 = n a^4 + n a^4 = 2n a^4 \]

Now substituting this back into the variance formula:

\[ \sigma^2 = \frac{1}{2n} (2n a^4) = a^4 \]

### Step 4: Use the Given Standard Deviation
We know that the standard deviation \(\sigma\) is given as 2, so:

\[ \sigma = \sqrt{a^4} = |a^2| \]

Thus,

\[ |a^2| = 2 \]

### Step 5: Solve for \(|a|\)
Taking the square root of both sides gives:

\[ |a| = \sqrt{2} \]

### Final Answer
Thus, the value of \(|a|\) is:

\[ \boxed{\sqrt{2}} \]