If `sum_(i=1)^n (x_i -a) =n` and `sum_(i=1)^n (x_i - a)^2 =na` then the standard deviation of variate `x_i`

Options

A. `sqrt(a^2-1)`
B. `sqrt(a-1)`
C. `sqrt(n^2a-1)`
D. `sqrt(n^2a^2-n)`

Answer

To find the standard deviation of the variate \( x_i \) given the conditions:

1. \( \sum_{i=1}^n (x_i - a) = n \)
2. \( \sum_{i=1}^n (x_i - a)^2 = na \)

We can follow these steps:

### Step 1: Understand the Standard Deviation Formula
The standard deviation (SD) is given by the formula:
\[ SD = \sqrt{\frac{\sum_{i=1}^n x_i^2}{n} - \left(\frac{\sum_{i=1}^n x_i}{n}\right)^2} \]

### Step 2: Rewrite the Given Sums
From the first condition, we can express the sum of \( x_i \):
\[ \sum_{i=1}^n (x_i - a) = n \implies \sum_{i=1}^n x_i - na = n \implies \sum_{i=1}^n x_i = n + na = n(1 + a) \]

### Step 3: Calculate the Mean
Now, we can find the mean \( \bar{x} \):
\[ \bar{x} = \frac{\sum_{i=1}^n x_i}{n} = \frac{n(1 + a)}{n} = 1 + a \]

### Step 4: Substitute into the Standard Deviation Formula
Now, we need to find \( \sum_{i=1}^n x_i^2 \). We know:
\[ \sum_{i=1}^n (x_i - a)^2 = na \]
Expanding this gives:
\[ \sum_{i=1}^n (x_i^2 - 2ax_i + a^2) = na \]
This can be rearranged to:
\[ \sum_{i=1}^n x_i^2 - 2a\sum_{i=1}^n x_i + na^2 = na \]
Substituting \( \sum_{i=1}^n x_i = n(1 + a) \):
\[ \sum_{i=1}^n x_i^2 - 2a(n(1 + a)) + na^2 = na \]
This simplifies to:
\[ \sum_{i=1}^n x_i^2 - 2an - 2a^2 + na^2 = na \]
Rearranging gives:
\[ \sum_{i=1}^n x_i^2 = na + 2an + 2a^2 = na + 2a(n + a) \]

### Step 5: Substitute into the Standard Deviation Formula
Now we can substitute \( \sum_{i=1}^n x_i^2 \) and \( \bar{x} \) into the SD formula:
\[ SD = \sqrt{\frac{\sum_{i=1}^n x_i^2}{n} - \left(\frac{\sum_{i=1}^n x_i}{n}\right)^2} \]
Substituting the values:
\[ SD = \sqrt{\frac{na + 2a(n + a)}{n} - (1 + a)^2} \]
This simplifies to:
\[ SD = \sqrt{a + 2a + \frac{2a^2}{n} - (1 + 2a + a^2)} \]
\[ SD = \sqrt{a - 1 + \frac{2a^2}{n}} \]

### Step 6: Final Expression
Thus, we can conclude that the standard deviation is:
\[ SD = \sqrt{a - 1} \]