Let n `ge ` 3. A list of numbers `0 le x_1 le x_2 le … le x_n` has mean `mu` and standard deviatiion `sigma`. A new list of numbers is made as follows : `y_1 = 0 , y_2 = x_2, …, y_(n-1) = x_(n-1) , y_(n) = x_1 + x_(n)`. The mean and the standard deviation of the new list are `hat(mu) and hat(sigma)`. Which of the following is necessarily true ?

Options

A. `mu= hat(mu), sigma le hat(sigma)`
B. `mu= hat(mu) , sigma le hat(sigma)`
C. `sigma = hat(sigma)`
D. `mu` may or may not be equal to ` hat(mu)`.

Answer

To solve the problem, we need to analyze the relationship between the original list of numbers \( x_1, x_2, \ldots, x_n \) and the new list of numbers \( y_1, y_2, \ldots, y_n \).

### Step 1: Calculate the Mean of the Original List
The mean \( \mu \) of the original list is given by:
\[ \mu = \frac{x_1 + x_2 + \ldots + x_n}{n} \]

### Step 2: Define the New List
The new list is defined as:
- \( y_1 = 0 \)
- \( y_2 = x_2 \)
- \( y_3 = x_3 \)
- ...
- \( y_{n-1} = x_{n-1} \)
- \( y_n = x_1 + x_n \)

### Step 3: Calculate the Mean of the New List
The mean \( \hat{\mu} \) of the new list is calculated as follows:
\[ \hat{\mu} = \frac{y_1 + y_2 + \ldots + y_n}{n} = \frac{0 + x_2 + x_3 + \ldots + x_{n-1} + (x_1 + x_n)}{n} \]
This simplifies to:
\[ \hat{\mu} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} = \mu \]

### Step 4: Calculate the Standard Deviation of the Original List
The standard deviation \( \sigma \) of the original list is given by:
\[ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} x_i^2 - \mu^2 \]

### Step 5: Calculate the Standard Deviation of the New List
The standard deviation \( \hat{\sigma} \) of the new list is calculated as follows:
\[ \hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^{n} y_i^2 - \hat{\mu}^2 \]
Substituting the values of \( y_i \):
\[ \hat{\sigma}^2 = \frac{1}{n} \left(0^2 + x_2^2 + x_3^2 + \ldots + x_{n-1}^2 + (x_1 + x_n)^2\right) - \mu^2 \]
Expanding the square:
\[ \hat{\sigma}^2 = \frac{1}{n} \left(0 + x_2^2 + x_3^2 + \ldots + x_{n-1}^2 + x_1^2 + x_n^2 + 2x_1x_n\right) - \mu^2 \]
Rearranging gives:
\[ \hat{\sigma}^2 = \frac{1}{n} \left(x_1^2 + x_2^2 + \ldots + x_n^2 + 2x_1x_n\right) - \mu^2 \]

### Step 6: Compare the Standard Deviations
We can relate this back to the original standard deviation:
\[ \hat{\sigma}^2 = \sigma^2 + \frac{2x_1x_n}{n} \]
Since \( x_1 \) and \( x_n \) are both non-negative, \( \frac{2x_1x_n}{n} \geq 0 \). Therefore:
\[ \hat{\sigma}^2 \geq \sigma^2 \implies \hat{\sigma} \geq \sigma \]

### Conclusion
From the calculations, we conclude:
1. The mean of the new list \( \hat{\mu} \) is equal to the mean of the original list \( \mu \).
2. The standard deviation of the new list \( \hat{\sigma} \) is greater than or equal to the standard deviation of the original list \( \sigma \).

Thus, the final result is:
\[ \hat{\mu} = \mu \quad \text{and} \quad \hat{\sigma} \geq \sigma \]