Let's start by defining the mean and variance for the original set of observations x_1, x_2,..., x_n. The mean, denoted by x, is the sum of all observations divided by the number of observations n. The variance, denoted by ², is the average of the squared differences from the mean.

Let's start by defining the mean and variance for the original set of observations x_1, x_2,..., x_n. The mean, denoted by x, is the sum of all observations divided by the number of observations n. The variance, denoted by ², is the average of the squared differences from the mean.

We can simplify the expression for the new mean y. By distributing the summation and recognizing that Σ_i=1^(n) a = na, we find that the new mean is simply the original mean plus 'a'.

Now, let's calculate the variance for the new set of observations, _y². This is defined as the average of the squared differences between each new observation y_i and the new mean y.

Since the formula for _y² is identical to the formula for ², we have shown that the variance remains unchanged when each observation is increased by a constant 'a'.

If each of the observation $x_{1}, x_{2}, ..., x_{n}$ is increased by ' $a$ ', where $a$ is a negative or positive number, show that the variance remains unchanged.

Answer: The variance remains unchanged.

Step-by-step solution

Step 1: Define original mean and variance

Let's start by defining the mean and variance for the original set of observations $x_1, x_2, ..., x_n$ . The mean, denoted by $\bar{x}$ , is the sum of all observations divided by the number of observations $n$ . The variance, denoted by $\sigma^2$ , is the average of the squared differences from the mean.

Step 2: Define new observations and their mean

Now, let's consider the new set of observations, $y_1, y_2, ..., y_n$ , where each observation $x_i$ is increased by 'a'. So, $y_i = x_i + a$ . We then calculate the mean of these new observations, $\bar{y}$ .

Step 3: Simplify the new mean

We can simplify the expression for the new mean $\bar{y}$ . By distributing the summation and recognizing that $\sum_{i=1}^{n} a = na$ , we find that the new mean is simply the original mean plus 'a'.

Step 4: Calculate the new variance

Now, let's calculate the variance for the new set of observations, $\sigma_y^2$ . This is defined as the average of the squared differences between each new observation $y_i$ and the new mean $\bar{y}$ .

Step 5: Substitute and simplify the new variance

Substitute $y_i = x_i + a$ and $\bar{y} = \bar{x} + a$ into the variance formula. Notice that the 'a' terms cancel out inside the parenthesis. This leaves us with the exact same expression as the original variance.

Step 6: Conclusion

Since the formula for $\sigma_y^2$ is identical to the formula for $\sigma^2$ , we have shown that the variance remains unchanged when each observation is increased by a constant 'a'.

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If each of the observation x1,x2,...,xnx_{1}, x_{2}, ..., x_{n}x1​,x2​,...,xn​ is increased by ' aaa ', where aaa is a negative or positive number, show that the variance remains unchanged.