Understanding Effects of Transformations on Average and Standard Deviation

Let's analyze how adding a constant to each entry or multiplying each entry by a constant affects the average (mean) and standard deviation of a list of numbers. Consider a list of $n$ entries: $x_1, x_2, \dots, x_n$.

Original Average and Standard Deviation

The original average (mean), $\bar{x}$, is given by:

$$\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$$

The original variance, $s^2$, is given by:

$$s^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2$$

The original standard deviation, $s$, is the square root of the variance:

$$s = \sqrt{\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2}$$

Analyzing Statement 1: Adding a Constant to Each Entry

Statement 1 says: If 10 is added to each entry on a list, then the average increases by 10.

Let the new list entries be $y_i = x_i + 10$. The new average, $\bar{y}$, is:

$$\bar{y} = \frac{1}{n} \sum_{i=1}^n y_i = \frac{1}{n} \sum_{i=1}^n (x_i + 10)$$

We can split the sum:

$$\bar{y} = \frac{1}{n} \left( \sum_{i=1}^n x_i + \sum_{i=1}^n 10 \right)$$

$$\bar{y} = \frac{1}{n} \left( \sum_{i=1}^n x_i + n \times 10 \right)$$

$$\bar{y} = \frac{1}{n} \sum_{i=1}^n x_i + \frac{n \times 10}{n}$$

$$\bar{y} = \bar{x} + 10$$

The new average is indeed the original average plus 10. Thus, statement 1 is correct.

Analyzing Statement 2: Adding a Constant to Standard Deviation

Statement 2 says: IF 10 is added to each entry on a list, then the standard deviation increases by 10.

Using the new list entries $y_i = x_i + 10$, the new average is $\bar{y} = \bar{x} + 10$. The deviations from the new mean are:

$$y_i - \bar{y} = (x_i + 10) - (\bar{x} + 10) = x_i + 10 - \bar{x} - 10 = x_i - \bar{x}$$

The deviations from the mean remain unchanged. The new variance, $s_y^2$, is:

$$s_y^2 = \frac{1}{n} \sum_{i=1}^n (y_i - \bar{y})^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 = s^2$$

The new variance is equal to the original variance. Therefore, the new standard deviation $s_y = \sqrt{s_y^2} = \sqrt{s^2} = s$.

Adding a constant to each entry shifts the entire distribution but does not change its spread. The standard deviation measures spread, so it remains unchanged. Thus, statement 2 is incorrect.

Analyzing Statement 3: Doubling Each Entry

Statement 3 says: if each entry on a list is doubled then the average doubles.

Let the new list entries be $z_i = 2x_i$. The new average, $\bar{z}$, is:

$$\bar{z} = \frac{1}{n} \sum_{i=1}^n z_i = \frac{1}{n} \sum_{i=1}^n (2x_i)$$

We can factor out the constant 2 from the sum:

$$\bar{z} = \frac{1}{n} \times 2 \sum_{i=1}^n x_i$$

$$\bar{z} = 2 \left( \frac{1}{n} \sum_{i=1}^n x_i \right)$$

$$\bar{z} = 2 \bar{x}$$

The new average is double the original average. Thus, statement 3 is correct.

Summary of Correct Statements

Based on the analysis:

• Statement 1: Correct (Average increases by the constant added)
• Statement 2: Incorrect (Standard deviation remains unchanged when a constant is added)
• Statement 3: Correct (Average is multiplied by the constant when entries are multiplied)

Therefore, statements 1 and 3 are correct.

Let's summarize the effects of transformations on mean and standard deviation in a table:

Comparing our findings to the statements:

• Statement 1 matches the effect of adding a constant on the average.
• Statement 2 contradicts the effect of adding a constant on the standard deviation.
• Statement 3 matches the effect of multiplying by a constant on the average (with $a=2$).

Thus, only statements 1 and 3 are correct.