Understanding Data Transformation and Statistical Measures

This question asks how the mean and standard deviation of a dataset change when each observation undergoes a linear transformation. We are given the initial mean and standard deviation for 100 observations and a specific transformation: subtracting 5 from each observation and then dividing the result by 4.

Let the original observations be denoted by $x_1, x_2, \dots, x_{100}$.

The original mean is given as $\bar{x} = 50$.

The original standard deviation is given as $\sigma_x = 10$.

The new observations, denoted by $y_i$, are obtained by the transformation:

$\qquad y_i = \frac{x_i - 5}{4}$

This transformation can be rewritten as:

$\qquad y_i = \frac{1}{4} x_i - \frac{5}{4}$

This is a linear transformation of the form $y_i = a + b x_i$, where $a = -\frac{5}{4}$ and $b = \frac{1}{4}$.

Calculating the New Mean After Transformation

The mean of a dataset is affected by both addition/subtraction and multiplication/division. If each observation $x_i$ is transformed to $y_i = a + b x_i$, the new mean $\bar{y}$ is related to the original mean $\bar{x}$ by the formula:

$\qquad \bar{y} = a + b \bar{x}$

In our case, $a = -\frac{5}{4}$ and $b = \frac{1}{4}$, and the original mean $\bar{x} = 50$.

Let's calculate the new mean:

$\qquad \bar{y} = -\frac{5}{4} + \left(\frac{1}{4}\right) \times 50$

$\qquad \bar{y} = -1.25 + \frac{50}{4}$

$\qquad \bar{y} = -1.25 + 12.5$

$\qquad \bar{y} = 11.25$

So, the new mean is 11.25.

Calculating the New Standard Deviation After Transformation

The standard deviation measures the spread or dispersion of the data. It is affected by multiplication and division, but not by addition or subtraction of a constant.

If each observation $x_i$ is transformed to $y_i = a + b x_i$, the new standard deviation $\sigma_y$ is related to the original standard deviation $\sigma_x$ by the formula:

$\qquad \sigma_y = |b| \times \sigma_x$

Note that the constant $a$ does not affect the standard deviation.

In our case, $b = \frac{1}{4}$ and the original standard deviation $\sigma_x = 10$.

Let's calculate the new standard deviation:

$\qquad \sigma_y = \left|\frac{1}{4}\right| \times 10$

$\qquad \sigma_y = \frac{1}{4} \times 10$

$\qquad \sigma_y = \frac{10}{4}$

$\qquad \sigma_y = 2.5$

So, the new standard deviation is 2.5.

Summary of Results

After subtracting 5 from each observation and dividing by 4:

• The new mean is 11.25.
• The new standard deviation is 2.5.

Therefore, the new mean and the new standard deviation are 11.25 and 2.5 respectively.

Revision Table: Key Concepts in Data Transformation

Additional Information: Why Standard Deviation is Unaffected by Addition/Subtraction

The standard deviation is calculated based on the deviations of each observation from the mean, i.e., $(x_i - \bar{x})$. When a constant $c$ is added to each observation, the new observation is $x'_i = x_i + c$. The new mean is $\bar{x}' = \bar{x} + c$. The new deviation is $(x'_i - \bar{x}') = (x_i + c) - (\bar{x} + c) = x_i + c - \bar{x} - c = x_i - \bar{x}$.

Since the deviations remain the same, the sum of squared deviations $\sum (x_i - \bar{x})^2$ and thus the variance and standard deviation also remain unchanged.

When multiplied by a constant $k$, the new observation is $x'_i = kx_i$ and the new mean is $\bar{x}' = k\bar{x}$. The new deviation is $(x'_i - \bar{x}') = kx_i - k\bar{x} = k(x_i - \bar{x})$. The new variance is $\sigma_{x'}^2 = \frac{\sum (k(x_i - \bar{x}))^2}{n-1} = \frac{\sum k^2 (x_i - \bar{x})^2}{n-1} = k^2 \frac{\sum (x_i - \bar{x})^2}{n-1} = k^2 \sigma_x^2$. Taking the square root, the new standard deviation is $\sigma_{x'} = \sqrt{k^2 \sigma_x^2} = |k|\sigma_x$. The absolute value is used because standard deviation is always non-negative.