Question

A population has a \(\mu=50\) and \(\sigma=10\). If these scores are transformed into z-scores, the population of z-scores will have a mean and standard deviation of

• A. \(\mu = 50\) and \(\sigma = 10\)
• B. \(\mu = 50\) and \(\sigma = 1.96\)
• C. \(\mu = 1\) and \(\sigma = 0\)
• D. \(\mu = 0\) and \(\sigma = 1\)

Hint:

Standardization (z-transformation) always produces a standardized normal distribution with \(\mu = 0\) and \(\sigma = 1\), regardless of the original mean and standard deviation.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Standardizing a set of scores involves transforming each score into a z-score.
This process centers the distribution at 0 and scales the variance to 1.
Detailed Explanation:
Let us mathematically demonstrate the mean and standard deviation of any standardized distribution:
Let \( X \) be a variable with mean \( \mu \) and standard deviation \( \sigma \).
The z-score transformation is defined as:
\[ z = \frac{X - \mu}{\sigma} \] 1. Expected Value (Mean) of the standardized variable \( z \):
\[ E[z] = E\left[ \frac{X - \mu}{\sigma} \right] = \frac{E[X] - \mu}{\sigma} = \frac{\mu - \mu}{\sigma} = 0 \] The mean of any z-score distribution is always \( 0 \).
2. Variance of the standardized variable \( z \):
\[ \text{Var}(z) = \text{Var}\left[ \frac{X - \mu}{\sigma} \right] = \frac{1}{\sigma^2} \text{Var}(X - \mu) = \frac{1}{\sigma^2} \text{Var}(X) = \frac{\sigma^2}{\sigma^2} = 1 \] Since the variance is \( 1 \), the standard deviation is also \( \sqrt{1} = 1 \).
This outcome is constant regardless of the original population parameters.
Therefore, the transformed population of z-scores always has a mean of 0 and a standard deviation of 1.

Step 2: Final Answer:
The standardized population will have a mean of 0 and a standard deviation of 1.