Question

A population has a mean (\(\mu\)) of 45 and a standard deviation (\(\sigma\)) of 8. After 5 points are added to every score in the population, what are the new values for the mean and standard deviation?

• A. \(\mu = 45\) and \(\sigma = 8\)
• B. \(\mu = 45\) and \(\sigma = 13\)
• C. \(\mu = 50\) and \(\sigma = 8\)
• D. \(\mu = 50\) and \(\sigma = 13\)

Hint:

- Adding or subtracting a constant shifts the mean by that constant but leaves the standard deviation unchanged.
- Multiplying or dividing by a constant scales both the mean and the standard deviation by that factor.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem addresses how linear transformations (specifically, adding a constant) affect measures of central tendency (mean) and measures of dispersion (standard deviation).
Detailed Explanation:
Let \( X \) be a random variable representing the original population scores, with:
- \( \mu_X = 45 \)
- \( \sigma_X = 8 \)
Let \( Y \) be the transformed variable after adding a constant \( c = 5 \) to every score:
\[ Y = X + 5 \] 1. Effect on the Mean:
Using the linearity of expectation:
\[ \mu_Y = E[Y] = E[X + 5] = E[X] + 5 = \mu_X + 5 \] Substituting the given values:
\[ \mu_Y = 45 + 5 = 50 \] 2. Effect on the Standard Deviation:
Using the properties of variance:
\[ \sigma^2_Y = \text{Var}(Y) = \text{Var}(X + 5) = \text{Var}(X) = \sigma^2_X \] Adding a constant to every score shifts the entire distribution but does not change the distance between scores.
Therefore, the dispersion (standard deviation) remains unchanged:
\[ \sigma_Y = \sigma_X = 8 \] The new population parameters are \( \mu = 50 \) and \( \sigma = 8 \).

Step 2: Final Answer:
The new values are \(\mu = 50\) and \(\sigma = 8\).