Question

If the standard deviation of six numbers \(x_1,x_2,x_3,x_4,x_5,x_6\) is \(4\), then the variance of \(2x_1+3,2x_2+3,2x_3+3,2x_4+3,2x_5+3,2x_6+3\) is

• A. \(64\)
• B. \(67\)
• C. \(16\)
• D. \(19\)
• E. \(8\)

Hint:

If every observation is multiplied by \(a\), the variance gets multiplied by \(a^2\). Adding or subtracting a constant does not change the variance.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem deals with the properties of variance and standard deviation when the data is transformed linearly. We need to understand how shifting (adding a constant) and scaling (multiplying by a constant) affect these measures of dispersion.
Step 2: Key Formula or Approach:
Let \(\sigma_x\) and \(\text{Var}(x)\) be the standard deviation and variance of a variable x.
If we have a new variable \(y = ax + b\), where a and b are constants, then:
The new standard deviation is \(\sigma_y = |a|\sigma_x\).
The new variance is \(\text{Var}(y) = a^2 \text{Var}(x)\).
Adding a constant (shifting) does not change the spread of the data, so it does not affect the variance or standard deviation. Multiplying by a constant scales the spread.
Step 3: Detailed Explanation:
We are given the standard deviation of the numbers \(x_1, \dots, x_6\). Let's denote this by \(\sigma_x\).
\[ \sigma_x = 4 \] The variance is the square of the standard deviation.
\[ \text{Var}(x) = \sigma_x^2 = 4^2 = 16 \] Now consider the new set of numbers \(y_i = 2x_i + 3\).
This is a linear transformation of the form \(y = ax+b\) with \(a=2\) and \(b=3\).
We want to find the variance of this new set, \(\text{Var}(y)\).
Using the property \(\text{Var}(ax+b) = a^2 \text{Var}(x)\):
\[ \text{Var}(y) = \text{Var}(2x+3) = 2^2 \text{Var}(x) \] \[ \text{Var}(y) = 4 \times \text{Var}(x) \] Substitute the variance of the original data:
\[ \text{Var}(y) = 4 \times 16 = 64 \] Step 4: Final Answer:
The variance of the new set of numbers is 64. This corresponds to option (A).