Question

The variance of 240, 260, 270, 280 is ________.

• A. $\frac{475}{4}$
• B. $\frac{475}{2}$
• C. $\frac{475}{8}$
• D. $\frac{475}{16}$
• E. $\frac{875}{4}$

Hint:

Subtract a common constant to simplify calculations; variance remains unchanged.

Answer 1

Answer: (E)

Step 1: Understanding the Concept
Variance (\(\sigma^2\)) is a measure of how spread out a set of data is. It is the average of the squared differences from the Mean. To simplify calculations with large numbers, we can use the "step-deviation" or "coding" method.
Step 2: Key Formula or Approach
Let the data points be \(x_i\). The variance is given by:
\[ \sigma^2 = \frac{\sum (x_i - \bar{x})^2}{N} = \frac{\sum x_i^2}{N} - (\bar{x})^2 \] where \(\bar{x}\) is the mean and N is the number of data points.
Coding Method: If we transform the data by \(u_i = \frac{x_i - A}{h}\), where A is an assumed mean and h is a common factor, then the variance of x is related to the variance of u by:
\[ \sigma_x^2 = h^2 \sigma_u^2 = h^2 \left[ \frac{\sum u_i^2}{N} - (\bar{u})^2 \right] \] This method simplifies the arithmetic.
Step 3: Detailed Explanation
Method 1: Direct Calculation
1. Calculate the mean (\(\bar{x}\)).
The data points are 240, 260, 270, 280. N=4.
\[ \bar{x} = \frac{240 + 260 + 270 + 280}{4} = \frac{1050}{4} = 262.5 \] 2. Calculate the variance.
\[ \sigma^2 = \frac{(240-262.5)^2 + (260-262.5)^2 + (270-262.5)^2 + (280-262.5)^2}{4} \] \[ \sigma^2 = \frac{(-22.5)^2 + (-2.5)^2 + (7.5)^2 + (17.5)^2}{4} \] \[ \sigma^2 = \frac{506.25 + 6.25 + 56.25 + 306.25}{4} \] \[ \sigma^2 = \frac{875}{4} \] Method 2: Coding Method (Easier)
1. Transform the data.
The data points are 240, 260, 270, 280. A common factor is h=10. Let's choose an assumed mean A=260.
Let \(u_i = \frac{x_i - 260}{10}\).
\(u_1 = \frac{240-260}{10} = -2\)
\(u_2 = \frac{260-260}{10} = 0\)
\(u_3 = \frac{270-260}{10} = 1\)
\(u_4 = \frac{280-260}{10} = 2\)
The new data set (u) is \(-2, 0, 1, 2\).
2. Calculate the variance of u (\(\sigma_u^2\)).
Mean of u: \(\bar{u} = \frac{-2+0+1+2}{4} = \frac{1}{4}\).
\[ \sigma_u^2 = \frac{\sum u_i^2}{N} - (\bar{u})^2 = \frac{(-2)^2 + 0^2 + 1^2 + 2^2}{4} - \left(\frac{1}{4}\right)^2 \] \[ \sigma_u^2 = \frac{4+0+1+4}{4} - \frac{1}{16} = \frac{9}{4} - \frac{1}{16} = \frac{36}{16} - \frac{1}{16} = \frac{35}{16} \] 3. Calculate the variance of x (\(\sigma_x^2\)).
\[ \sigma_x^2 = h^2 \sigma_u^2 = (10)^2 \times \frac{35}{16} = 100 \times \frac{35}{16} = 25 \times \frac{35}{4} = \frac{875}{4} \] Step 4: Final Answer
The variance is \(\frac{875}{4}\).