The mean and variance of 7 observations are 8 and 16 respectively. If two observations are 6 and 8, then the variance of the remaining 5 observations is :
Show Hint
When adding or removing observations, always track \(\sum x_i\) and \(\sum x_i^2\). These two quantities allow you to reconstruct the mean and variance for any new subset of the data.
Step 1: Understanding the Concept:
We are given the mean and variance for a set of 7 data points. By identifying specific values and their contribution to the total sum and total sum of squares, we can find the statistics for the subset of remaining observations. Step 2: Key Formula or Approach:
1. Mean \(\bar{x} = \frac{\sum x_i}{n}\).
2. Variance \(\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2\). Step 3: Detailed Explanation:
For \(n=7\): \(\bar{x} = 8\) and \(\sigma^2 = 16\).
Total sum \(\sum_{i=1}^7 x_i = 7 \times 8 = 56\).
Total sum of squares \(\frac{\sum x_i^2}{7} - 8^2 = 16 \implies \sum_{i=1}^7 x_i^2 = 7 \times (16 + 64) = 7 \times 80 = 560\).
Let the remaining 5 observations be \(y_1, y_2, ..., y_5\). Given \(x_6 = 6\) and \(x_7 = 8\).
Sum of remaining 5 observations: \(\sum y_j = 56 - (6 + 8) = 56 - 14 = 42\).
New Mean \(\bar{y} = \frac{42}{5}\).
Sum of squares of remaining 5 observations: \(\sum y_j^2 = 560 - (6^2 + 8^2) = 560 - (36 + 64) = 460\).
New Variance \(\sigma_{new}^2 = \frac{\sum y_j^2}{5} - (\bar{y})^2\):
\[ \sigma_{new}^2 = \frac{460}{5} - \left(\frac{42}{5}\right)^2 = 92 - \frac{1764}{25} \]
\[ \sigma_{new}^2 = \frac{92 \times 25 - 1764}{25} = \frac{2300 - 1764}{25} = \frac{536}{25} \] Step 4: Final Answer:
The variance of the remaining 5 observations is \(\frac{536}{25}\).
