To find the sample variance of the original dataset, we need to correct the mistakenly recorded value and then recalculate the variance.

Given:

• Mean ($\bar{x}$) of the dataset: 14
• Sample standard deviation (s): 10
• Number of observations (n): 9
• Incorrect value: 6 (should be 11)

First, let's determine the sum of the original dataset using the mean: $\bar{x} = \frac{\sum x_i}{n} \implies \sum x_i = \bar{x} \times n = 14 \times 9 = 126$

Now, let's find the sum of squares of the original dataset using the variance formula: $s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2 \implies \sum_{i=1}^{n} (x_i - \bar{x})^2 = s^2 \times (n-1) = 10^2 \times 8 = 800$

Next, we need to adjust this sum of squares to account for the incorrect value: $\text{Original sum of squares} = \sum_{i=1}^{9} x_i^2$

The incorrect dataset sum of squares can be expressed as: $\sum_{i=1}^{9} x_i^2 = 800 + n \cdot \bar{x}^2 = 800 + 9 \cdot 14^2 = 800 + 1764 = 2564$

Identify and correct the contribution of the incorrect value:

• Incorrect value: 6
• Correct value: 11

The difference in the squares due to the correction is: $11^2 - 6^2 = 121 - 36 = 85$

The corrected sum of squares: $\text{Corrected sum of squares} = 2564 + 85 = 2649$

Recalculate the sample variance with the corrected sum of squares: $s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i^2) = \frac{2649 - 126^2 / 9}{8} = \frac{2649 - 1764}{8} = \frac{885}{8} = 110.625$

Thus, the sample variance of the original dataset, corrected to two decimal places, is: $110.63$

Would you like further details or have any questions? Here are some related questions you might find useful:

1. How do you calculate the sample mean from a dataset?
2. What is the difference between sample variance and population variance?
3. How do you find the sum of squares in a dataset?
4. What steps are involved in correcting a dataset with an error?
5. How do changes in individual data points affect the overall variance?

Tip: Always double-check your dataset for accuracy before performing statistical calculations to ensure valid results.