Key Concepts and Formulas

To solve this problem, we rely on the fundamental definitions and formulas for the mean and variance of a dataset:

• Mean ($\bar{x}$): The average of $n$ observations $x_1, x_2, \ldots, x_n$. $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$ From this, the sum of observations can be expressed as $\sum x_i = n \bar{x}$.

• Variance ($\sigma^2$): A measure of the spread of data points around the mean. The most computationally convenient formula for variance is: $\sigma^2 = \frac{\sum_{i=1}^{n} x_i^2}{n} - \left(\frac{\sum_{i=1}^{n} x_i}{n}\right)^2$ This can be concisely written as: $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ This formula is particularly useful for correction problems, as it directly relates variance to the sum of observations ($\sum x_i$) and the sum of squares of observations ($\sum x_i^2$). We can rearrange it to find the sum of squares: $\sum x_i^2 = n \left( \sigma^2 + (\bar{x})^2 \right)$

Step-by-Step Solution

Our strategy involves using the initial (incorrect) mean and variance to find the initial sum of observations and sum of squares. Then, we will correct these sums based on the given errors. Finally, we will use the corrected sums to compute the new mean and variance.

Step 1: Calculate the Initial (Incorrect) Sum of Observations and Sum of Squares of Observations

We are given the following information for 12 observations:

• Number of observations, $n = 12$.
• Initial (incorrect) Mean, $\bar{x}_{\text{old}} = \frac{9}{2}$.
• Initial (incorrect) Variance, $\sigma^2_{\text{old}} = 4$.

We need to determine the initial sum of observations ($\sum x_{\text{old}}$) and the initial sum of squares of observations ($\sum x^2_{\text{old}}$) because these are the components that will be corrected.

• 1.1 Calculate the initial sum of observations ($\sum x_{\text{old}}$): Using the formula $\sum x_i = n \bar{x}$: $\sum x_{\text{old}} = n \times \bar{x}_{\text{old}} = 12 \times \frac{9}{2}$ $\sum x_{\text{old}} = 6 \times 9 = 54$ So, the initial sum of observations is 54.

• 1.2 Calculate the initial sum of squares of observations ($\sum x^2_{\text{old}}$): Using the rearranged variance formula $\sum x_i^2 = n \left( \sigma^2 + (\bar{x})^2 \right)$: $\sum x^2_{\text{old}} = 12 \left( \sigma^2_{\text{old}} + (\bar{x}_{\text{old}})^2 \right)$ Substituting the given values: $\sum x^2_{\text{old}} = 12 \left( 4 + \left(\frac{9}{2}\right)^2 \right)$ $\sum x^2_{\text{old}} = 12 \left( 4 + \frac{81}{4} \right)$ To combine the terms inside the parenthesis, we find a common denominator: $\sum x^2_{\text{old}} = 12 \left( \frac{16}{4} + \frac{81}{4} \right)$ $\sum x^2_{\text{old}} = 12 \left( \frac{97}{4} \right)$ $\sum x^2_{\text{old}} = 3 \times 97 = 291$ Thus, the initial sum of squares of observations is 291.

Step 2: Correct the Sum of Observations ($\sum x_{\text{new}}$)

The problem states that two observations were incorrectly recorded. To find the correct sum of observations, we subtract the values that were wrongly included and add the values that should have been included.

• Wrong observations: 9 and 10. Their sum is $9 + 10 = 19$.
• Correct observations: 7 and 14. Their sum is $7 + 14 = 21$.

The formula for the corrected sum is: $\sum x_{\text{new}} = \sum x_{\text{old}} - (\text{sum of wrong values}) + (\text{sum of correct values})$ $\sum x_{\text{new}} = 54 - (9 + 10) + (7 + 14)$ $\sum x_{\text{new}} = 54 - 19 + 21$ $\sum x_{\text{new}} = 35 + 21 = 56$ The corrected sum of observations is 56.

Step 3: Correct the Sum of Squares of Observations ($\sum x^2_{\text{new}}$)

Similarly, to correct the sum of squares, we subtract the squares of the wrong observations and add the squares of the correct observations. This is crucial for accurately calculating the new variance.

• Squares of wrong observations: $9^2 = 81$ and $10^2 = 100$. Their sum is $81 + 100 = 181$.
• Squares of correct observations: $7^2 = 49$ and $14^2 = 196$. Their sum is $49 + 196 = 245$.

The formula for the corrected sum of squares is: $\sum x^2_{\text{new}} = \sum x^2_{\text{old}} - (\text{sum of squares of wrong values}) + (\text{sum of squares of correct values})$ $\sum x^2_{\text{new}} = 291 - (9^2 + 10^2) + (7^2 + 14^2)$ $\sum x^2_{\text{new}} = 291 - (81 + 100) + (49 + 196)$ $\sum x^2_{\text{new}} = 291 - 181 + 245$ $\sum x^2_{\text{new}} = 110 + 245 = 355$ The corrected sum of squares of observations is 355.

Step 4: Calculate the Correct Mean ($\bar{x}_{\text{new}}$)

With the corrected sum of observations, we can now find the correct mean. The number of observations ($n=12$) remains unchanged.

$\bar{x}_{\text{new}} = \frac{\sum x_{\text{new}}}{n} = \frac{56}{12}$ Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 4: $\bar{x}_{\text{new}} = \frac{14}{3}$ The correct mean is $\frac{14}{3}$.

Step 5: Calculate the Correct Variance ($\sigma^2_{\text{new}}$)

Finally, we use the corrected sum of squares ($\sum x^2_{\text{new}}$), the number of observations ($n=12$), and the corrected mean ($\bar{x}_{\text{new}}$) to calculate the correct variance.

$\sigma^2_{\text{new}} = \frac{\sum x^2_{\text{new}}}{n} - (\bar{x}_{\text{new}})^2$ Substitute the calculated values: $\sigma^2_{\text{new}} = \frac{355}{12} - \left(\frac{14}{3}\right)^2$ $\sigma^2_{\text{new}} = \frac{355}{12} - \frac{196}{9}$ To subtract these fractions, we find the least common multiple (LCM) of the denominators 12 and 9, which is 36. $\sigma^2_{\text{new}} = \frac{355 \times 3}{12 \times 3} - \frac{196 \times 4}{9 \times 4}$ $\sigma^2_{\text{new}} = \frac{1065}{36} - \frac{784}{36}$ $\sigma^2_{\text{new}} = \frac{1065 - 784}{36}$ $\sigma^2_{\text{new}} = \frac{281}{36}$ The correct variance is $\frac{281}{36}$.

Step 6: Determine m, n, and their Sum

The problem states that the correct variance is $\frac{m}{n}$, where $m$ and $n$ are coprime. From our calculation, $\sigma^2_{\text{new}} = \frac{281}{36}$. So, $m = 281$ and $n = 36$.

To confirm they are coprime, we check their greatest common divisor.

• Factors of $36$: $1, 2, 3, 4, 6, 9, 12, 18, 36$.
• To check if 281 is prime, we can test divisibility by prime numbers up to $\sqrt{281} \approx 16.7$. Primes to check are 2, 3, 5, 7, 11, 13.
  • 281 is not divisible by 2 (odd).
  • Sum of digits $2+8+1=11$, not divisible by 3.
  • Does not end in 0 or 5, so not divisible by 5.
  • $281 = 7 \times 40 + 1$, not divisible by 7.
  • $281 = 11 \times 25 + 6$, not divisible by 11.
  • $281 = 13 \times 21 + 8$, not divisible by 13. Since 281 is not divisible by any of these primes, it is a prime number. As 281 is prime and 36 is not a multiple of 281, $m=281$ and $n=36$ are indeed coprime.

Finally, we calculate $m+n$: $m+n = 281 + 36 = 317$

Common Mistakes & Tips

• Squaring vs. Not Squaring: A very common mistake is to subtract/add the values of observations instead of their squares when correcting $\sum x^2$. Always remember to square the individual observations before adding or subtracting them for $\sum x^2$.
• Arithmetic with Fractions: Be meticulous when performing calculations involving fractions, especially when finding common denominators for addition or subtraction.
• Coprime Check: Don't forget to verify that $m$ and $n$ are coprime as specified in the question. This ensures the fraction is in its simplest form.

Summary

This problem required us to correct the mean and variance of a dataset after identifying errors in two observations. We started by calculating the initial (incorrect) sum of observations and sum of squares using the given mean and variance. Then, we systematically adjusted these sums by subtracting the incorrect values/squares and adding the correct values/squares. With the corrected sums, we re-calculated the mean and then the variance. The final correct variance was found to be $\frac{281}{36}$, leading to $m=281$ and $n=36$. Their sum, $m+n$, is $317$.

The final answer is $\boxed{317}$, which corresponds to option (A).