Key Concepts and Formulas

• Mean ($\bar{x}$): The arithmetic average of a set of observations. For $n$ observations $x_1, x_2, \ldots, x_n$, it is given by: $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$
• Variance ($V$ or $\sigma^2$): A measure of the dispersion of observations around their mean. It is the average of the squared differences from the mean. A computationally convenient formula is: $V = \frac{\sum_{i=1}^{n} x_i^2}{n} - (\bar{x})^2$
• Mean Deviation about the Mean (MD): The average of the absolute differences between each observation and the mean. It quantifies the average absolute distance of data points from the central value. $MD(\bar{x}) = \frac{\sum_{i=1}^{n} |x_i - \bar{x}|}{n}$

Step-by-Step Solution

Step 1: Use the given mean to establish a relationship between $x$ and $y$. We are provided with 6 observations: $1, 2, 4, 5, x,$ and $y$. Their mean is given as 5. The definition of the mean helps us form our first equation.

• Why: The mean is the sum of all observations divided by the number of observations. By using the given mean, we can find the sum of the unknown variables.
• Calculation: $\bar{x} = \frac{1 + 2 + 4 + 5 + x + y}{6}$ Substitute the given mean $\bar{x} = 5$: $5 = \frac{12 + x + y}{6}$ Multiply both sides by 6: $30 = 12 + x + y$ Subtract 12 from both sides to isolate $x+y$: $x + y = 18 \quad \ldots(1)$ This equation provides the sum of the two unknown observations.

Step 2: Use the given variance to establish another relationship between $x$ and $y$. The variance of the observations is given as 10. We will use the computational formula for variance, as it is generally more efficient when the mean is already known.

• Why: Variance quantifies the spread of data. By using the given variance and mean, we can determine the sum of the squares of all observations, leading to an equation for $x^2+y^2$.
• Calculation: $V = \frac{\sum x_i^2}{n} - (\bar{x})^2$ Substitute the given variance $V = 10$, number of observations $n = 6$, and mean $\bar{x} = 5$: $10 = \frac{1^2 + 2^2 + 4^2 + 5^2 + x^2 + y^2}{6} - (5)^2$ $10 = \frac{1 + 4 + 16 + 25 + x^2 + y^2}{6} - 25$ $10 = \frac{46 + x^2 + y^2}{6} - 25$ Add 25 to both sides: $35 = \frac{46 + x^2 + y^2}{6}$ Multiply both sides by 6: $210 = 46 + x^2 + y^2$ Subtract 46 from both sides to isolate $x^2+y^2$: $x^2 + y^2 = 164 \quad \ldots(2)$ This equation provides the sum of the squares of the two unknown observations.

Step 3: Solve the system of equations to find the values of $x$ and $y$. We now have a system of two equations:

1. $x + y = 18$
2. $x^2 + y^2 = 164$

• Why: With two independent equations involving $x$ and $y$, we can solve for their individual values. We can use algebraic identities to simplify this process.
• Calculation: Recall the algebraic identity: $(x+y)^2 = x^2 + y^2 + 2xy$. Substitute the values from equations (1) and (2) into this identity: $(18)^2 = 164 + 2xy$ $324 = 164 + 2xy$ Subtract 164 from both sides: $160 = 2xy$ Divide by 2: $xy = 80$ Now we have two pieces of information: $x+y=18$ and $xy=80$. We know that $x$ and $y$ are the roots of a quadratic equation of the form $t^2 - (x+y)t + xy = 0$. Substituting the values: $t^2 - 18t + 80 = 0$ To solve this quadratic equation, we can factor it. We need two numbers that multiply to 80 and add up to -18. These numbers are -8 and -10. $(t - 8)(t - 10) = 0$ This yields two possible values for $t$: $t = 8$ or $t = 10$. Therefore, the two unknown observations are $8$ and $10$. (It does not matter which is $x$ and which is $y$, as the set of observations remains the same.) Our complete set of observations is now: $1, 2, 4, 5, 8, 10$.

Step 4: Calculate the Mean Deviation about the Mean. With all observations identified and the mean known, we can now calculate the mean deviation about the mean.

• Why: The mean deviation measures the average absolute distance of each data point from the mean. We need all observations to perform this calculation.
• Calculation: The observations are $x_i \in \{1, 2, 4, 5, 8, 10\}$. The mean is $\bar{x} = 5$. Calculate the absolute difference $|x_i - \bar{x}|$ for each observation:
  • $|1 - 5| = |-4| = 4$
  • $|2 - 5| = |-3| = 3$
  • $|4 - 5| = |-1| = 1$
  • $|5 - 5| = |0| = 0$
  • $|8 - 5| = |3| = 3$
  • $|10 - 5| = |5| = 5$ Sum these absolute differences: $\sum |x_i - \bar{x}| = 4 + 3 + 1 + 0 + 3 + 5 = 16$ Finally, apply the Mean Deviation formula: $MD(\bar{x}) = \frac{\sum |x_i - \bar{x}|}{n} = \frac{16}{6}$ Simplify the fraction: $MD(\bar{x}) = \frac{8}{3}$

Common Mistakes & Tips

• Absolute Values in MD: A common error is forgetting to take the absolute value of the differences $(x_i - \bar{x})$ when calculating Mean Deviation.
• Algebraic Accuracy: Be careful with arithmetic and algebraic manipulations, especially when solving the system of equations for $x$ and $y$. A small error can propagate.
• Verification: After finding $x$ and $y$, it's good practice to quickly verify if they satisfy the initial conditions for mean and variance. For instance, sum $1,2,4,5,8,10 = 30$, so mean is $30/6=5$ (Correct). Sum of squares $1^2+2^2+4^2+5^2+8^2+10^2 = 1+4+16+25+64+100 = 210$. Variance is $210/6 - 5^2 = 35 - 25 = 10$ (Correct).

Summary

We began by utilizing the given mean and variance to form a system of two equations involving the unknown observations $x$ and $y$. Solving this system, we determined that the unknown observations are $8$ and $10$. With all six observations now known ($1, 2, 4, 5, 8, 10$), we proceeded to calculate the mean deviation about the mean. This involved finding the absolute difference of each observation from the mean, summing these differences, and dividing by the total number of observations. The calculated mean deviation is $\frac{8}{3}$.

The final answer is $\boxed{\frac{8}{3}}$, which corresponds to option (B).