1. Key Concepts and Formulas

• Mean and Variance of a Discrete Data Set: For a set of $N$ observations $\{x_1, x_2, \dots, x_N\}$:
  • Mean ($E[X]$ or $\mu_X$): $E[X] = \frac{\sum_{i=1}^{N} x_i}{N}$
  • Variance ($\text{Var}(X)$ or $\sigma_X^2$): $\text{Var}(X) = E[X^2] - (E[X])^2 = \frac{\sum_{i=1}^{N} x_i^2}{N} - \left(\frac{\sum_{i=1}^{N} x_i}{N}\right)^2$
• Properties of Mean and Variance under Linear Transformation: If a new variable $Y$ is linearly related to $X$ by $Y = aX + b$ (where $a$ and $b$ are constants):
  • Mean of Y: $E[Y] = aE[X] + b$
  • Variance of Y: $\text{Var}(Y) = a^2\text{Var}(X)$ This property is crucial as it allows us to relate the statistics of $Y$ back to $X$ without listing all elements of $Y$.
• Summation Formulas for Natural Numbers: For the first $N$ natural numbers:
  • Sum: $\sum_{k=1}^{N} k = \frac{N(N+1)}{2}$
  • Sum of Squares: $\sum_{k=1}^{N} k^2 = \frac{N(N+1)(2N+1)}{6}$

2. Step-by-Step Solution

We are given the set $X = \{x \in \mathbb{N} : 1 \le x \le 17\}$, which means $X = \{1, 2, 3, \dots, 17\}$. The number of elements in $X$ is $N = 17$. The set $Y$ is defined by the linear transformation $Y = \{ax + b : x \in X \text{ and } a, b \in \mathbb{R}, a > 0\}$. We are provided with the mean of $Y$, $E[Y] = 17$, and the variance of $Y$, $\text{Var}(Y) = 216$. Our objective is to find the value of $a+b$.

Step 1: Calculate the Mean of X ($E[X]$)

• What we are doing: We need to find the mean of the original data set $X$. This is the first piece of information required from set $X$ to use the linear transformation property for the mean.
• Why we are doing it: The relationship $E[Y] = aE[X] + b$ requires the value of $E[X]$.
• Calculation: The sum of the first $N=17$ natural numbers is given by the formula $\sum_{k=1}^{N} k = \frac{N(N+1)}{2}$. $\sum_{x=1}^{17} x = \frac{17(17+1)}{2} = \frac{17 \times 18}{2} = 17 \times 9 = 153$ Now, we calculate the mean of $X$: $E[X] = \frac{\sum_{x=1}^{17} x}{N} = \frac{153}{17} = 9$ So, $E[X] = 9$.

Step 2: Calculate the Variance of X ($\text{Var}(X)$)

• What we are doing: We need to find the variance of the original data set $X$. This is the second crucial piece of information from set $X$ to use the linear transformation property for the variance.
• Why we are doing it: The relationship $\text{Var}(Y) = a^2\text{Var}(X)$ requires the value of $\text{Var}(X)$. We will use the formula $\text{Var}(X) = E[X^2] - (E[X])^2$. We already have $E[X]=9$, so we first need to calculate $E[X^2]$.
• Calculation: First, find the sum of the squares of the elements in $X$ using the formula $\sum_{k=1}^{N} k^2 = \frac{N(N+1)(2N+1)}{6}$: $\sum_{x=1}^{17} x^2 = \frac{17(17+1)(2 \times 17 + 1)}{6} = \frac{17 \times 18 \times (34+1)}{6} = \frac{17 \times 18 \times 35}{6}$ $\sum_{x=1}^{17} x^2 = 17 \times 3 \times 35 = 1785$ Now, calculate $E[X^2]$: $E[X^2] = \frac{\sum_{x=1}^{17} x^2}{N} = \frac{1785}{17} = 105$ Finally, calculate $\text{Var}(X)$: $\text{Var}(X) = E[X^2] - (E[X])^2 = 105 - (9)^2 = 105 - 81 = 24$ So, $\text{Var}(X) = 24$.

Step 3: Use the Given Mean of Y to Form an Equation

• What we are doing: We are using the known mean of $Y$ and the calculated mean of $X$ to establish an equation involving the unknown constants $a$ and $b$.
• Why we are doing it: This step directly applies one of the key linear transformation properties to relate the given information to our unknowns.
• Calculation: Using the property $E[Y] = aE[X] + b$: $17 = a(9) + b$ $9a + b = 17 \quad \text{(Equation 1)}$

Step 4: Use the Given Variance of Y to Form Another Equation and Find 'a'

• What we are doing: We are using the known variance of $Y$ and the calculated variance of $X$ to directly solve for the constant $a$.
• Why we are doing it: The variance transformation property $\text{Var}(Y) = a^2\text{Var}(X)$ is particularly useful because the constant $b$ does not affect the variance, allowing us to isolate and solve for $a$ independently.
• Calculation: Using the property $\text{Var}(Y) = a^2\text{Var}(X)$: $216 = a^2(24)$ Solve for $a^2$: $a^2 = \frac{216}{24} = 9$ Taking the square root, we get $a = \pm \sqrt{9} = \pm 3$. The problem statement specifies that $a > 0$. Therefore, we choose the positive value: $a = 3$

Step 5: Solve for $b$ and then Calculate $a+b$

• What we are doing: Now that we have the value of $a$, we substitute it into Equation 1 to find $b$. Finally, we calculate the sum $a+b$.
• Why we are doing it: This completes the process of finding both unknown constants and then computes the desired final value.
• Calculation: Substitute $a=3$ into Equation 1: $9(3) + b = 17$ $27 + b = 17$ $b = 17 - 27$ $b = -10$ Finally, calculate $a+b$: $a+b = 3 + (-10) = -7$

3. Common Mistakes & Tips

• Misapplying Variance Property: A common error is to write $\text{Var}(aX+b) = a\text{Var}(X)+b$ or $a^2\text{Var}(X)+b^2$. Remember that $\text{Var}(aX+b) = a^2\text{Var}(X)$ because the additive constant $b$ only shifts the data, not its spread.
• Forgetting $a>0$ Condition: When solving $a^2 = k$, always consider both positive and negative roots. The problem statement often provides a condition (like $a>0$) to uniquely determine the value of $a$.
• Calculation Errors in Summation Formulas: Be careful with arithmetic when using the summation formulas for natural numbers and their squares. Double-check your calculations, especially with larger numbers.
• Confusing $E[X^2]$ and $(E[X])^2$: Ensure you correctly apply the variance formula $\text{Var}(X) = E[X^2] - (E[X])^2$. These two terms are distinct and rarely equal.

4. Summary

This problem effectively demonstrates the power of using properties of mean and variance under linear transformations. Instead of calculating the mean and variance of the transformed set $Y$ directly, we first computed the mean and variance of the simpler base set $X$. Then, by applying the transformation rules $E[Y] = aE[X] + b$ and $\text{Var}(Y) = a^2\text{Var}(X)$, we set up a system of equations. Solving these equations, along with the given condition $a>0$, allowed us to find the values of $a$ and $b$, and subsequently their sum.

The final answer is $\boxed{-7}$, which corresponds to option (C).