Let X = {x ∈ N : 1 ≤ x ≤ 19} and for some a, b ∈ R, Y = {ax + b : x ∈ X}. If the mean and variance o

Q: Wait, why does the variance not change when we add $b$?

That is a fantastic question! Think of variance as a measure of the 'spread' or the average squared distance of data points from their mean. When you add a constant $b$ to every single data point, you are essentially shifting the entire set along the number line. The distance between any two points $x_i$ and $x_j$ remains $(x_i + b) - (x_j + b) = x_i - x_j$. Since the relative distances between points are preserved, the spread—and therefore the variance—remains completely unaffected by the shift $b$.

Q: Why is the variance formula $\sigma_y^2 = a^2 \sigma_x^2$ and not $a \sigma_x^2$?

Remember that variance is defined as the average of the *squared* deviations from the mean. If you multiply every element by $a$, the deviation $(x_i - \bar{x})$ becomes $(ax_i - a\bar{x}) = a(x_i - \bar{x})$. When you square this new deviation to calculate the variance, you get $a^2(x_i - \bar{x})^2$. Thus, the variance is scaled by the square of the factor $a$. It is a common trap to forget the square, so always keep that in mind!

Q: Why do we get two values for $a$? Is one of them wrong?

Neither is wrong! When we solve $a^2 = 25$, we are looking for all real numbers that, when squared, yield 25. Both $5$ and $-5$ satisfy this condition. In the context of linear transformations, a positive $a$ stretches the data while maintaining its order, whereas a negative $a$ stretches the data and reflects it across the mean. Both are perfectly valid transformations that result in the same variance. Always solve for both roots to ensure you capture all possible values of $b$.

Question

Let $X = \{x \in \mathbb{N} : 1 \le x \le 19\}$ and for some $a, b \in \mathbb{R}, Y = \{ax + b : x \in X\}$. If the mean and variance of the elements of $Y$ are 30 and 750, respectively, then the sum of all possible values of $b$ is

Options:

(A) 60

(B) 100

(C) 80

(D) 20

Sigma Percentile · Accepted Answer

### The Geometry of Data: A Journey Through Linear Transformations

Imagine you are standing on a number line, looking at a collection of nineteen points: the set $X = \{1, 2, 3, \dots, 19\}$. These points are perfectly ordered, like soldiers in a line.

Our goal is to understand what happens when we subject this set to a linear transformation, $Y = aX + b$. This isn't just an algebraic exercise; it is a study of how data behaves when it is stretched, flipped, and shifted.

### Phase 1: The Heartbeat of Set $X$

Before we can transform our set, we must understand its nature. We need two vital statistics: the mean (the center of gravity) and the variance (the measure of spread).

For the first $n$ natural numbers, the mean is given by:

$$\bar{x} = \frac{n+1}{2}$$

With $n = 19$, we find $\bar{x} = \frac{19+1}{2} = 10$. This is the anchor of our set.

Next, we look at the variance, $\sigma_x^2$. The formula for the variance of the first $n$ natural numbers is:

$$\sigma_x^2 = \frac{n^2-1}{12}$$

Plugging in our value of $n = 19$, we calculate:

$$\sigma_x^2 = \frac{19^2-1}{12} = \frac{361-1}{12} = \frac{360}{12} = 30$$

Now we know the soul of our set: it is centered at $10$ with a spread of $30$.

### Phase 2: The Transformation Dance

Now, we introduce the transformation $Y = aX + b$. This is where the magic happens. When we transform our data, the mean and variance respond in very specific, elegant ways.

The mean is sensitive to both the scaling factor $a$ and the shift $b$, following the rule $\bar{y} = a\bar{x} + b$. We are given that the new mean $\bar{y} = 30$, so we have our first equation:

$$10a + b = 30$$

But what about the variance? This is the trap where many students stumble. The shift $b$ does not change the spread of the data—it just moves the whole cluster.

However, the scaling factor $a$ stretches the distances between points, and because variance involves squared distances, it scales by $a^2$. Thus, $\sigma_y^2 = a^2 \sigma_x^2$. Given $\sigma_y^2 = 750$ and $\sigma_x^2 = 30$, we get:

$$750 = a^2 \cdot 30$$

### Phase 3: Solving the Mystery

With our two equations, the path is clear. First, we solve for $a$:

$$a^2 = \frac{750}{30} = 25$$

This gives us two possibilities: $a = 5$ or $a = -5$. Do not discard the negative root! A negative $a$ simply means the set has been reflected across the mean, which is a perfectly valid transformation.

Now, we find the corresponding values of $b$:

1. **Case 1 ($a = 5$):** Substituting into $10a + b = 30$, we get $10(5) + b = 30$, which simplifies to $50 + b = 30$, yielding $b = -20$.

2. **Case 2 ($a = -5$):** Substituting into $10a + b = 30$, we get $10(-5) + b = 30$, which simplifies to $-50 + b = 30$, yielding $b = 80$.

### The Final Celebration

The problem asks for the sum of all possible values of $b$. We have found $b = -20$ and $b = 80$.

Adding these together, we get:

$$-20 + 80 = 60$$

We have navigated the transformation, respected the properties of variance, and arrived at the **final answer of 60**. Statistics is not just about formulas; it is about understanding how data moves through space.

Animated Solution for Mathematics - Statistics: Let $X = {x \in N : 1 \leq x \leq 19}$ and for some $a, b \in R, Y = {a x + b : x \in X}$ . If the mean and variance of the elements of $Y$ are 30 and 750, respectively, then the sum of all possible values of $b$ is

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Visualized Solution

Analyze Set $X$ and Find its Mean

Calculate the Variance of Set $X$

Introduce Transformation $Y = a X + b$

Apply Mean Property for Transformation

Apply Variance Property for Transformation

Solve for Parameter $a$

Case 1: Find $b$ when $a = 5$

Case 2: Find $b$ when $a = - 5$

Final Step: Sum of all values of $b$

The Sigma Insight: Variance and Standard Deviation

The Geometry of Data

Phase 1

Phase 2

Phase 3

The Final Celebration

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Visualized Solution

Analyze Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X and Find its Mean

Calculate the Variance of Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X

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