Let the observations xi (1 ≤ i ≤ 10) satisfy the equations, Σ(xi - 5) = 10 and Σ(xi - 5)² = 40. If μ

Q: Wait, why can't I just solve for each $x_i$ individually? There are 10 variables and 2 equations.

That is a classic trap! You have 10 unknowns but only 2 equations. It is mathematically impossible to find the individual values of $x_1, x_2, \dots, x_{10}$. The beauty of statistics is that we don't need the individual values to find the mean and variance. We only need the sum of the observations and the sum of their squares. Trust the properties of the mean and variance!

Q: Does variance really not change when we subtract 3 from every data point?

Yes, absolutely! Think of variance as a measure of how spread out the data points are from their mean. If you shift every single data point by the same amount (subtracting 3), the entire 'cloud' of data moves, but the distance between the points remains exactly the same. Therefore, the spread—and the variance—remains unchanged.

Q: What if the question asked for the variance of $2x_i - 3$ instead? Would it still be 3?

Great question! No, it would not. While variance is independent of the change of origin (the '-3' part), it is NOT independent of the change of scale (the '2' part). If you multiply every observation by a constant $k$, the variance is multiplied by $k^2$. So, the variance of $2x_i - 3$ would be $2^2 \times \text{Var}(x) = 4 \times 3 = 12$.

Question

Let the observations $x_i (1 \le i \le 10)$ satisfy the equations, $\sum_{i=1}^{10} (x_i - 5) = 10$ and $\sum_{i=1}^{10} (x_i - 5)^2 = 40$. If $\mu$ and $\lambda$ are the mean and the variance of observations, $(x_1 - 3), (x_2 - 3) \dots (x_{10} - 3)$, then the ordered pair $(\mu, \lambda)$ is equal to:

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a field with ten markers placed at various positions. You are told the sum of their positions relative to a central point $5$ is $10$, and the sum of the squares of these relative positions is $40$.

We begin by defining a new variable to simplify our calculations. Let $u_i = x_i - 5$. This represents our shift of origin.

The problem provides the following sums for $n = 10$:
$$\sum_{i=1}^{10} u_i = 10$$
$$\sum_{i=1}^{10} u_i^2 = 40$$

### Decoding the Statistical Parameters

The mean of $u$ is calculated as:
$$\bar{u} = \frac{\sum u_i}{n} = \frac{10}{10} = 1$$

For the variance of $u$, we use the fundamental formula $	ext{Var}(u) = \frac{\sum u_i^2}{n} - (\bar{u})^2$. Substituting our known values:
$$	ext{Var}(u) = \frac{40}{10} - (1)^2 = 4 - 1 = 3$$

This value represents the spread of our data relative to the point $5$.

### The Mean's Sensitivity

Now, let us find the mean of our original data, $x_i$. Since $u_i = x_i - 5$, it follows that $x_i = u_i + 5$.

By the linearity of the mean:
$$\bar{x} = \bar{u} + 5$$
$$\bar{x} = 1 + 5 = 6$$

The mean is sensitive; it follows the shift of the origin perfectly. If you shift the data, the mean shifts with it.

### The Variance's Stoicism

We want the mean and variance of $y_i = x_i - 3$. Let us look at the variance first.

The variance is a measure of dispersion—how far the points are from each other. If you subtract $3$ from every single data point, you are simply sliding the entire distribution along the number line. The relative distances between the points do not change.

Thus, the variance remains unchanged:
$$	ext{Var}(x - 3) = 	ext{Var}(x) = 3$$

The variance is the stoic observer; it remains unmoved by the change of origin.

### The Final Synthesis

Finally, we calculate the new mean $\mu$ for our observations $y_i = x_i - 3$. Using the property of the mean:
$$\bar{y} = \bar{x} - 3$$
$$\mu = 6 - 3 = 3$$

We have found our values: the new mean $\mu = 3$ and the new variance $\sigma^2 = 3$. The ordered pair $(\mu, \sigma^2)$ is **$(3, 3)$**.

Through this journey, we have seen that while the mean is a follower, shifting with the data, the variance is a leader, standing firm regardless of the shift. Keep this distinction in your toolkit, and no statistics problem will ever intimidate you again.

Enter Numerical Value:

Visualized Solution

Understanding the Given Data

Defining the Deviation Variable $u_{i}$

Calculating the Mean of $u_{i}$

Calculating the Variance of $u_{i}$

Finding the Mean of $x_{i}$

Variance Property: Change of Origin

Defining the New Observations $y_{i}$

Finding the New Mean $μ$

Finding the New Variance $λ$

Final Result

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

Decoding the Statistical Parameters

The Mean's Sensitivity

The Variance's Stoicism

The Final Synthesis

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Enter Numerical Value:

Visualized Solution

Understanding the Given Data

Defining the Deviation Variable .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ui​

Calculating the Mean of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ui​

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

Finding the Mean of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xi​