Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respe

Q: Wait, why does the variance stay the same when we subtract 3 or add 2? Shouldn't it change?

This is a classic point of confusion! Remember that variance measures the 'spread' or the average squared distance of data points from their mean. If you shift every single data point by the same constant $k$, the distance between any two points $x_i$ and $x_j$ remains $(x_i + k) - (x_j + k) = x_i - x_j$. Since the relative distances between points are unchanged, the dispersion remains identical. Thus, variance is invariant under translation.

Q: Can't I just take the average of the two variances to get the combined variance?

That is a very common trap! Averaging the variances only works if the means of the two sets are identical. When the means are different, the combined variance must account for the 'spread' created by the difference between the individual means and the overall combined mean. That is why we must use the sum of squares method to correctly calculate the combined variance.

Q: Why do we need to calculate the sum of squares? Is there a faster way?

While there is a formula for combined variance involving the difference of means, calculating the sum of squares is the most robust and fundamental way to avoid errors. It forces you to account for the mean and variance of each set explicitly, ensuring you don't miss the contribution of the mean difference to the total variance.

Question

Let sets $A$ and $B$ have 5 elements each. Let the mean of the elements in sets $A$ and $B$ be 5 and 8 respectively and the variance of the elements in sets $A$ and $B$ be 12 and 20 respectively. A new set $C$ of 10 elements is formed by subtracting 3 from each element of $A$ and adding 2 to each element of $B$. Then the sum of the mean and variance of the elements of $C$ is

Options:

(A) 40

(B) 32

(C) 38

(D) 36

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are diving into a problem that tests your fundamental grasp of statistics. It is not just about plugging numbers into formulas; it is about understanding how data behaves when we manipulate it.

Imagine you have two groups of students, Set A and Set B, each with five members. You know their average scores and how much those scores vary. Now, you decide to change their scores—subtracting 3 from everyone in Group A and adding 2 to everyone in Group B.

### The Transformation

The first step is to understand the effect of translation on our statistical measures. When you shift every element in a set by a constant $k$, the mean shifts by exactly that amount.

If the original mean of Set A is $\mu_A = 5$, then the new mean $\mu_{A'}$ becomes:
$$\mu_{A'} = 5 - 3 = 2$$

Similarly, for Set B, the new mean $\mu_{B'}$ becomes:
$$\mu_{B'} = 8 + 2 = 10$$

The crucial insight is that variance is a measure of how spread out the data is. If you move the entire group by a fixed amount, the relative distance between any two students remains exactly the same. Therefore, the variance is invariant under translation.

The variance of Set A remains:
$$\sigma_{A'}^2 = 12$$

The variance of Set B remains:
$$\sigma_{B'}^2 = 20$$

### The Combined Mean

Now that we have the new means, we need to find the mean of the combined Set C, which contains all 10 elements. The combined mean $\mu_C$ is simply the weighted average of the individual means.

Since both sets have 5 elements, it is a simple average:
$$\mu_C = \frac{5(2) + 5(10)}{10} = \frac{10 + 50}{10} = 6$$

### The Variance Trap

This is where many students stumble. You cannot simply average the variances. To find the combined variance, we must return to the definition:
$$\sigma^2 = \frac{\sum x_i^2}{n} - \mu^2$$

Rearranging this gives us the sum of squares:
$$\sum x_i^2 = n(\sigma^2 + \mu^2)$$

We calculate this for each set:
For Set A', $\sum (x_i')^2 = 5(12 + 2^2) = 5(12 + 4) = 80$.
For Set B', $\sum (y_i')^2 = 5(20 + 10^2) = 5(20 + 100) = 600$.

### Final Calculation

Now, we combine these to find the variance of the total set C. The combined variance is:
$$\sigma_C^2 = \frac{\sum (x_i')^2 + \sum (y_i')^2}{n_A + n_B} - \mu_C^2$$

Plugging in our values, we get:
$$\sigma_C^2 = \frac{80 + 600}{10} - 6^2 = 68 - 36 = 32$$

Finally, the question asks for the sum of the mean and variance of Set C:
$$6 + 32 = 38$$

You have navigated the complexity, avoided the traps, and arrived at the elegant solution. The final answer is **38**.

Select Answer:

Visualized Solution

Initial Data for Sets $A$ and $B$

Effect of Translation on Mean

Calculating New Means $μ_{A^{'}}$ and $μ_{B^{'}}$

Effect of Translation on Variance

Combined Mean of Set $C$

Formula for Sum of Squares

Sum of Squares for Set $A^{'}$

Sum of Squares for Set $B^{'}$

Calculating Combined Variance $σ_{C}^{2}$

Final Result: Sum of Mean and Variance

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Transformation

The Combined Mean

The Variance Trap

Final Calculation

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