If the mean and variance of five observations are 24/5 and 194/25 respectively and the mean of first

Q: Wait, can't I just subtract the variance of the fifth observation from the total variance?

That is a very common trap! Variance is a measure of spread, not a simple additive quantity. You cannot subtract variances directly because they depend on the mean, which changes when you remove an observation. You must work with the sum of squares, which is additive, to correctly partition the data.

Q: Why do we use the formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{X})^2$ instead of $\frac{\sum(x_i - \bar{X})^2}{n}$?

This is the 'computational formula.' It is much more efficient here because we do not know the individual values of $x_1, x_2, x_3, x_4$. We only know their aggregate properties. This formula allows us to work with the sum of squares directly, which is the bridge between the two sets of data.

Q: Is there a faster way to do this without calculating the sum of squares?

While it might seem tedious, calculating the sum of squares is the most robust method. In JEE Advanced problems, you are often testing your ability to manipulate these aggregate statistical properties. Trust the process—once you find the sum of squares, the rest of the problem collapses into a simple arithmetic step.

Question

If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively and the mean of first four observations is $\frac{7}{2}$, then the variance of the first four observations in equal to

Options:

(A) $\frac{4}{5}$

(B) $\frac{77}{12}$

(C) $\frac{5}{4}$

(D) $\frac{105}{4}$

Sigma Percentile · Accepted Answer

### The Detective Work of Statistics

Welcome, future engineer! Today, we are going to peel back the layers of a statistical problem that, at first glance, might seem like a dry collection of numbers. Imagine you are a detective with a set of five mysterious observations, and you are given clues about their collective behavior—their mean and their variance.

Your mission is to uncover the hidden properties of just the first four observations. This is not just algebra; this is the art of data reconstruction.

### Phase 1: The Mean as a Sum

We have five observations: $x_1, x_2, x_3, x_4, x_5$. The problem provides the mean of all five as $\bar{X}_5 = \frac{24}{5}$.

Recall the definition of the mean: the sum of all observations divided by the count. Mathematically, this is expressed as:

$$\bar{X} = \frac{\sum x_i}{n}$$

When we multiply the mean by the number of observations, we get the total sum. For our five observations, the sum is:

$$5 	imes \frac{24}{5} = 24$$

This is our first major clue, which we will call Equation 1:

$$\sum_{i=1}^5 x_i = 24$$

Now, consider the first four observations. Their mean is $\bar{X}_4 = \frac{7}{2}$. Using the same logic, their sum is:

$$4 	imes \frac{7}{2} = 14$$

This is Equation 2:

$$\sum_{i=1}^4 x_i = 14$$

The difference between the total sum and the partial sum must be the fifth observation. Subtracting Equation 2 from Equation 1, we find:

$$x_5 = 24 - 14 = 10$$

We have successfully identified the "fifth element" of our data set.

### Phase 2: The Variance Mystery

The variance $\sigma^2$ is defined as the average of the squared deviations from the mean. To calculate this efficiently, we use the "computational formula":

$$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{X})^2$$

This formula allows us to work with the sum of squares, $\sum x_i^2$, which is much easier to manipulate. For our five observations, we have:

$$\frac{\sum_{i=1}^5 x_i^2}{5} - \left(\frac{24}{5}ight)^2 = \frac{194}{25}$$

Expanding the square of the mean, we get $\frac{576}{25}$. Our equation becomes:

$$\frac{\sum_{i=1}^5 x_i^2}{5} - \frac{576}{25} = \frac{194}{25}$$

Moving the mean term to the right, we obtain:

$$\frac{\sum_{i=1}^5 x_i^2}{5} = \frac{194 + 576}{25} = \frac{770}{25}$$

Simplifying this, we find:

$$\frac{\sum_{i=1}^5 x_i^2}{5} = \frac{154}{5} \implies \sum_{i=1}^5 x_i^2 = 154$$

### Phase 3: The Final Synthesis

We know the total sum of squares for all five observations is $154$. Since the fifth observation is $10$, the sum of squares of the first four observations plus the square of the fifth must equal the total sum of squares:

$$\sum_{i=1}^4 x_i^2 + x_5^2 = 154$$

Substituting $x_5 = 10$, we get:

$$\sum_{i=1}^4 x_i^2 + 100 = 154 \implies \sum_{i=1}^4 x_i^2 = 54$$

Now, we apply the variance formula one last time, specifically for the first four observations:

$$\sigma_4^2 = \frac{\sum_{i=1}^4 x_i^2}{4} - (\bar{X}_4)^2$$

Plugging in our values:

$$\sigma_4^2 = \frac{54}{4} - \left(\frac{7}{2}ight)^2 = \frac{54}{4} - \frac{49}{4} = \frac{5}{4}$$

The variance of the first four observations is **$\frac{5}{4}$** (or **$1.25$**). You have navigated the maze of statistics, isolated the unknown, and reconstructed the properties of the data.

Animated Solution for Mathematics - Statistics: If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively and the mean of first four observations is $\frac{7}{2}$ , then the variance of the first four observations in equal to

Select Answer:

Visualized Solution

Given Data for Observations

Sum of Five Observations

Sum of First Four Observations

Calculating the Fifth Observation $x_{5}$

Variance Formula for Five Observations

Simplifying the Variance Equation

Isolating Sum of Squares of First Four

Setting up Variance of First Four

Final Calculation

The Sigma Insight: Measures of Dispersion

The Detective Work of Statistics

Phase 1

Phase 2

Phase 3

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