The mean of five observations is 5 and their variance is 9.20. If three of the given five observatio

Q: Wait, why do we use the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ instead of the definition $\frac{\sum (x_i - \bar{x})^2}{n}$?

Great question! Both formulas are mathematically equivalent. However, the formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ is a powerful tool in competitive exams because it allows us to isolate the sum of squares ($\sum x_i^2$) directly without needing to know the individual deviations from the mean first. It saves time and reduces the chance of calculation errors.

Q: Can we solve this without a quadratic equation?

You could try substitution, but it often leads to more complex algebraic manipulation. Forming the quadratic equation $t^2 - (sum)t + (product) = 0$ is the most elegant and standard method in JEE. It turns a potentially messy system of equations into a simple factorization problem.

Q: Does the order of $x_1$ and $x_2$ matter for the ratio?

Not at all! Whether you label the first missing number as $x_1$ or $x_2$, you will end up with the same set of two numbers. The ratio will simply be the ratio of those two numbers, regardless of which one you put in the numerator.

Question

The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is :

Options:

(A) 4 : 9

(B) 6 : 7

(C) 5 : 8

(D) 10 : 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are a data scientist tasked with reconstructing a dataset where some values have been lost to time. You are given five observations in total, but only three are visible: $1$, $3$, and $8$.

The other two, let us call them $x_1$ and $x_2$, are shrouded in mystery. You have two powerful clues: the mean of the entire set is $\bar{x} = 5$, and the variance is $\sigma^2 = 9.20$.

### The Power of the Mean

The mean is the balancing point of any dataset. Mathematically, it is the sum of all observations divided by the count. Given the mean $\bar{x} = 5$ and the count $n = 5$, the sum of all observations must be:

$$5 	imes 5 = 25$$

Our equation becomes $1 + 3 + 8 + x_1 + x_2 = 25$. Simplifying this, we find $12 + x_1 + x_2 = 25$, which gives us our first vital piece of information:

$$x_1 + x_2 = 13$$

### Unlocking the Variance

Variance measures the spread of data. The formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ is our primary tool here. Substituting the known values $\sigma^2 = 9.20$ and $\bar{x} = 5$:

$$9.20 = \frac{1^2 + 3^2 + 8^2 + x_1^2 + x_2^2}{5} - 5^2$$

Squaring the mean gives $25$, and adding it to the variance gives $34.20$. Multiplying by $5$ yields $171$. The sum of the squares of the knowns is $1^2 + 3^2 + 8^2 = 74$.

Thus, $171 = 74 + x_1^2 + x_2^2$, leading us to our 'sum of squares constraint':

$$x_1^2 + x_2^2 = 97$$

### The Algebraic Bridge

Now we have two equations: $x_1 + x_2 = 13$ and $x_1^2 + x_2^2 = 97$. To find the individual values, we utilize the algebraic identity:

$$(x_1 + x_2)^2 = x_1^2 + x_2^2 + 2x_1x_2$$

Substituting our knowns: $13^2 = 97 + 2x_1x_2$, which means $169 = 97 + 2x_1x_2$. Subtracting $97$ from $169$ gives $72$, so $2x_1x_2 = 72$, or:

$$x_1x_2 = 36$$

### The Quadratic Finale

We are looking for two numbers that add up to $13$ and multiply to $36$. These are the roots of the quadratic equation $t^2 - (	ext{sum})t + (	ext{product}) = 0$:

$$t^2 - 13t + 36 = 0$$

Factoring this, we get $(t - 9)(t - 4) = 0$. The roots are $4$ and $9$.

The two missing observations are **$4$ and $9$**. We have successfully reconstructed the dataset.

Animated Solution for Mathematics - Statistics: The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is :

Select Answer:

Visualized Solution

Define the Unknowns

The Mean Formula

Raw Setup for Mean

Sum of Unknowns

The Variance Formula

Substitute into Variance

Simplify Variance Equation

Sum of Squares

Algebraic Identity

Find the Product

Product of Unknowns

Forming the Quadratic

Solving for Observations

Final Ratio and Summary

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Power of the Mean

Unlocking the Variance

The Algebraic Bridge

The Quadratic Finale

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