If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then a + b +

Q: Why can't I just use the definition of variance $\frac{\sum (x_i - \bar{x})^2}{n}$? It seems more intuitive.

You absolutely can! However, the computational formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ is a powerful shortcut. Expanding $(x_i - \bar{x})^2$ for every term with variables $a$ and $b$ is prone to arithmetic errors. The computational formula keeps the algebra clean and elegant.

Q: Wait, do I need to solve for $a$ and $b$ individually to find $a + b + ab$?

That is a great question! While you could solve the quadratic equation to find the individual values of $a$ and $b$, it is unnecessary. We only need the symmetric sums $a+b$ and $ab$. Always look for the 'shortcut'—in competitive exams, time is your most precious resource.

Q: Is there a risk that $a$ or $b$ might be complex numbers?

In the context of statistics problems involving mean and variance of a data set, we assume $a$ and $b$ are real numbers. If you find that your discriminant for the quadratic equation is negative, double-check your arithmetic in the variance calculation!

Q: How do I know if I made a mistake in the sum of squares?

The most common trap is forgetting to square the mean when using the computational formula. Always remember: Variance is the mean of the squares minus the square of the mean. If your variance comes out negative, you likely missed the $(\bar{x})^2$ term.

Q: Why is the number of observations $n=8$ even though there are two variables?

The variables $a$ and $b$ are simply placeholders for two of the eight data points. The total count of items in the set remains 8. Never let the presence of variables distract you from the fundamental count of the data set!

Question

If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then $a + b + ab$ is equal to :

Options:

(A) 105

(B) 103

(C) 100

(D) 106

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence. Today, we aren't just solving a statistics problem; we are performing a detective investigation.

We have a set of numbers, but two of them are cloaked in mystery: $a$ and $b$. We are given the mean and the variance—the two most fundamental 'fingerprints' of any data set—and our mission is to uncover the value of $a + b + ab$.

### The Balance of the Mean

Imagine the mean as the center of gravity of your data. We are given the set: $6, 4, a, 8, b, 12, 10, 13$. With $n = 8$ observations, the mean $\bar{x} = 9$ acts as our anchor.

The definition of the mean is:
$$\bar{x} = \frac{\sum x_i}{n}$$

By summing our knowns and our unknowns, we get the equation:
$$\frac{6 + 4 + a + 8 + b + 12 + 10 + 13}{8} = 9$$

When we aggregate the constants, we find $53 + a + b = 72$. Just like that, the fog clears, and we have our first vital clue:
**$a + b = 19$**

### The Power of Variance

Now, we turn to the variance, $\sigma^2 = 9.25$. While the definition $\frac{\sum (x_i - \bar{x})^2}{n}$ is conceptually beautiful, we prefer the computational efficiency of the following formula:
$$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$$

Substituting our knowns, we write:
$$9.25 = \frac{6^2 + 4^2 + a^2 + 8^2 + b^2 + 12^2 + 10^2 + 13^2}{8} - 9^2$$

As we calculate the squares, we see the numbers grow: $36, 16, a^2, 64, b^2, 144, 100, 169$. Summing the constants gives us $529$. Our equation transforms into:
$$9.25 = \frac{529 + a^2 + b^2}{8} - 81$$

By adding $81$ to both sides and multiplying by $8$, we isolate the sum of the squares of our unknowns:
$$90.25 	imes 8 = 529 + a^2 + b^2$$
$$722 = 529 + a^2 + b^2$$

Thus, we arrive at:
**$a^2 + b^2 = 193$**

### The Algebraic Finale

We now stand at the summit. We know $a + b = 19$ and $a^2 + b^2 = 193$. We need to find $a + b + ab$.

We invoke the classic identity:
$$(a + b)^2 = a^2 + b^2 + 2ab$$

Substituting our values, we get:
$$(19)^2 = 193 + 2ab$$
$$361 = 193 + 2ab$$

Solving for $ab$, we find $2ab = 168$, so $ab = 84$. Finally, we combine our findings:
$$a + b + ab = 19 + 84 = 103$$

The final result is:
**$103$**

Animated Solution for Mathematics - Statistics: If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then $a + b + ab$ is equal to :

Select Answer:

Visualized Solution

Given Data

Formula for Mean

Simplifying the Mean Equation

Value of $a + b$

Formula for Variance

Calculating the Sum of Squares

Simplifying the Variance Equation

Value of $a^{2} + b^{2}$

Algebraic Identity

Solving for $ab$

Final Calculation

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Balance of the Mean

The Power of Variance

The Algebraic Finale

Similar Questions

If the mean and variance of six observations 7, 10, 11, 15, a, b are 10 and $\frac{20}{3}$ , respectively, then the value of $∣ a - b ∣$ is equal to :

The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following gives possible values of a and b?

If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, $α, β, 60$ where $α > β$ are 56 and 66.2 respectively, then $α^{2} + β^{2}$ is equal to

The mean and variance of a data of 10 observations are 10 and 2, respectively. If an observations $α$ in this data is replaced by $β$ , then the mean and variance become 10.1 and 1.99, respectively. Then $α + β$ equals

Let the six numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ be in A.P. and $a_{1} + a_{3} = 10$ . If the mean of these six numbers is $\frac{19}{2}$ and their variance is $σ^{2}$ , then $8 σ^{2}$ is equal to

If the mean of the data : 7, 8, 9, 7, 8, 7, $λ$ , 8 is 8, then the variance of this data is :-

The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and $a$ and $b$ are respectively mean and variance of remaining 6 observation, then $a + 3 b - 5$ is equal to ______.

If the mean and variance of eight numbers $3, 7, 9, 12, 13, 20, x$ and $y$ be 10 and 25 respectively then $x y$ is equal to ________.

Let the Mean and Variance of five observations $x_{1} = 1, x_{2} = 3, x_{3} = a, x_{4} = 7$ and $x_{5} = b, a > b$ , be 5 and 10 respectively. Then the Variance of the observations $n + x_{n}, n = 1, 2, \dots, 5$ is

Let the mean and the variance of $5$ observations $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be $\frac{24}{5}$ and $\frac{194}{25}$ respectively. If the mean and variance of the first $4$ observation are $\frac{7}{2}$ and $a$ respectively, then $(4 a + x_{5})$ is equal to:

Animated Solution for Mathematics - Statistics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then a+b+ab is equal to :