If the mean and the variance of the data (Class: 4-8, 8-12, 12-16, 16-20; Frequency: 3, λ, 4, 7) are

Q: Why can't we just use the standard deviation formula $\sigma = \sqrt{\frac{\sum f_i(x_i - \mu)^2}{N}}$?

You absolutely can! However, in competitive exams like JEE, time is your most precious currency. The computational formula $\sigma^2 = \frac{\sum f_i x_i^2}{N} - \mu^2$ is algebraically more efficient because it avoids calculating the deviation $(x_i - \mu)$ for every single class interval. It turns a multi-step subtraction process into a single, clean algebraic expression.

Q: How do I know if I should reject one of the roots of the quadratic equation?

Great question! In physics and statistics, math is just the language; the physical reality dictates the rules. Here, $\lambda$ represents a frequency. By definition, a frequency must be a non-negative integer. If your quadratic equation yields a fraction or a negative number for $\lambda$, that solution is physically impossible. Always check your roots against the constraints of the problem!

Q: Is there a shortcut to avoid the massive algebra?

Yes! The 'pro' move is the substitution $u = 14 + \lambda$. Whenever you see a complex denominator like $14 + \lambda$ appearing repeatedly, replace it with a single variable. This transforms a terrifying rational equation into a simple quadratic one. It is not just a shortcut; it is a way to reduce cognitive load and prevent silly calculation errors.

Question

If the mean and the variance of the data \begin{array}{|c|c|c|c|c|} \hline Class & 4-8 & 8-12 & 12-16 & 16-20 \ \hline Frequency & 3 & $\lambda$ & 4 & 7 \ \hline \end{array} are $\mu$ and 19 respectively, then the value of $\lambda + \mu$ is

Options:

(A) 21

(B) 19

(C) 20

(D) 18

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The first step in any statistical journey is to find the representative value for each interval. We cannot perform arithmetic on a range like $4-8$, so we find the midpoint $x_i$.

For the intervals $4-8$, $8-12$, $12-16$, and $16-20$, the midpoints are $6, 10, 14,$ and $18$ respectively.

Now, we define our total frequency $N$. Summing the frequencies $3, \lambda, 4, 7$, we get:

$$N = 3 + \lambda + 4 + 7 = 14 + \lambda$$

This $N$ is the anchor of our calculation and serves as the denominator for all subsequent statistical measures.

### The Mean and the Variance

We know the mean $\mu$ is defined as $\mu = \frac{\sum f_i x_i}{N}$. Calculating the numerator, we get:

$$\sum f_i x_i = 3(6) + \lambda(10) + 4(14) + 7(18) = 18 + 10\lambda + 56 + 126 = 200 + 10\lambda$$

Thus, the mean is expressed as:

$$\mu = \frac{200 + 10\lambda}{14 + \lambda}$$

For the variance, we use the computational formula $\sigma^2 = \frac{\sum f_i x_i^2}{N} - \mu^2$. We calculate the sum of squares:

$$\sum f_i x_i^2 = 3(36) + \lambda(100) + 4(196) + 7(324) = 108 + 100\lambda + 784 + 2268 = 3160 + 100\lambda$$

Given that $\sigma^2 = 19$, we have established the foundation for our algebraic solution.

### The Algebraic Masterstroke

To avoid drowning in complex quadratic expansions, we introduce a substitution: $u = 14 + \lambda$. This implies $\lambda = u - 14$.

The mean transforms into a much simpler form:

$$\mu = \frac{200 + 10(u - 14)}{u} = \frac{60 + 10u}{u} = \frac{60}{u} + 10$$

Similarly, the sum of squares becomes $3160 + 100(u - 14) = 1760 + 100u$.

### The Final Resolution

Substituting these into our variance equation $19 = \frac{\sum f_i x_i^2}{N} - \mu^2$, we get:

$$19 = \frac{1760 + 100u}{u} - \left( \frac{60}{u} + 10 \right)^2$$

Multiplying by $u^2$ to clear the denominators, we obtain:

$$19u^2 = u(1760 + 100u) - (60 + 10u)^2$$

Expanding this expression:

$$19u^2 = 1760u + 100u^2 - (3600 + 1200u + 100u^2)$$

The $100u^2$ terms cancel out, leaving us with the quadratic equation:

$$19u^2 - 560u + 3600 = 0$$

Solving this quadratic, we find $u = 20$ or $u = \frac{180}{19}$. Since $\lambda$ must be an integer, $u$ must be an integer, leading us to $u = 20$.

Consequently, $\lambda = 20 - 14 = 6$. The mean is then $\mu = \frac{200 + 60}{20} = 13$.

The final result is **$\lambda + \mu = 6 + 13 = 19$**.

Animated Solution for Mathematics - Statistics: If the mean and the variance of the data \begin{array}{|c|c|c|c|c|} \hline Class & 4-8 & 8-12 & 12-16 & 16-20 \\ \hline Frequency & 3 & $λ$ & 4 & 7 \\ \hline \end{array} are $μ$ and 19 respectively, then the value of $λ + μ$ is

Select Answer:

Visualized Solution

Identify Midpoints and Frequencies

Calculate Total Frequency $N$

Define the Mean $μ$

Calculate $\sum f_{i} x_{i}^{2}$

Set up the Variance Equation

Simplify using Substitution

Transform the Variance Equation

Form the Quadratic Equation

Solve for $u$ and $λ$

Find $μ$ and the Final Answer

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Mean and the Variance

The Algebraic Masterstroke

The Final Resolution

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Visualized Solution

Identify Midpoints and Frequencies

Calculate Total Frequency .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

Define the Mean .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } μ

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑fi​xi2​