Let the mean of the data x: 1, 3, 5, 7, 9; Frequency (f): 4, 24, 28, α, 8 be 5. If m and σ² are resp

Q: Why do we use absolute values for Mean Deviation but squares for Variance?

Great question! Mean deviation is about the 'average distance' from the mean, so we treat all deviations as positive magnitudes to see how far, on average, data points lie from the center. Variance, however, is a statistical tool that penalizes larger deviations more heavily by squaring them. This property makes variance mathematically more tractable for advanced calculus and probability theory, even if it makes the units 'squared' (like $units^2$).

Q: Is there a shortcut for Variance?

Yes, you could use the computational formula $\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2$. However, in JEE problems with small, discrete values like these, calculating $(x_i - \bar{x})^2$ directly is often less prone to calculation errors. Stick to the method that keeps your numbers small and manageable. Don't hunt for shortcuts if the direct path is clear and safe.

Q: What if I get a negative value for $\alpha$?

That would be a red flag! Frequencies represent counts of occurrences, so they must always be non-negative integers. If you get a negative $\alpha$, re-check your cross-multiplication in the mean equation. It is almost certainly a sign of a simple algebraic slip in the expansion of the denominator or the numerator.

Question

Let the mean of the data $$\begin{array}{|c|c|c|c|c|c|} \hline x & 1 & 3 & 5 & 7 & 9 \ \hline 	ext{Frequency } (f) & 4 & 24 & 28 & \alpha & 8 \ \hline \end{array}$$ be 5. If $m$ and $\sigma^2$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3\alpha}{m+\sigma^2}$ is equal to _______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the JEE journey. Today, we are not just solving a statistics problem; we are peeling back the layers of a data distribution to understand its heartbeat.

We are given a frequency distribution table with observations $x_i \in \{1, 3, 5, 7, 9\}$ and corresponding frequencies $f_i \in \{4, 24, 28, \alpha, 8\}$. We are given that the mean $\bar{x} = 5$.

The mean is defined as the weighted average:
$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$

### The Detective Work

First, let us find the total frequency $N$:
$$N = 4 + 24 + 28 + \alpha + 8 = 64 + \alpha$$

Next, we calculate the weighted sum of the observations:
$$\sum f_i x_i = (4 	imes 1) + (24 	imes 3) + (28 	imes 5) + (\alpha 	imes 7) + (8 	imes 9) = 288 + 7\alpha$$

Setting up the equation for the mean:
$$\frac{288 + 7\alpha}{64 + \alpha} = 5$$

Cross-multiplying yields:
$$288 + 7\alpha = 320 + 5\alpha \Rightarrow 2\alpha = 32 \Rightarrow \alpha = 16$$

With $\alpha = 16$, the total frequency becomes $N = 64 + 16 = 80$.

### The Anatomy of Dispersion

Now, we calculate the Mean Deviation ($m$), defined as:
$$m = \frac{\sum f_i |x_i - \bar{x}|}{N}$$

Using $\bar{x} = 5$, we compute the absolute deviations $|x_i - 5|$:

*   For $x=1$: $4 	imes |1-5| = 16$
*   For $x=3$: $24 	imes |3-5| = 48$
*   For $x=5$: $28 	imes |5-5| = 0$
*   For $x=7$: $16 	imes |7-5| = 32$
*   For $x=9$: $8 	imes |9-5| = 32$

Summing these values gives $\sum f_i |x_i - 5| = 16 + 48 + 0 + 32 + 32 = 128$. Thus:
$$m = \frac{128}{80} = \frac{8}{5}$$

### The Weight of Variance

Next, we calculate the Variance ($\sigma^2$), defined as:
$$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{N}$$

We compute the squared deviations $(x_i - 5)^2$:

*   For $x=1$: $4 	imes (1-5)^2 = 64$
*   For $x=3$: $24 	imes (3-5)^2 = 96$
*   For $x=5$: $28 	imes (5-5)^2 = 0$
*   For $x=7$: $16 	imes (7-5)^2 = 64$
*   For $x=9$: $8 	imes (9-5)^2 = 128$

Summing these values gives $\sum f_i (x_i - 5)^2 = 64 + 96 + 0 + 64 + 128 = 352$. Thus:
$$\sigma^2 = \frac{352}{80} = \frac{22}{5}$$

### Final Calculation

We evaluate the expression $\frac{3\alpha}{m + \sigma^2}$ using $\alpha = 16$, $m = \frac{8}{5}$, and $\sigma^2 = \frac{22}{5}$.

The denominator is:
$$m + \sigma^2 = \frac{8}{5} + \frac{22}{5} = \frac{30}{5} = 6$$

The numerator is:
$$3\alpha = 3 	imes 16 = 48$$

The final result is:
$$\frac{48}{6} = \mathbf{8}$$

Animated Solution for Mathematics - Statistics: Let the mean of the data $x Frequency (f) 14324528 7 α 98$ be 5. If $m$ and $σ^{2}$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3 α}{m + σ ^{2}}$ is equal to _______.

Enter Numerical Value:

Visualized Solution

Analyze the Given Frequency Data

Apply the Formula for Mean $\overset{x}{ˉ}$

Substitute and Simplify the Mean Equation

Solve for the Unknown Frequency $α$

Define Mean Deviation about Mean ( $m$ )

Calculate Absolute Deviations $f_{i} ∣ x_{i} - 5∣$

Compute the Value of Mean Deviation $m$

Define Variance $σ^{2}$

Calculate Squared Deviations $f_{i} (x_{i} - 5)^{2}$

Compute the Value of Variance $σ^{2}$

Setup the Final Expression

Final Calculation and Result

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

The Detective Work

The Anatomy of Dispersion

The Weight of Variance

Final Calculation

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Enter Numerical Value:

Visualized Solution

Analyze the Given Frequency Data

Apply the Formula for Mean .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xˉ

Substitute and Simplify the Mean Equation

Solve for the Unknown Frequency .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Define Mean Deviation about Mean ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m)

Calculate Absolute Deviations .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } fi​∣xi​−5∣

Compute the Value of Mean Deviation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Define Variance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ2

Calculate Squared Deviations .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } fi​(xi​−5)2

Compute the Value of Variance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ2