Let the mean and variance of the frequency distribution x: x1=2, x2=6, x3=8, x4=9; f: 4, 4, α, β be

Q: Why can't we just use the shortcut formula for variance \sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2?

You absolutely can! However, in competitive exams like JEE, time is your most precious currency. Using the raw score formula involves squaring larger numbers (like 8^2=64 and 9^2=81), which increases the probability of arithmetic errors. The deviation method keeps the numbers small and manageable, which is a hallmark of a smart, efficient problem-solver.

Q: What if the mean wasn't an integer? Would the deviation method still be easy?

That is a fantastic question. If the mean were a fraction, the deviation method would introduce fractions into every single term, making the algebra messy and prone to error. In that specific scenario, the raw score formula is often safer. Always assess the 'niceness' of your mean before choosing your weapon!

Q: Do I really need to recalculate the total frequency when x_3 changes?

Yes, always check! In this specific problem, the frequencies \alpha and \beta are attached to the values. Changing the value x_3 from 8 to 7 changes the data points, but the weights (frequencies) remain the same. The total frequency N remains unchanged, but the sum of products \sum f_i x_i definitely changes. Never assume anything stays the same—verify it!

Question

Let the mean and variance of the frequency distribution $$\begin{array}{ccccc} x : & x_1 = 2 & x_2 = 6 & x_3 = 8 & x_4 = 9 \ f : & 4 & 4 & \alpha & \beta \end{array}$$ 
be 6 and 6.8 respectively. If $x_3$ is changed from 8 to 7, then the mean for the new data will be:

Options:

(A) 4

(B) 5

(C) $\frac{17}{3}$

(D) $\frac{16}{3}$

Sigma Percentile · Accepted Answer

### The Data Detective: Unlocking the Mystery of Distributions

Welcome, future engineer. Today, we are not just solving a statistics problem; we are stepping into the shoes of a data detective. In the world of JEE Advanced, statistics is often viewed as a 'formula-heavy' topic, but I want you to see it differently. It is a study of balance.

When we talk about mean and variance, we are talking about the center of gravity and the spread of a system. Let us unravel this puzzle together.

### Phase 1: The Mystery of the Missing Frequencies

Imagine you are looking at a frequency distribution table. You see values $x_1=2$, $x_2=6$, $x_3=8$, $x_4=9$, but the frequencies for the last two are hidden behind the Greek letters $\alpha$ and $\beta$. We are given the mean $\bar{x} = 6$ and the variance $\sigma^2 = 6.8$.

In algebra, whenever you have two unknowns, you need two independent constraints to find them. Here, the mean and the variance are those two keys.

Let us start with the mean. The definition is:
$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$

The total frequency $N$ is $4 + 4 + \alpha + \beta = 8 + \alpha + \beta$. The sum of products is:
$$4(2) + 4(6) + 8\alpha + 9\beta = 32 + 8\alpha + 9\beta$$

Setting this equal to the mean of $6$, we get:
$$6 = \frac{32 + 8\alpha + 9\beta}{8 + \alpha + \beta}$$

Cross-multiplying gives us $48 + 6\alpha + 6\beta = 32 + 8\alpha + 9\beta$, which simplifies beautifully to our first linear equation:
$$2\alpha + 3\beta = 16$$

Keep this close; it is the foundation of our solution.

### Phase 2: The Variance Challenge

Now, we tackle the variance. Since the mean is $6$, a perfect integer, the deviation method is incredibly elegant:
$$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$$

Let us calculate the squared deviations $(x_i - \bar{x})^2$:
*   For $2$: $(2-6)^2 = 16$
*   For $6$: $(6-6)^2 = 0$
*   For $8$: $(8-6)^2 = 4$
*   For $9$: $(9-6)^2 = 9$

Multiplying these by their respective frequencies, we get the numerator:
$$4(16) + 4(0) + \alpha(4) + \beta(9) = 64 + 4\alpha + 9\beta$$

Equating this to the variance $6.8$, we have:
$$6.8 = \frac{64 + 4\alpha + 9\beta}{8 + \alpha + \beta}$$

### Phase 3: The Algebraic Dance

Solving this second equation requires patience. Cross-multiplying gives:
$$6.8(8 + \alpha + \beta) = 64 + 4\alpha + 9\beta$$

Expanding this, we get $54.4 + 6.8\alpha + 6.8\beta = 64 + 4\alpha + 9\beta$. Rearranging the terms, we arrive at:
$$2.8\alpha - 2.2\beta = 9.6$$

To clear the decimals, multiply the entire equation by $5$, yielding:
$$14\alpha - 11\beta = 48$$

Now, we have a system: $2\alpha + 3\beta = 16$ and $14\alpha - 11\beta = 48$. Multiplying the first equation by $7$ gives $14\alpha + 21\beta = 112$.

Subtracting the second equation from this, the $\alpha$ terms vanish, leaving $32\beta = 64$, so $\beta = 2$. Substituting back, we find $\alpha = 5$. We have successfully cracked the code!

### Phase 4: The Twist

Finally, the problem throws a curveball. We change $x_3$ from $8$ to $7$. The frequencies $\alpha=5$ and $\beta=2$ remain unchanged.

Our new dataset is $2, 6, 7, 9$ with frequencies $4, 4, 5, 2$. The total frequency $N$ remains $15$. The new sum of products is:
$$4(2) + 4(6) + 5(7) + 2(9) = 8 + 24 + 35 + 18 = 85$$

The new mean is simply:
$$\bar{x}_{new} = \frac{85}{15} = \frac{17}{3}$$

See how the logic flows? We didn't just calculate; we adapted. That is the essence of JEE physics and math. Stay curious, stay rigorous, and keep solving.

Animated Solution for Mathematics - Statistics: Let the mean and variance of the frequency distribution $x : f : x_{1} = 2 4 x_{2} = 6 4 x_{3} = 8 α x_{4} = 9 β$ be 6 and 6.8 respectively. If $x_{3}$ is changed from 8 to 7, then the mean for the new data will be:

Select Answer:

Visualized Solution

Understanding the Frequency Distribution

Applying the Mean Formula

Deriving the First Equation

Understanding Variance with Deviations

Setting up the Variance Equation

Simplifying the Variance Equation

Solving the System of Equations

Setting Up the New Data

Calculating the New Mean

The Sigma Insight: Measures of Dispersion

The Data Detective

Phase 1

Phase 2

Phase 3

Phase 4

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